I'm pretty sure it has to do with my core.js file with the ajax hashing url. But I'm trying to submit a form, but it's not submitting like I want it to. This is the core.js file:
// call init
$(init);
function init() {
ajax_page_handler();
page_load($(window.location).attr("hash")); // goto first page if #! is available
}
function page_load($href) {
if($href != undefined && $href.substring(0, 2) == '#/') {
// replace body the #content with loaded html
$('#content').load($href.substring(2), function () {
$('#content').hide().fadeIn('slow');
});
}
}
function ajax_page_handler() {
$(window).bind('hashchange', function () {
$href = $(window.location).attr("hash");
page_load($href);
});
// this allow you to reload by clicking the same link
$('a[href^="#/"]').live('click', function() {
$curhref = $(window.location).attr("hash");
$href = $(this).attr('href');
if($curhref == $href) {
page_load($href);
}
});
}
The live viewing is over at www.krissales.com. The form is here: http://www.krissales.com/#/media/5.Testing-1
Hit the link "Post Comment", then you'll type info in, then hit comment, but it just refreshes, but doesn't submit it.
The steps I've taken to solve it was in the comment file, in the form action field, I inserted the tag name="#content" simply because that's the name of my div that I'm submitting to.
The original stuff is on http://blog.krissales.com/article/7.Testing-3-man ( where you can actually post a comment, and it'll work)
But apparently it's not working. Do you guys have a clue as to what it is that I'm doing wrong? thanks for your help in advance!
<script type="text/javascript">
tinyMCE.init({
mode : "textareas",
theme : "simple"
});
</script>
<form action="#/media/article.php" name="#content" method="POST">
Name:
<br />
<input type="text" name="name" class="userpass"/>
<br /><br />
Comment:
<br />
<textarea id="elm1" name="comment" rows="7" cols="30" style="width: 500px;">
</textarea>
<br />
<input type="submit" name="submit" value="Comment" class="button" />
<input type="reset" name="submit" value="Reset" class="button" />
</form>
I noticed that you are not setting the ajax type on the file 'comment.php'.
you need...
$.ajax({
type: 'POST',
url: 'comment_ajax.php',
data: { form_name: name, form_comment: comment },
success: function(data) {
$('#new_comment').prepend(data);
// close modal box
// do other shit
// kthnxbai
}
});
If type is not specified, it defaults to a GET request which will not post data. :)
Your current core.js handles changes in the URL hash, and it reroutes any links with a hash to load that relative path into #content. What's missing is code to redirect form submits to do the same thing (add this to ajax_page_handler):
$('form').live('submit', function(e) {
var $action = $(this).attr('action');
if($action.substring(0, 2) == '#/') {
// replace the #content with result of the form post
$.ajax({
url: $action.substring(2),
type: $(this).attr('method'),
data: $(this).serialize(),
success: function(data) {
$('#content').html(data);
$('#content').hide().fadeIn('slow');
}
});
// stop the real form submit from happening
e.preventDefault();
}
});
You should change the action attribute of your form like this :
<form action="script-handling-comment-data.php#/media/article.php" name="#content" method="POST">
For the moment, you're sending the comment data to http://www.krissales.com/ and i think the main page doesn't handle the comment posting.
You seem to be handling links properly, but form submission isn't a link, you probably want to handle submission using $(form).submit(function(){ ... })
In your case, if you gave your form the id form1
$('#form1').submit(function(){
var keyValues = $(this).serializeArray();
var map = {};
for(i in keyValues)
{
var value = keyValues.value;
var name = keyValues.name;
map[name] = value;
}
$.post($(this).attr('action'),map,function(){
alert("Submitted values: " + $(this).serialize());
});
return false;
})
See serializeArray, $.post and .submit for more information
Related
I have a classifieds website, and on the page where ads are showed, I am creating a "Send a tip to a friend" form...
So anybody who wants can send a tip of the ad to some friends email-adress.
I am guessing the form must be submitted to a php page right?
<form name="tip" method="post" action="tip.php">
Tip somebody:
<input
name="tip_email"
type="text"
size="30"
onfocus="tip_div(1);"
onblur="tip_div(2);"
/>
<input type="submit" value="Skicka Tips" />
<input type="hidden" name="ad_id" />
</form>
When submitting the form, the page gets reloaded... I don't want that...
Is there any way to make it not reload and still send the mail?
Preferrably without ajax or jquery...
I've found what I think is an easier way.
If you put an Iframe in the page, you can redirect the exit of the action there and make it show up.
You can do nothing, of course. In that case, you can set the iframe display to none.
<iframe name="votar" style="display:none;"></iframe>
<form action="tip.php" method="post" target="votar">
<input type="submit" value="Skicka Tips">
<input type="hidden" name="ad_id" value="2">
</form>
You'll need to submit an ajax request to send the email without reloading the page. Take a look at http://api.jquery.com/jQuery.ajax/
Your code should be something along the lines of:
$('#submit').click(function() {
$.ajax({
url: 'send_email.php',
type: 'POST',
data: {
email: 'email#example.com',
message: 'hello world!'
},
success: function(msg) {
alert('Email Sent');
}
});
});
The form will submit in the background to the send_email.php page which will need to handle the request and send the email.
You either use AJAX or you
create and append an iframe to the document
set the iframes name to 'foo'
set the forms target to 'foo'
submit
have the forms action render javascript with 'parent.notify(...)' to give feedback
optionally you can remove the iframe
Fastest and easiest way is to use an iframe.
Put a frame at the bottom of your page.
<iframe name="frame"></iframe>
And in your form do this.
<form target="frame">
</form>
and to make the frame invisible in your css.
iframe{
display: none;
}
SUBMITTING THE FORM WITHOUT RELOADING THE PAGE AND GET THE RESULT OF SUBMITTED DATA ON THE SAME PAGE.
Here's some of the code I found on the internet that solves this problem :
1.) IFRAME
When the form is submitted, The action will be executed and target the specific iframe to reload.
index.php
<iframe name="content" style="">
</iframe>
<form action="iframe_content.php" method="post" target="content">
<input type="text" name="Name" value="">
<input type="submit" name="Submit" value="Submit">
</form>
iframe_content.php
<?php
$Submit = isset($_POST['Submit']) ? $_POST['Submit'] : false;
$Name = isset($_POST['Name']) ? $_POST['Name'] : '';
if($Submit){
echo $Name;
}
?>
2.) AJAX
Index.php:
<form >
<input type="" name="name" id="name">
<input type="" name="descr" id="descr">
<input type="submit" name="" value="submit" onclick="return clickButton();">
</form>
<p id="msg"></p>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script type="text/javascript">
function clickButton(){
var name=document.getElementById('name').value;
var descr=document.getElementById('descr').value;
$.ajax({
type:"post",
url:"server_action.php",
data:
{
'name' :name,
'descr' :descr
},
cache:false,
success: function (html)
{
alert('Data Send');
$('#msg').html(html);
}
});
return false;
}
</script>
server_action.php
<?php
$name = isset($_POST['name']) ? $_POST['name'] : '';
$descr = isset($_POST['descr']) ? $_POST['descr'] : '';
echo $name;
echo $descr;
?>
Tags: phpajaxjqueryserversidehtml
A further possibility is to make a direct javascript link to your function:
<form action="javascript:your_function();" method="post">
...
It's a must to take help of jquery-ajax in this case. Without ajax, there is currently no solution.
First, call a JavaScript function when the form is submitted. Just set onsubmit="func()". Even if the function is called, the default action of the submission would be performed. If it is performed there would be no way of stoping the page from refreshing or redirecting. So, next task is to prevent the default action. Insert the following line at the start of func().
event.preventDefault()
Now, there will be no redirecting or refreshing. So, you simply make an ajax call from func() and do whatever you want to do when call ends.
Example:
Form:
<form id="form-id" onsubmit="func()">
<input id="input-id" type="text">
</form>
Javascript:
function func(){
event.preventDefault();
var newValue = $('#input-field-id').val();
$.ajax({
type: 'POST',
url: '...',
data: {...},
datatype: 'JSON',
success: function(data){...},
error: function(){...},
});
}
this is exactly how it CAN work without jQuery and AJAX and it's working very well using a simple iFrame. I LOVE IT, works in Opera10, FF3 and IE6. Thanks to some of the above posters pointing me the right direction, that's the only reason I am posting here:
<select name="aAddToPage[65654]"
onchange="
if (bCanAddMore) {
addToPage(65654,this);
}
else {
alert('Could not add another, wait until previous is added.');
this.options[0].selected = true;
};
" />
<option value="">Add to page..</option>
[more options with values here]</select>
<script type="text/javascript">
function addToPage(iProduct, oSelect){
iPage = oSelect.options[oSelect.selectedIndex].value;
if (iPage != "") {
bCanAddMore = false;
window.hiddenFrame.document.formFrame.iProduct.value = iProduct;
window.hiddenFrame.document.formFrame.iAddToPage.value = iPage;
window.hiddenFrame.document.formFrame.submit();
}
}
var bCanAddMore = true;</script>
<iframe name="hiddenFrame" style="display:none;" src="frame.php?p=addProductToPage" onload="bCanAddMore = true;"></iframe>
the php code generating the page that is being called above:
if( $_GET['p'] == 'addProductToPage' ){ // hidden form processing
if(!empty($_POST['iAddToPage'])) {
//.. do something with it..
}
print('
<html>
<body>
<form name="formFrame" id="formFrameId" style="display:none;" method="POST" action="frame.php?p=addProductToPage" >
<input type="hidden" name="iProduct" value="" />
<input type="hidden" name="iAddToPage" value="" />
</form>
</body>
</html>
');
}
This should solve your problem.In this code after submit button click we call jquery ajax and we pass url to posttype POST/GET
data: data information you can select input fields or any other.
sucess: callback if everything is ok from server
function parameter text, html or json, response from server
in sucess you can write write warnings if data you got is in some kind of state and so on. or execute your code what to do next.
<form id='tip'>
Tip somebody: <input name="tip_email" id="tip_email" type="text" size="30" onfocus="tip_div(1);" onblur="tip_div(2);"/>
<input type="submit" id="submit" value="Skicka Tips"/>
<input type="hidden" id="ad_id" name="ad_id" />
</form>
<script>
$( "#tip" ).submit(function( e ) {
e.preventDefault();
$.ajax({
url: tip.php,
type:'POST',
data:
{
tip_email: $('#tip_email').val(),
ad_id: $('#ad_id').val()
},
success: function(msg)
{
alert('Email Sent');
}
});
});
</script>
You can try setting the target attribute of your form to a hidden iframe, so the page containing the form won't get reloaded.
I tried it with file uploads (which we know can't be done via AJAX), and it worked beautifully.
Have you tried using an iFrame? No ajax, and the original page will not load.
You can display the submit form as a separate page inside the iframe, and when it gets submitted the outer/container page will not reload. This solution will not make use of any kind of ajax.
function Foo(){
event.preventDefault();
$.ajax( {
url:"<?php echo base_url();?>Controllername/ctlr_function",
type:"POST",
data:'email='+$("#email").val(),
success:function(msg) {
alert('You are subscribed');
}
} );
}
I tried many times for a good solution and answer by #taufique helped me to arrive at this answer.
NB : Don't forget to put event.preventDefault(); at the beginning of the body of the function .
I did something similar to the jquery above, but I needed to reset my form data and graphic attachment canvases.
So here is what I came up with:
<script>
$(document).ready(function(){
$("#text_only_radio_button_id").click(function(){
$("#single_pic_div").hide();
$("#multi_pic_div").hide();
});
$("#pic_radio_button_id").click(function(){
$("#single_pic_div").show();
$("#multi_pic_div").hide();
});
$("#gallery_radio_button_id").click(function(){
$("#single_pic_div").hide();
$("#multi_pic_div").show();
});
$("#my_Submit_button_ID").click(function() {
$("#single_pic_div").hide();
$("#multi_pic_div").hide();
var url = "script_the_form_gets_posted_to.php";
$.ajax({
type: "POST",
url: url,
data: $("#html_form_id").serialize(),
success: function(){
document.getElementById("html_form_id").reset();
var canvas=document.getElementById("canvas");
var canvasA=document.getElementById("canvasA");
var canvasB=document.getElementById("canvasB");
var canvasC=document.getElementById("canvasC");
var canvasD=document.getElementById("canvasD");
var ctx=canvas.getContext("2d");
var ctxA=canvasA.getContext("2d");
var ctxB=canvasB.getContext("2d");
var ctxC=canvasC.getContext("2d");
var ctxD=canvasD.getContext("2d");
ctx.clearRect(0, 0,480,480);
ctxA.clearRect(0, 0,480,480);
ctxB.clearRect(0, 0,480,480);
ctxC.clearRect(0, 0,480,480);
ctxD.clearRect(0, 0,480,480);
} });
return false;
}); });
</script>
That works well for me, for your application of just an html form, we can simplify this jquery code like this:
<script>
$(document).ready(function(){
$("#my_Submit_button_ID").click(function() {
var url = "script_the_form_gets_posted_to.php";
$.ajax({
type: "POST",
url: url,
data: $("#html_form_id").serialize(),
success: function(){
document.getElementById("html_form_id").reset();
} });
return false;
}); });
</script>
I don't know JavaScript and I just started to learn PHP, so what helped for me from all those responses was:
Create inedx.php and insert:
<iframe name="email" style=""></iframe>
<form action="email.php" method="post" target="email">
<input type="email" name="email" >
<input type="submit" name="Submit" value="Submit">
</form>
Create email.php and insert this code to check if you are getting the data (you should see it on index.php in the iframe):
<?php
if (isset($_POST['Submit'])){
$email = $_POST['email'];
echo $email;
}
?>
If everything is ok, change the code on email.php to:
<?php
if (isset($_POST['Submit'])){
$to = $_POST['email'];
$subject = "Test email";
$message = "Test message";
$headers = "From: test#test.com \r\n";
$headers .= "Reply-To: test#test.com \r\n";
$headers .= "MIME-Version: 1.0\r\n";
$headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n";
mail($to, $subject, $message, $headers);
}
?>
Hope this helps for all other rookies like me :)
Modern Answer without XHR or jQuery
It's 2022, we don't need to use old tools like XHR or jQuery when we have the Fetch API and the FormData API!
The first thing we need to do is prevent the default form submission behavior from occurring with event.preventDefault():
form.addEventListener("submit", function(event){
event.preventDefault();
// ...
});
Now we need to replace the submission behavior with our own AJAX request. The Fetch API makes it pretty simple to post form data - just create a new FormData object, populating it with the form's values, and use it as the body of a fetch request:
fetch(form.action, {
method: "post",
body: new URLSearchParams(new FormData(form))
});
Note that this submits an HTTP request using the multipart/form-data format. If you need to post the data using application/x-www-form-urlencoded, create a new URLSearchParams object from the FormData object and use that as the fetch's body.
fetch(form.action, {
method: "post",
body: new URLSearchParams(new FormData(form))
});
Here's a full code example:
let form = document.querySelector("form");
form.addEventListener("submit", function(event){
event.preventDefault();
fetch(form.action, {
method: "post",
body: //new FormData(form) // for multipart/form-data
new URLSearchParams(new FormData(form)) //for application/x-www-form-urlencoded
});
});
<form method="POST">
<input name="name" placeholder="Name" />
<input name="phone" type="tel" placeholder="Phone" />
<input name="email" type="email" placeholder="Email" />
<input name="submit" type="submit" />
</form>
The page will get reloaded if you don't want to use javascript
You will need to use JavaScript without resulting to an iframe (ugly approach).
You can do it in JavaScript; using jQuery will make it painless.
I suggest you check out AJAX and Posting.
if you're submitting to the same page where the form is you could write the form tags with out an action and it will submit, like this
<form method='post'> <!-- you can see there is no action here-->
Here is some jQuery for posting to a php page and getting html back:
$('form').submit(function() {
$.post('tip.php', function(html) {
// do what you need in your success callback
}
return false;
});
I have a classifieds website, and on the page where ads are showed, I am creating a "Send a tip to a friend" form...
So anybody who wants can send a tip of the ad to some friends email-adress.
I am guessing the form must be submitted to a php page right?
<form name="tip" method="post" action="tip.php">
Tip somebody:
<input
name="tip_email"
type="text"
size="30"
onfocus="tip_div(1);"
onblur="tip_div(2);"
/>
<input type="submit" value="Skicka Tips" />
<input type="hidden" name="ad_id" />
</form>
When submitting the form, the page gets reloaded... I don't want that...
Is there any way to make it not reload and still send the mail?
Preferrably without ajax or jquery...
I've found what I think is an easier way.
If you put an Iframe in the page, you can redirect the exit of the action there and make it show up.
You can do nothing, of course. In that case, you can set the iframe display to none.
<iframe name="votar" style="display:none;"></iframe>
<form action="tip.php" method="post" target="votar">
<input type="submit" value="Skicka Tips">
<input type="hidden" name="ad_id" value="2">
</form>
You'll need to submit an ajax request to send the email without reloading the page. Take a look at http://api.jquery.com/jQuery.ajax/
Your code should be something along the lines of:
$('#submit').click(function() {
$.ajax({
url: 'send_email.php',
type: 'POST',
data: {
email: 'email#example.com',
message: 'hello world!'
},
success: function(msg) {
alert('Email Sent');
}
});
});
The form will submit in the background to the send_email.php page which will need to handle the request and send the email.
You either use AJAX or you
create and append an iframe to the document
set the iframes name to 'foo'
set the forms target to 'foo'
submit
have the forms action render javascript with 'parent.notify(...)' to give feedback
optionally you can remove the iframe
Fastest and easiest way is to use an iframe.
Put a frame at the bottom of your page.
<iframe name="frame"></iframe>
And in your form do this.
<form target="frame">
</form>
and to make the frame invisible in your css.
iframe{
display: none;
}
SUBMITTING THE FORM WITHOUT RELOADING THE PAGE AND GET THE RESULT OF SUBMITTED DATA ON THE SAME PAGE.
Here's some of the code I found on the internet that solves this problem :
1.) IFRAME
When the form is submitted, The action will be executed and target the specific iframe to reload.
index.php
<iframe name="content" style="">
</iframe>
<form action="iframe_content.php" method="post" target="content">
<input type="text" name="Name" value="">
<input type="submit" name="Submit" value="Submit">
</form>
iframe_content.php
<?php
$Submit = isset($_POST['Submit']) ? $_POST['Submit'] : false;
$Name = isset($_POST['Name']) ? $_POST['Name'] : '';
if($Submit){
echo $Name;
}
?>
2.) AJAX
Index.php:
<form >
<input type="" name="name" id="name">
<input type="" name="descr" id="descr">
<input type="submit" name="" value="submit" onclick="return clickButton();">
</form>
<p id="msg"></p>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
<script type="text/javascript">
function clickButton(){
var name=document.getElementById('name').value;
var descr=document.getElementById('descr').value;
$.ajax({
type:"post",
url:"server_action.php",
data:
{
'name' :name,
'descr' :descr
},
cache:false,
success: function (html)
{
alert('Data Send');
$('#msg').html(html);
}
});
return false;
}
</script>
server_action.php
<?php
$name = isset($_POST['name']) ? $_POST['name'] : '';
$descr = isset($_POST['descr']) ? $_POST['descr'] : '';
echo $name;
echo $descr;
?>
Tags: phpajaxjqueryserversidehtml
A further possibility is to make a direct javascript link to your function:
<form action="javascript:your_function();" method="post">
...
It's a must to take help of jquery-ajax in this case. Without ajax, there is currently no solution.
First, call a JavaScript function when the form is submitted. Just set onsubmit="func()". Even if the function is called, the default action of the submission would be performed. If it is performed there would be no way of stoping the page from refreshing or redirecting. So, next task is to prevent the default action. Insert the following line at the start of func().
event.preventDefault()
Now, there will be no redirecting or refreshing. So, you simply make an ajax call from func() and do whatever you want to do when call ends.
Example:
Form:
<form id="form-id" onsubmit="func()">
<input id="input-id" type="text">
</form>
Javascript:
function func(){
event.preventDefault();
var newValue = $('#input-field-id').val();
$.ajax({
type: 'POST',
url: '...',
data: {...},
datatype: 'JSON',
success: function(data){...},
error: function(){...},
});
}
this is exactly how it CAN work without jQuery and AJAX and it's working very well using a simple iFrame. I LOVE IT, works in Opera10, FF3 and IE6. Thanks to some of the above posters pointing me the right direction, that's the only reason I am posting here:
<select name="aAddToPage[65654]"
onchange="
if (bCanAddMore) {
addToPage(65654,this);
}
else {
alert('Could not add another, wait until previous is added.');
this.options[0].selected = true;
};
" />
<option value="">Add to page..</option>
[more options with values here]</select>
<script type="text/javascript">
function addToPage(iProduct, oSelect){
iPage = oSelect.options[oSelect.selectedIndex].value;
if (iPage != "") {
bCanAddMore = false;
window.hiddenFrame.document.formFrame.iProduct.value = iProduct;
window.hiddenFrame.document.formFrame.iAddToPage.value = iPage;
window.hiddenFrame.document.formFrame.submit();
}
}
var bCanAddMore = true;</script>
<iframe name="hiddenFrame" style="display:none;" src="frame.php?p=addProductToPage" onload="bCanAddMore = true;"></iframe>
the php code generating the page that is being called above:
if( $_GET['p'] == 'addProductToPage' ){ // hidden form processing
if(!empty($_POST['iAddToPage'])) {
//.. do something with it..
}
print('
<html>
<body>
<form name="formFrame" id="formFrameId" style="display:none;" method="POST" action="frame.php?p=addProductToPage" >
<input type="hidden" name="iProduct" value="" />
<input type="hidden" name="iAddToPage" value="" />
</form>
</body>
</html>
');
}
This should solve your problem.In this code after submit button click we call jquery ajax and we pass url to posttype POST/GET
data: data information you can select input fields or any other.
sucess: callback if everything is ok from server
function parameter text, html or json, response from server
in sucess you can write write warnings if data you got is in some kind of state and so on. or execute your code what to do next.
<form id='tip'>
Tip somebody: <input name="tip_email" id="tip_email" type="text" size="30" onfocus="tip_div(1);" onblur="tip_div(2);"/>
<input type="submit" id="submit" value="Skicka Tips"/>
<input type="hidden" id="ad_id" name="ad_id" />
</form>
<script>
$( "#tip" ).submit(function( e ) {
e.preventDefault();
$.ajax({
url: tip.php,
type:'POST',
data:
{
tip_email: $('#tip_email').val(),
ad_id: $('#ad_id').val()
},
success: function(msg)
{
alert('Email Sent');
}
});
});
</script>
You can try setting the target attribute of your form to a hidden iframe, so the page containing the form won't get reloaded.
I tried it with file uploads (which we know can't be done via AJAX), and it worked beautifully.
Have you tried using an iFrame? No ajax, and the original page will not load.
You can display the submit form as a separate page inside the iframe, and when it gets submitted the outer/container page will not reload. This solution will not make use of any kind of ajax.
function Foo(){
event.preventDefault();
$.ajax( {
url:"<?php echo base_url();?>Controllername/ctlr_function",
type:"POST",
data:'email='+$("#email").val(),
success:function(msg) {
alert('You are subscribed');
}
} );
}
I tried many times for a good solution and answer by #taufique helped me to arrive at this answer.
NB : Don't forget to put event.preventDefault(); at the beginning of the body of the function .
I did something similar to the jquery above, but I needed to reset my form data and graphic attachment canvases.
So here is what I came up with:
<script>
$(document).ready(function(){
$("#text_only_radio_button_id").click(function(){
$("#single_pic_div").hide();
$("#multi_pic_div").hide();
});
$("#pic_radio_button_id").click(function(){
$("#single_pic_div").show();
$("#multi_pic_div").hide();
});
$("#gallery_radio_button_id").click(function(){
$("#single_pic_div").hide();
$("#multi_pic_div").show();
});
$("#my_Submit_button_ID").click(function() {
$("#single_pic_div").hide();
$("#multi_pic_div").hide();
var url = "script_the_form_gets_posted_to.php";
$.ajax({
type: "POST",
url: url,
data: $("#html_form_id").serialize(),
success: function(){
document.getElementById("html_form_id").reset();
var canvas=document.getElementById("canvas");
var canvasA=document.getElementById("canvasA");
var canvasB=document.getElementById("canvasB");
var canvasC=document.getElementById("canvasC");
var canvasD=document.getElementById("canvasD");
var ctx=canvas.getContext("2d");
var ctxA=canvasA.getContext("2d");
var ctxB=canvasB.getContext("2d");
var ctxC=canvasC.getContext("2d");
var ctxD=canvasD.getContext("2d");
ctx.clearRect(0, 0,480,480);
ctxA.clearRect(0, 0,480,480);
ctxB.clearRect(0, 0,480,480);
ctxC.clearRect(0, 0,480,480);
ctxD.clearRect(0, 0,480,480);
} });
return false;
}); });
</script>
That works well for me, for your application of just an html form, we can simplify this jquery code like this:
<script>
$(document).ready(function(){
$("#my_Submit_button_ID").click(function() {
var url = "script_the_form_gets_posted_to.php";
$.ajax({
type: "POST",
url: url,
data: $("#html_form_id").serialize(),
success: function(){
document.getElementById("html_form_id").reset();
} });
return false;
}); });
</script>
I don't know JavaScript and I just started to learn PHP, so what helped for me from all those responses was:
Create inedx.php and insert:
<iframe name="email" style=""></iframe>
<form action="email.php" method="post" target="email">
<input type="email" name="email" >
<input type="submit" name="Submit" value="Submit">
</form>
Create email.php and insert this code to check if you are getting the data (you should see it on index.php in the iframe):
<?php
if (isset($_POST['Submit'])){
$email = $_POST['email'];
echo $email;
}
?>
If everything is ok, change the code on email.php to:
<?php
if (isset($_POST['Submit'])){
$to = $_POST['email'];
$subject = "Test email";
$message = "Test message";
$headers = "From: test#test.com \r\n";
$headers .= "Reply-To: test#test.com \r\n";
$headers .= "MIME-Version: 1.0\r\n";
$headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n";
mail($to, $subject, $message, $headers);
}
?>
Hope this helps for all other rookies like me :)
Modern Answer without XHR or jQuery
It's 2022, we don't need to use old tools like XHR or jQuery when we have the Fetch API and the FormData API!
The first thing we need to do is prevent the default form submission behavior from occurring with event.preventDefault():
form.addEventListener("submit", function(event){
event.preventDefault();
// ...
});
Now we need to replace the submission behavior with our own AJAX request. The Fetch API makes it pretty simple to post form data - just create a new FormData object, populating it with the form's values, and use it as the body of a fetch request:
fetch(form.action, {
method: "post",
body: new URLSearchParams(new FormData(form))
});
Note that this submits an HTTP request using the multipart/form-data format. If you need to post the data using application/x-www-form-urlencoded, create a new URLSearchParams object from the FormData object and use that as the fetch's body.
fetch(form.action, {
method: "post",
body: new URLSearchParams(new FormData(form))
});
Here's a full code example:
let form = document.querySelector("form");
form.addEventListener("submit", function(event){
event.preventDefault();
fetch(form.action, {
method: "post",
body: //new FormData(form) // for multipart/form-data
new URLSearchParams(new FormData(form)) //for application/x-www-form-urlencoded
});
});
<form method="POST">
<input name="name" placeholder="Name" />
<input name="phone" type="tel" placeholder="Phone" />
<input name="email" type="email" placeholder="Email" />
<input name="submit" type="submit" />
</form>
The page will get reloaded if you don't want to use javascript
You will need to use JavaScript without resulting to an iframe (ugly approach).
You can do it in JavaScript; using jQuery will make it painless.
I suggest you check out AJAX and Posting.
if you're submitting to the same page where the form is you could write the form tags with out an action and it will submit, like this
<form method='post'> <!-- you can see there is no action here-->
Here is some jQuery for posting to a php page and getting html back:
$('form').submit(function() {
$.post('tip.php', function(html) {
// do what you need in your success callback
}
return false;
});
Hi I have been at this for days now and I just cant figure out why this isn't working, please could someone take a look.
index.php
<form id = "update_status" method = "POST">
<textarea id="shadow" name ="user_status" placeholder = "Share what is on your mind!" cols="97" rows="1" title="Share what's on your mind"></textarea>
<input style="float:right; margin-top: 0.3%;margin-right : 0% !important;" id = "btnStatus_update" name = "btnStatus_udate" type="button" value="Add Post" title="Your posts will be made public"></input></form>
swift.php
$(document).ready(function(){
$("#btnStatus_update").click(function(e) {
e.preventDefault();
//alert("Confirming function works");
$.ajax({
cache: false,
type: 'POST',
url: './datacenter.php',
data: $("#user_status"),
success: function(d) {
$("#successMesg").html(d);
}
});
});
});
datacenter.php
if (isset($_POST['user_status'])) {
var_dump($_POST['user_status']);
foreach ($_POST as $ak => $av) {
if (empty(trim($_POST['user_status']))) {
echo 'Say what is on your mind'.' ';
} else {
$userstatus = $_POST['user_status'];
add_user_status($user_data['id'], $userstatus);
}
}
}
using var_dump for testing I was hoping it did return the data, but instead i get NULL, i need the data so i can pass it into the add_user_status function to be added to the database, but it seems there is something missing or off about the code denying me my satisfaction. Please help
There are a few things that appear to be missing:
Index.php:
You need to add an action="something" attribute to the form tag, this tells the form what to do when you submit it. (unless you are manually handling this is JS somewhere else?)
<form id = "update_status" action ="data.php" method = "POST">
Also, unless you are using JavaScript on the index page to handle the actual submitting of the form, your <input> should include a type="submit" attribute. (this will also make it a button) and when clicked it will automatically submit the form to the action location above.
Swift.php:
the code posted is JS, and does what the first line of the previous paragraph mentioned (handles the submit button). Do you include this file inside the index.php? if the index.php cannot see it, then it wont run. It must also be in the html somewhere in a proper <script> block.
I believe the correct way to send form data using ajax is to serialize the data:
$('#update_status').serialize() instead of just sending the one input field.
You will also be required to reference the jQuery libraries, preferably in the index, but could also go in swift.php. I am also assuming that the code posted appears in the necessary <script> block.
Data.php:
should this be datacenter.php? your Swift.php is sending the ajax request to ./datacenter.php
On a side note, if you need it to use Ajax then you actually don't need the action ="data.php" method = "POST" in the form (Ajax does all that for you)
The way it could be done would be something like this:
Index.php:
// HTML beginning stuff
<head>
// Either reference the script in its own JS file or:
// Need to also include jquery library
<script type="text/javascript">
$(document).ready(function(){
$("#btnStatus_update").click(function(e) {
e.preventDefault();
//alert("Confirming function works");
$.ajax({
cache: false,
type: 'POST',
url: './datacenter.php',
data: $('#update_status').serialize(),
success: function(d) {
$("#successMesg").html(d);
}
});
});
});
</script>
</head>
<body>
<form id = "update_status">
<textarea id="shadow" name ="user_status" placeholder = "Share what is on your mind!" cols="97" rows="1" title="Share what's on your mind"></textarea>
<input style="float:right; margin-top: 0.3%;margin-right : 0% !important;" id = "btnStatus_update" name = "btnStatus_udate" type="button" value="Add Post" title="Your posts will be made public"></input>
</form>
</body>
datacenter.php:
<?php
var UserStatus = user_status
if (!empty(UserStatus)) {
var_dump($_POST['user_status']);
// process as necessary
}
?>
But, including the
<form id = "update_status" action ="datacenter.php" method = "POST">
and changing the button to
<input style="float:right; margin-top: 0.3%;margin-right : 0% !important;" id = "btnStatus_update" name = "btnStatus_udate" type="submit" value="Add Post" title="Your posts will be made public">
will allow users to still post information if they have JavaScript disabled.
the selector for the textarea is incorrect, the id is not "user_status", its "shadow"
data: $("#user_status"),
should really be:
data: $("textarea[name=user_status]")
or
data: $("#shadow")
I'm new to jQuery / AJAX.
I'm trying to send single input with jquery/ajax/php.
LIVE EXAMPLE
But, after pressing submit nothing is happening, where is my error?
Any help much appreciated.
HTML:
<form action="submit.php">
<input id="number" name="number" type="text" />
<input id="submit" name="submit" type="submit" />
</form>
JQUERY / AJAX:
$(document).ready(function(e) {
$('input#submit').click(function() {
var number = $('input[name=number]');
var data = 'number=' + number.val();
$.ajax({
url: "submit.php",
type: "GET",
data: data,
cache: false,
success: function(html) {
if (html == 1) {
alert('wyslane');
}
else {
alert('error');
}
}
});
return false;
});
});
PHP:
<?php
$mailTo = 'email#gmail.com';
$mailFrom = 'email#gmail.com';
$subject = 'Call Back';
$number = ($_GET['number']) ? $_GET['number'] : $_POST['number'];
mail($mailTo, $subject, $number, "From: ".$mailFrom);
?>
HTML:
<form id=submit action="">
<input id="number" name="number" type="text" />
<input name="submit" type="submit" />
</form>
The action URL is irrelevant as you want to submit your data via AJAX. Add the submit id to the form and override the default submit behavior, instead of overriding the onclick handler of the submit button. I'll explain in the JS section.
JS:
var number = $('input[name="number"]');
Quotes were missing.
$(document).ready(function(e) {
$('#submit').submit(function() {
var number = $('input[name=number]');
var data = 'number=' + number.val();
$.ajax({
url: "submit.php",
type: "GET",
data: data,
cache: false,
success: function(html) {
if (html == 1) {
alert('wyslane');
}
else {
alert('error');
}
}
});
return false;
});
});
I don't really understand your success callback, why do you expect that html should be equal to 1?
Atleast I got 404 error when pressed your submit button:
Not Found
The requested URL /index.php was not found on this server.
Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.
When you get it to work, remember to add mysql_real_escape_string function to avoid SQL injections http://php.net/manual/en/function.mysql-real-escape-string.php
Since you are also using ID for number, you could just use: var data = 'number=' + $('#number').val()
Also if you add ID to your form, you can use:
$('#formId').submit(function(){
});
instead of that click. This function will launch when that form is submitted. This is better way because users can submit the form with other ways aswell than just clicking the submit button (enter).
var number = $('input[name=number]');
is wrong. It's
var number = $('input[name="number"]');
I have a feedback form in a pop-up div that otherwise works fine but processes SQL twice when the form results in error at first instance.
This is the html form:
<div id="stylized" class="myform">
<form id="form" method="post" name="form">
<p>Report:
<select id="fbtype" name="fbtype">
<option>Bug</option>
<option>Suggestion</option>
<option>Discontentment</option>
<option>Appreciation</option>
</select>
</p>
<p>Brief description:
<textarea name="comments" id="comments" cols="45" rows="10"></textarea>
</p>
<span class="error" style="display:none">Please describe your feedback.</span>
<span class="success" style="display:none">We would like to thank you for your valuable input.</span>
<input type="button" value="Submit" class="submit" onclick="feedback_form_submit()"/>
</form>
</div>
The feedback_form_submit() function is:
function feedback_form_submit() {
$(function() {
$(".submit").click(function() {
var fbtype = $("#fbtype").val();
var comments = $("#comments").val();
var dataString = 'fbtype='+ fbtype + '&comments=' + comments;
if(fbtype=='' || comments=='' )
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "processfeedback.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
});
}
And the processfeedback.php has:
include "./include/session_check.php";
include_once "./include/connect_to_mysql.php";
if (isset($_POST['fbtype'])){
$userid =$_SESSION['id'];
$fbtype=$_POST['fbtype'];
$comments=$_POST['comments'];
$sql = mysql_query("INSERT INTO feedback (userid, type, comments)
VALUES('$userid','$fbtype','$comments')") or die (mysql_error());
}
Could anyone figure out why does the form submits twice? And any suggestion to control this behaviour?
If this is actually the code you're using, you seem to have wrapped your onclick function around the $.click event-adding function:
function feedback_form_submit() {
$(function() {
// This adds a click handler each time you run feedback_form_submit():
$(".submit").click(function() {
// ... //
return false;
});
});
}
When I tried this on jsFiddle.net, I clicked Submit the first time and nothing happened, then the second time it posted twice, the third click posted three times, etc.
You should just keep it simple: take out the onclick attribute:
<input type="button" value="Submit" class="submit" />
and remove the feedback_form_submit() wrapper:
$(function() {
$(".submit").click(function() {
// ... //
return false;
});
});
This way the $.click handler function will be applied just once, when the page loads, and will only run once when Submit is clicked.
EDIT:
If your form is loaded via AJAX in a popup DIV, you have two options:
Keep your onclick but remove the $.click wrapper instead:
function feedback_form_submit() {
// ... //
}
and
<input type="button" value="Submit" class="submit" onclick="feedback_form_submit()" />
Note that you only need to return false if you're using <input type="submit" ... >; when using <input type="button" ... >, the browser does not watch the return value of onclick to determine whether to post the form or not. (The return value may affect event propagation of the click, however ...).
Alternatively, you can use jQuery's $.live function:
$(function() {
$('.submit').live('click',function() {
// ... //
});
});
and
<input type="button" value="Submit" class="submit" />
This has the effect of watching for new DOM elements as they are added dynamically; in your case, new class="submit" elements.
Your feedback_form_submit function doesn't return false and on submit click you're also posting to the server. There is no need to have onClick in:
<input type="button" value="Submit" class="submit" onclick="feedback_form_submit()"/>
Change that to:
<input type="button" value="Submit" class="submit"/>
And change your code to:
// Update: Since you're loading via AJAX, bind it in the success function of the
// AJAX request.
// Let's make a function that handles what should happen when the popup div is rendered
function renderPopupDiv() {
// Bind a handler to submit's click event
$(".submit").click(function(event) {
var fbtype = $("#fbtype").val();
var comments = $("#comments").val();
var dataString = 'fbtype='+ fbtype + '&comments=' + comments;
if(fbtype=='' || comments=='' )
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "processfeedback.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
// This is the success function for the AJAX request that loads the popup div
success: function() {
...
// Run our "onRender" function
renderPopupDiv();
}