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.
Related
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;
});
So, I have a form with one input text field(search_term) and a submit button.
What I'm trying to do is type in a keyword into the input text field, press Submit, the keyword gets sent to a php script that will json_encode it and send it back to the form page where an alert box should appear and show the keyword.
Instead I keep getting null shown in the alert box after pressing the submit.
PHP script works fine if I actually type it out in the url with keyword passed too:
localhost/filter.php?search_term=hey
JavaScript to submit the form and get the value back asynchroniously(without reloading the page):
$('#filter_form').on('submit', function(e){
var filtered_data = null;
e.preventDefault();
$.ajax({
type: 'GET',
dataType: 'json',
url: 'filter.php',
async: false,
success: function(json)
{
filtered_data = json;
}
});
alert(filtered_data);
});
filter.php:
$search_term = $_GET['search_term'];
echo json_encode($search_term);
You need to POST the value of the input to the processing page.
$('#filter_form').on('submit', function(e) {
e.preventDefault();
$inputValue = $('#search_term').val(); //if the id of the input is "search_term"
$.ajax({
type: 'POST',
data: {
'inputValue': $inputValue
},
url: 'filter.php',
success: function(json) {
alert(json);
}
});
});
filter.php should be changed to:
$search_term = $_POST['inputValue'];
echo json_encode($search_term);
I want to enhance my tool's page where as soon use click a button. Request goes to server and depending upon return type (fail/pass) i change color of button. No Refresh/page reload
Page has multiple buttons : some what like below.
Name 9-11 - 11-2 2-5
Resource1 - Button - Button - Button
Resource2 - Button - Button - Button
Resource1 - Button - Button - Button
I am a c++ programmer so you might feel i asked a simple question
Here's a sample of jQuery Ajax posting a Form. Personally, I'm unfamiliar with PHP but Ajax is the same no matter what. You just need to post to something that can return Success = true or false. This POST happens asynchronously so you don't get a page refresh unless you do something specific in the success: section.
$("document").ready(function () {
$('form').submit(function () {
if ($(this).valid()) {
$.ajax({
url: yourUrlHere,
dataType: "json",
cache: false,
type: 'POST',
data: $(this).serialize(),
success: function (result) {
if(result.Success) {
// do nothing
}
}
});
}
return false;
});
});
Of course you don't have to be doing a POST either, it could be a GET
type: 'GET',
And if you don't need to pass any data just leave data: section out. But if you want to specify the data you can with data: { paramName: yourValue },
The cache: false, line can be left out if you want to cache the page. Seeing as how you aren't going to show any changes you can remove that line. jQuery appends a unique value to the Url so as to keep it from caching. Specifying type: "json", or whatever your specific type is, is always a good idea but not necessary.
Try using the $.post or $.get functions in jquery
$.post("url",$("#myform").serialize());
Adding a callback function as Fabrício Matté suggested
$.post("url",$("#myform").serialize(),function(data){alert(data);$("#myform").hide()//?Do something with the returned data here});
Here you go. You will find an example of a form, a button a the necessary ajax processing php page. Try it out and let us know how it goes:
<form action="" method="post" name="my_form" id="my_form">
<input type="submit" name="my_button" id="my_button" value="Submit">
</form>
<script type="text/javascript">
$("document").ready(function () {
$('#my_form').submit(function () {
$.ajax({
url: "ajaxpage.php",
dataType: "json",
type: "POST",
data: $(this).serialize(),
success: function (result)
{
//THere was an error
if(result.error)
{
//So apply 'red' color to button
$("#my_button").addClass('red');
}
else
{
//there was no error. So apply 'green' color
$("#my_button").addClass('green');
}
}
});
return false;
});
});
</script>
<?php
//ajaxpage.php
//Do your processing here
if ( $processed )
{
$error = false;
}
else
{
$error = true;
}
print json_encode(array('error' => $error));
die();
?>
this is an ajax method that inserts the data into a db and should supposedly display the new content.
<script type = "text/javascript">
$(document).ready(function() {
$('#submit').live('click', function(eve) {
eve.preventDefault() ;
var form_data = {
title: $('#title').val()
};
$.ajax({
url: "http://localhost/ci/index.php/chat/comment",
type: 'POST',
data: form_data,
success: function(msg) {
alert(msg);
}
});
});
});
</script>
However in my /chat/comment, i am loading the view again, i.e, user submits a comment, load the view again and the comment should be there. My response from server is the view's HTML. However the view comes with all the divs and there are many of them. I need to retrieve only part of the div, say, #commentspace from the ajax on success.
Look at the jQuery $.load() function?
Example
Inside "firstpage.html"
$('#content').load('secondpage.html #content');
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);
}
});