I have the following 2 files :
<?php
?>
<script src="http://code.jquery.com/jquery-1.10.2.min.js"></script>
<script>
$(document).ready(function () {
$("form").submit(function () {
var textu = $("#textu").val();
$.post("index.php", function (data) {
alert(data);
});
return false;
});
});
<script>
<form method="post" action="">
<input type="text" id="textu" name="textu">
<input type="submit" value="send" id="send">
</form>
Now I want to style the data that it's returned from the index.php where I sent the values.
The problem is that if I use something like <h1> or any html tag in the file index.php the test file will just return it as plain text for example <h1>text</h1>.
I want to style the data that comes from index.php but I can't manage to do it because it keeps returning me the plain text and I can't use any html tags or anything apart from PHP in that file. How can I do that ? Any ideas ?
try this:
html
<input type="text" id="textu" name="textu">
<input value="send" id="send">
jQuery:
$("#send").click(function(){
var textu = $("#textu").val();
$.post("index.php",{"textu":textu},function(data){
alert(data);
});
});
Add datatype html as third parameter to $.post - http://api.jquery.com/jQuery.post/
$.post("index.php", function (data) {
alert(data);
}, 'html');
Related
I just want to know how i can send a "callback" message for "success" or "error".
I really don't know much about jquery/ajax, but, i tried to do this:
I have a basic form with some informations and i sent the informations for a "test.php" with POST method.
My send (not input) have this id: "#send". And here is my JS in the index.html
$(document).ready(function() {
$("#send").click(function(e) {
e.preventDefault();
$(".message").load('teste.php');
});
});
And, in my PHP (test.php) have this:
<?php
$name = $_POST['name'];
if($name == "Test")
{
echo "Success!";
}
else{
echo "Error :(";
}
?>
When i click in the button, the message is always:
Notice: Undefined index: name in /Applications/XAMPP/xamppfiles/htdocs/sites/port/public/test.php on line 3
Error :(
Help :'(
This is your new JS:
$(document).ready(function()
{
$("#send").click(function(e) {
e.preventDefault();
var form_data = $("#my_form").serialize();
$.post('teste.php', form_data, function(data){
$(".message").empty().append(data);
});
});
});
This is your new HTML:
<form id="my_form">
<input type="text" name="name" value="" />
<input type="button" id="send" value="Send" />
</form>
The problem is you have not passed name data to your PHP Use My Javascript Code.
Problem in understanding please reply
$(document).ready(function() {
$(document).on('click','#send',function(e)
{
var params={};
params.name="Your Name ";
$.post('test.php',params,function(response)
{
e.preventDefault();
alert(response); //Alert Response
$(".message").html(response); //Load Response in message class div span or anywhere
});
});
});
This is somewhat more complicated by you can use it more generally in your project. just add a new callback function for each of the forms that you want to use.
<form method="POST" action="test.php" id="nameForm">
<input name="name">
<input type="submit">
</form>
<script>
// wrap everything in an anonymous function
// as not to pollute the global namespace
(function($){
// document ready
$(function(){
$('#nameForm').on('submit', {callback: nameFormCallback },submitForm);
});
// specific code to your form
var nameFormCallback = function(data) {
alert(data);
};
// general form submit function
var submitForm = function(event) {
event.preventDefault();
event.stopPropagation();
var data = $(event.target).serialize();
// you could validate your form here
// post the form data to your form action
$.ajax({
url : event.target.action,
type: 'POST',
data: data,
success: function(data){
event.data.callback(data);
}
});
};
}(jQuery));
</script>
I have a form where I use a GET/IMPORT button to get values from other document and into the current form as shown below. Previously I had set fix value: $file = '/user/doc.xml'; inside meta.php so when I pressed the GET/IMPORT button it got the results from the /user/doc.xml file.
Now I have added <input name="file" value="" /> to this form and want to send path to $file using this filed. I know when I press GET/IMPORT button then meta.php file is called.
So my question is: How can I send $file value/path to meta.php using this input field.
Here is my script and the $file input field:
<form id=file method="POST" >
<input name="file" value="" />
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js">
</script>
<script type="text/javascript">
$(document).ready(function() {
$('.button').click(function() {
$.get('meta.php', function(data) {
result = $.parseJSON(data);
$("input[name='nick_name']").val(result.avaname);
$("#age").val(result.tavaage).attr("selected","selected");
});
});
});
</script>
<input class="button" type="button" value="GET/IMPORT" />
</form>
Thank you for all help.
Just get the value from the input field and add it to the querystring of the url. Then in meta.php you can read the filename from the request parameters and load that file.
Don't forget to validate the filename.
<input name="file" type="text" id="file">
<script type="text/javascript">
$(document).ready(function() {
$('.button').click(function() {
var val = $('#file').val();
$.get('meta.php', {file: val}, function(data) {
var result = $.parseJSON(data);
$('input[name="nick_name"]').val(result.avaname);
$('#age').val(result.tavaage).attr('selected', 'selected');
});
});
});
</script>
<input type="button" class="button" value="GET/IMPORT">
Then in your meta.php you can get the the filename with $_GET['file'].
Hope it helps
You can use the $.get method with "data" like this :
$(document).ready(function() {
$('.button').click(function() {
$.get('meta.php', {file: $('input[name="file"]').val()}; function(data) {
result = $.parseJSON(data);
$("input[name='nick_name']").val(result.avaname);
$("#age").val(result.tavaage).attr("selected","selected");
});
});
});
issues are still there...pls help
I am unable to load external file while using AJAX jquery. I want to use Jquery ajax to pop up form then validate, enter data in mysql. but starting from a simple ajax function. kindly let me know where i am going wrong
<link rel="stylesheet" type="text/css" media="all" href="test_style.css">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
});
});
</script>
</head>
<body>
<div id="contact_form">
<form id="ajax-contact-form" name="contact" action="">
<fieldset>
<label for="name" id="name_label">Name</label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" />
<label class="error" for="name" id="name_error">This field is required.</label>
<INPUT class="button" type="submit" name="submit" value="Send Message">
</fieldset>
</form>
</div>
</body>
</html>
and contact.php file is
<?php
echo "Hello";
?>
You need to return false; to prevent the form from submitting and refreshing the page and check if your $("#div1") is missing.
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax(
{
type: "POST",
url:"contact.php",
data: str,
success:function(result)
{
$("#div1").html(result);
}
});
return false;
});
});
Simple. Since you are posting your form through ajax, you must prevent the default form submit by returning a false inside the submit method. Below is the correct version:
<script>
$(document).ready(function(){
$("#ajax-contact-form").submit(function(){
var str = $(this).serialize();
$.ajax({
type: "POST",
url:"contact.php",
data: str,
success:function(result) {
$("#div1").html(result);
}
});
return false;
});
});
</script>
You can use a more simple form of post request as follows:
$.post("url",{var1: value1, var2: value2},function(data,status){
if(status=='success')
alert(data);
});
the second argument you can pass as many using this post request. The first argument url, if ofcourse relative to the document in which this js is loaded or you can give the exact url on the server.
According to your php file, data=='Hello'.
Similar is the procedure for any GET request also.
Make sure you are missing the div1
please use
<div id="div1"><div>
i am building one form with fancy box, form work fine but i could not get value from text box.
my java script is
$(function () {
$("#button").click(function () {
var yourName = $("input#yourName").val();
alert(yourName);
if (yourName == "") {
$('input#yourName').css({
backgroundColor: "#F90"
});
$("input#yourName").focus();
return false;
}
$.ajax({
type: "POST",
url: "bin/form1.php",
data: $("#login_form").serialize(),
context: document.body,
success: function (res) {
var success = '<?php echo sprintf(_m('Success name % s '), "" ); ?>';
success = success.replace(/^\s*|\s*$/g, "");
var RegularExpression = new RegExp(success);
if (RegularExpression.test(res)) {
alert('Message Sent');
$.fancybox.close();
} else {
alert('Error While processing');
}
},
});
return false;
});
});
now my html is
<div style="display:none; height:450px">
<form id="login_form" action="">
<p id="login_error">
Please, enter data
</p>
<input type="hidden" name="action" value="send_friend_post" />
<label for="yourName" style="margin-right:110px;">
<?php _e( 'Your name', 'newcorp') ; ?>
</label>
<input id="yourName" type="text" name="yourName" value="" class="text_new " />
<br/>
<input type="submit" value="Submit" id="button" class="text_new" style="background:#930; color:#FFF; width:250px; margin-left:170px" />
</form>
</div>
i am opening this fancy box popup from
<a id="tip5" title="" href="#login_form"><?php echo "Login" ; ?></a>
every thing work fine but i can't get value of yourName even alert is showing blank.
Thanks
instead of this
var yourName = $("input#yourName").val();
try
var yourName = $("input[name='yourName']:text").val();
this will make ur function read the value of text box .
Firstly, with your form, it should be better to use
$(document).on('submit', 'form', function () { ... })
instead of
$(document).on('click', '#myButton', function () { ... })
Secondly, you should encapsulate your code into $(document).ready(function () { .. });
Here is a working example ;) http://jsfiddle.net/aANmv/1/
Check that the ID of Your input is not changed by fancybox at the time of opening the popup.
Instead of directly catching up the click event try to live the event if using jQuery 1.7 > or on when using jQuery 1.7 <.
live:
$("#button").live('click', function() { ... });
on:
$("#button").on('click', function() { ... });
Also try to load the javascript after the DOM is loaded:
$(document).ready(function(){
$("#button").live('click', function() { ... });
...
});
Instead of that
$(function() {
try
$(document).ready() {
this way you are sure that code is executed only after the document has loaded completely. Oh, and you could use #yourName instead of input#yourName, if i'm not wrong it should be faster.
Could someone provide me with the most simple code for form submission with jquery. On the web is with all sorts of gizmo coding.
$('#your_form_id').submit(function(){
var dataString = $("#your_form_id").serialize();
$.ajax({
type: "POST",
url: "submit.php",
data: dataString,
success: function() {
alert('Sent!');
}
});
return false;
});
here is a another solution, not as simple as the Jquery Form Plugin, but it can be useful if you want to handle errors codes and messages by yourself
look at this HTML + Javascript sample :
<div>
<form method="post" id="fm-form" action ="">
<label>Name:</label>
<input type="text" id="fm-name" name="fm-name" value="" />
<label>Email:</label>
<input type="text" id="fm-email" name="fm-email" value="" />
<label>Birthdate:</label>
<input type="text" id="fm-birthdate" name="fm-birthdate" value="" />
<input type="submit" id="fm-submit" value="Save it">
</form>
</div>
<script type="text/javascript">
$(function() {
// disable the form submission
$("#fm-form").submit(function () { return false; });
// post the datas to "submit_form.php"
$("#fm-submit").click(function() {
$.post("/ajax/submit_form.php",
{ 'fm-name':$("#fm-name").val(),
'fm-email':$("#fm-email").val(),
'fm-birthdate':$("#fm-birthdate").val()
}
,function(xml) {
// submit_form.php will return an XML or JSON document
// check it for potential errors
});
});
});
</script>
What you want is jquery form plugin. It allows you to simply send normal 'form' using ajax - you can make a non-visible form and use this plugin to subnmit it. The option in Joel's answer is possible as well, it depends on the complexity of the thing you want to submit.
Take a look at the Form plugin:
$(function() {
$('#myForm').ajaxForm(function() {
alert("Thank you for your comment!");
});
});