edit - the info appears to be posting, but on form_data.php it doesn't seem to be retrieving the posted values
Here's the AJAX
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$("#submit_boxes").submit(function() { return false; });
$('input[type=submit]').click(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: $(this).serialize(),
success: function(data) {
$('#view_inputs').html(data); //view_inputs contains a PHP generated table with data that is processed from the post. Is this doable or does it have to be javascript?
});
return false;
});
};
</script>
</head>
Here is the form I'm trying to submit
<form action="#" id = "submit_boxes">
<input type= "submit" name="submit_value"/>
<input type="textbox" name="new_input">
</form>
Here is the form_data page that gets the info posted to
<?php
if($_POST['new_input']){
echo "submitted";
$value = $_POST['new_input'];
$add_to_box = new dynamic_box();
array_push($add_to_box->box_values,$value);
print_r($add_to_box->box_values);
}
?>
Your form is submitting because you have errors which prevents the code that stops the form from submiting from running. Specifically dataType: dataType and this.html(data) . Firstly dataType is undefined, if you don't know what to set the data type to then leave it out. Secondly this refers to the form element which has no html method, you probably meant $(this).html(data) although this is unlikely what you wanted, most likely its $(this).serialize() you want. So your code should look like
$('form#submit_boxes').submit(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: $(this).serialize(),
success: success
})
return false;
});
Additionally if you have to debug ajax in a form submit handler the first thing you do is prevent the form from submitting(returning false can only be done at the end) so you can see what errors occurred.
$('form#submit_boxes').submit(function(event) {
event.preventDefault();
...
});
You can use jQuery's .serialize() method to send form data
Some nice links below for you to understand that
jquery form.serialize and other parameters
http://www.tutorialspoint.com/jquery/ajax-serialize.htm
http://api.jquery.com/serialize/
One way to handle it...
Cancel the usual form submit:
$("#submit_boxes").submit(function() { return false; });
Then assign a click handler to your button:
$('input[type=submit]').click(function() {
$.ajax({
type: 'POST',
url: 'form_data.php',
data: this.html(data),
success: success,
dataType: dataType
})
return false;
});
Related
I'm learning AJAX by reading some online tutorials, so please understand I am very new to AJAX and programming in general. I have managed to do the following with 3 selectboxes:
populates selectbox 2 based on selection from selectbox 1
populates selectbox 3 based on selection from selectbox 2
Everything is working perfectly
Here is my code:
$(document).ready(function()
{
$(".sport").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_sport.php",
dataType : 'html',
data: dataString,
cache: false,
success: function(html)
{
$(".tournament").html(html);
}
});
});
$(".tournament").change(function()
{
var id=$(this).val();
var dataString = 'id='+ id;
$.ajax
({
type: "POST",
url: "get_round.php",
data: dataString,
cache: false,
success: function(html)
{
$(".round").html(html);
}
});
});
});
</script>
Here is an Example
What I want to do
I would like to send the value of the 3 selectboxes to 3 php variables without the form reloading.
My Problem
When the user clicks submit:
The form reloads (which I dont want)
The selectbox values does not get send to my php variables
my code to get the values after submit is clicked is as follows:
if(isset($_POST['submit'])){
$a = $_POST['sport'];
$b = $_POST['tournament'];
:
}
However my code is flawed as I mentioned above.
If any one can help me to explain how to send my form data to the 3 php variables without the form reloading it will be greatly appreciated
If you don't want to submit your form when you click the button, you need to set that input as button and not submit. You can, also, attach the submit event handler to the form and prevent it to submit:
$("form").on("submit", function(e){
e.preventDefault(); //This is one option
return false; //This is another option (and return true if you want to submit it).
});
So, being said this, you could probably do something like:
$("form").on("submit", function(e) {
var formData = $(this).serialize();
e.preventDefault();
$.ajax({
url: 'yoururl',
data: formData,
type: 'post', //Based on what you have in your backend side
success: function(data) {
//Whatever you want to do once the server returns a success response
}
});
});
In your backend:
if (isset($_POST["sport"])) {
//Do something with sport
}
if (isset($_POST["tournament"])) {
//Do something with torunament
}
echo "Successfull response!"; //You have to "write" something in your response and that is what the frontend is going to receive.
Hope this helps!
Try using the javascript function preventDefault().
See this SO question.
Use a <button>Submit</button> element instead of <input type="submit"/> since the submit automatically submits the form.
Edit: And you would have to use on.('click') instead of looking for submit event in your jQuery.
Apologies if this has been answered before (I couldn't find the answer when I searched the archives)
I've got a page protected by a password:
<?php
if($_POST['pw'] == 'pw')
{
//Page content
} else
{
//Display password form
}
?>
Within the page content, I've got another form, which I want to submit using jQuery, and have the following code:
<script type='text/javascript'>
var dataString = $('input#input1').val();
$(function() {
$('#submit').click(function()
{
$.ajax({
type: 'POST',
url: 'p2.php',
data: dataString,
dataType: html,
success: function(data2) {
$('#testResult').html(data2);
}
});
return false;
});
});
</script>
<form name='form1' id='form1' action=''>
<fieldset>
<label for='input1' id='input1_label'>Input 1</label>
<input type='text' name='input1' id='input1' size='30' />
<input type='submit' value='Update / reset' id='submit' class='buttons' />
</fieldset>
</form>
<div id='#testResult'></div>;
However, clicking submit then sends the form to p1.php?input1=test (i.e., the data string is being sent to p1.php, not p2.php). If I edit the code and remove dataType:html and the 2 references of data2, then this doesn't happen (infact, nothing happens, so I assume that jQuery is submitting the data to the form). I've also changed the type to 'GET', incase the 2 POST requests on the same page were causing problems, but this didn't change the result.
What am I missing to get the information from p2.php (i.e. data2) and displaying it?!
EDIT
Thanks to a comment pointing out a typo, I've changed dataType: html to dataType: 'html' - this now doesn't cause the page to redirect to p1.php?input1=test, but once again, it doesn't do anything (when it should still be returning the value of data2)
EDIT 2
I've updated the code so dataString is now:
var dataString = $('input#input1').val();
dataString = 'var1='+dataString;
but this hasn't made any difference
For clarification, my p2.php just contains the following:
<?php
echo "<p>HELLO!</p>";
?>
EDIT 3
I made the changes to my code has suggested by Damien below; I get the alert of "works!" but still nothing seems to be returned from p2.php, and nothing is inserted into the #testResult div.
var dataString = $('input#input1').val();
$(function() {
$('#submit').click(function(evt)
{
evt.preventDefault();
$.ajax({
type: 'POST',
url: 'p2.php',
data: "someval="+dataString,
dataType: 'html',
success: function(data2) {
$('#testResult').html(data2);
}
});
return false;
});
});
$(function() {
$('#submit').click(function()
{
var dataString = $('#form1').serialize();
$.ajax({
type: 'POST',
url: 'p2.php',
data: dataString,
success: function(data2) {
alert('works!'); // ADDED AFTER UPDATE
$('#testResult').html(data2);
},
/* ADDED AFTER UPDATE */
error:function(obj,status,error)
{
alert(error);
}
});
return false;
});
});
Edit:
In p2.php:
<?php
var_dump($_POST['pw']);
?>
In p2.php you then need to output ( using echo, for example) what you want to be returned as 'data2' in your ajax success call.
UPDATE:
Since you're Ajax request fires succesfully, that means either your post is not passed correctly, or you're not outputting anything. I've re-looked at your code and I saw this:
<input type='text' name='input1' id='input1' size='30' />
that means you're fetching the wrong $_POST variable!
Do this:
Since you're sending a name="input1", in your p2.php try with:
<?php
if(isset($_POST['input1'])
{
echo $_POST['input1'];
}
else
{
echo 'No post variable!';
}
And in your jquery success:
success: function(data2) {
alert(data2);
$('#testResult').html(data2);
},
That oughta work, if you follow it literally. In the remote possibility it won't work, forget AJAX, remove the javascript and do a normal post submitting with p2.php as an action of your form :)
I think you have to prevent the default action of the form.
Try this:
$('#submit').click(function(e)
{
e.preventDefault();
$.ajax({
type: 'POST',
url: 'p2.php',
data: dataString,
dataType: html,
success: function(data2) {
$('#testResult').html(data2);
}
});
return false;
});
});
The data should be formatted like this:
variable1=value1&variable2=varlue2
I also think you can remove the dataType property.
I'm currently learning PHP. I've made a simple script # http://hash.techho.me, only thing is, I want the form to submit then load the results via AJAX, without the user leaving the page. Possible?
post the form using ajax
$.ajax({
url:'yoururl',
data:$("form").serialize(),
type:'POST',
success:function(data){
alert("success");
},
error:function(jxhr){
alert(jxhr.responseText);
}
});
jQuery.ajax() – jQuery API
Posting to the same page should do the trick. No need to use ajax for that
> <?php
>
> //do stuff with $_POST
> ?>
>
> <html> <body> <form method="post">
>
> <?php echo $result ?>
>
> </form>
> </body>
Fike
use ajax for this, lets suppose try this one for your practice
var string = $("#string").val();
var dataString = 'string=' + string ;
if(string==''){
alert('enter any string');
}
else{
$.ajax({
type: "POST",
url: "path of php file",
data: dataString,
suceess: function(){
//do something
},
error: function(){
//do something
}
});
}
You can use jQuery or Prototype JS libraries to make an easy AJAX call. Example using jQuery would be:
$.ajax({
url:'hashed.php',
data:$("form").serialize(),
type:'POST',
success: function(data){
$('hashmd5').html(data.md5);
$('hashsha1').html(data.sha1);
},
error: function(jxhr){
alert(jxhr.responseText);
}
});
Don't use the same id value in HTML, never ever. They must be unique to correct perform JavaScript functions on elements.
yes it is possible. Write a javascript function that would trigger on submit, disable the submit button so user couldn't press it again, and finally request the server via ajax. on successful response update the content. Something like following in Jquery
$('.form-submit').click(function(event)) {
event.preventDefault();
if(form is valid and not empty) {
$.ajax({
type: "POST",
url: "path to script that will handle insetion",
data: "data from form", //like ({username : $('#username').val()}),
suceess: function(data){
//update the content or what. data is the response got from server. you can also do like this to show feedback etc...
$('.feedback').html("Data has been saved successfully");
},
error: function(){
$('.feedback').html("Data couldn't be saved");
}
});
}
}
Just started using AJAX today via JQuery and I am getting nowhere. As an example I have set up a job for it to do. Submit a form and then display the results. Obviously I haven't got it right.
The HTML.
<form id="PST_DT" name="PST_DT" method="post">
<input name="product_title_1682" id="product_title_1682" type="hidden" value="PADI Open Water">
<input name="product_title_1683" id="product_title_1683" type="hidden" value="PADI Advanced Open Water">
<input type="submit" name="submit" id="submit" value="Continue" onclick="product_analysis_global(); test();"/>
</form>
<span id="results"></span>
There are actually many more fields all loaded in dynamically. I plan to use ajax to submit to PHP for some simple maths and then return the results but we can worry about that later.
The JQuery
function test() {
//Get the data from all the fields
var alpha = $('#product_title_1682').val();
JQuery.ajax({
type: 'POST',
url: 'http://www.divethegap.com/update/functions/totals.php',
data: 'text=' + alpha,
beforeSend: function () {
$('#results').html('processing');
},
error: function () {
$('#results').html('failure');
},
timeout: 3000,
});
};
and the PHP
<?php
$alpha = $_POST['alpha'];
echo 'Marvellous',$alpha;
?>
That's my attempt and nothing happens. Any ideas?
Marvellous.
First of all, you're passing the $_POST variable as 'text' while your script is looking for $_POST['alpha']. If you update your PHP to $_POST['text'], you should see the proper text.
Also, if your form is going to have lots of inputs and you want to be sure to pass all of them to your AJAX Request, I'd recommend using jQuery's serialize() method.
data: $('#PST_DT').serialize(), // this will build query string based off the <form>
// eg: product_title_1682=PADI+Open+Water&product_title_1683=PADI+Advanced+Open+Water
In your PHP script you'd then need to use $_POST['product_title_1682'] and $_POST['product_title_1683'].
UPDATE Add a success callback to your $.ajax call.
function test() {
// serialize form data
var data= $('#PST_DT').serialize();
// ajax request
$.ajax({
type : 'POST',
url : 'http://www.divethegap.com/update/functions/totals.php',
data : data,
beforeSend : function() {
$('#results').html('processing');
},
error : function() {
$('#results').html('failure');
},
// success callback
success : function (response) {
$('#results').html(response);
},
timeout : 3000,
});
};
In your PHP script you can debug the information sent using:
var_dump($_POST);
In your AJAX request, you are sending the parameter foo.php?text=..., but in the PHP file, you're calling $_POST['alpha'], which looks for foo.php?alpha=....
Change $_POST['alpha'] to $_POST['text'] and see if that works.
There is a simpler method:
$("#PST_DT").submit(function(e){
e.preventDefault();
$.ajax({
data: $(this).serialize(),
type: "POST",
url: 'http://www.divethegap.com/update/functions/totals.php',
success: function(){
....do stuff.
}
});
return false;
});
This will allow you to process the variables like normal.
Currently my AJAX is working like this:
index.php
<a href='one.php' class='ajax'>One</a>
<div id="workspace">workspace</div>
one.php
$arr = array ( "workspace" => "One" );
echo json_encode( $arr );
ajax.js
jQuery(document).ready(function(){
jQuery('.ajax').live('click', function(event) {
event.preventDefault();
jQuery.getJSON(this.href, function(snippets) {
for(var id in snippets) {
jQuery('#' + id).html(snippets[id]);
}
});
});
});
Above code is working perfectly. When I click link 'One' then one.php is executed and String "One" is loaded into workspace DIV.
Question:
Now I want to submit a form with AJAX. For example I have a form in index.php like this.
<form id='myForm' action='one.php' method='post'>
<input type='text' name='myText'>
<input type='submit' name='myButton' value='Submit'>
</form>
When I submit the form then one.php should print the textbox value in workspace DIV.
$arr = array ( "workspace" => $_POST['myText'] );
echo json_encode( $arr );
How to code js to submit the form with AJAX/JSON.
Thanks
Here is my complete solution:
jQuery('#myForm').live('submit',function(event) {
$.ajax({
url: 'one.php',
type: 'POST',
dataType: 'json',
data: $('#myForm').serialize(),
success: function( data ) {
for(var id in data) {
jQuery('#' + id).html(data[id]);
}
}
});
return false;
});
Submitting the form is easy:
$j('#myForm').submit();
However that will post back the entire page.
A post via an Ajax call is easy too:
$j.ajax({
type: 'POST',
url: 'one.php',
data: {
myText: $j('#myText').val(),
myButton: $j('#myButton').val()
},
success: function(response, textStatus, XMLHttpRequest) {
$j('div.ajax').html(response);
}
});
If you then want to do something with the result you have two options - you can either explicitly set the success function (which I've done above) or you can use the load helper method:
$j('div.ajax').load('one.php', data);
Unfortunately there's one messy bit that you're stuck with: populating that data object with the form variables to post.
However it should be a fairly simple loop.
Have a look at the $.ajaxSubmit function in the jQuery Form Plugin. Should be as simple as
$('#myForm').ajaxSubmit();
You may also want to bind to the form submit event so that all submissions go via AJAX, as the example on the linked page shows.
You can submit the form with jQuery's $.ajax method like this:
$.ajax({
url: 'one.php',
type: 'POST',
data: $('#myForm').serialize(),
success:function(data){
alert(data);
}
});