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How to make an AJAX call to an html element?

What I want to do is pretty simple. I want to make an AJAX call to a specific html class, so that whenever the html page is loaded, jquery will make an AJAX call to that specific html div class.
For example:
<div class="targeted"></div>
In jquery:
$('.targeted')
I know that the syntax to make an AJAX call is:
$.ajax({
type: "GET",
url: "/api/something",
success: function(data) {
console.log(data);
}
});
But how do I implement this AJAX call to the $('.targeted') whenever the page is loaded?
Thanks

If you mean you want to display the result of the ajax call in the element, you update the element from within the success callback:
$.ajax({
type: "GET",
url: "/api/something",
success: function(data) {
$('.targeted').html(data);
}
});
That example assumes
You want to replace the content of the element (rather than adding to it); more options in the jQuery API.
data will be HTML. If it's plain text, use .text(data), not .html(data). If it's structured data, then of course you'll need to do more work to put the information in the desired form.

window.onload = function() {
yourFunction();
};
function yourFunction(){
$.ajax({
type: "GET",
url: "/api/something",
success: function(data) {
$('.targeted').html(data);
}
});
}
OR Drectly you can pass that method in document ready it will execute automatically
$(document).ready(function(){
//This will execute onload oof your web page what you required
yourFunction();
})
function yourFunction(){
$.ajax({
type: "GET",
url: "/api/something",
success: function(data) {
$('.targeted').html(data);
}
});
}

For when the page is loaded, you use:
$( document ).ready(function() {
console.log( "ready!" );
});
Inside the document ready, you put your AJAX call. If the result you get is in JSON format, you need to include the dataType as well like this:
$.ajax({
method: "GET",
url: "/api/something",
dataType: "json"
})
.done(function( data ) {
$('.targeted').append(JSON.stringify(data));
});
If the result is not JSON, then you can just append the data.
Also note:
The jqXHR.success(), jqXHR.error() and jqXHR.complete() callbacks are removed as of jQuery 3.0. You can use jqXHR.done(), jqXHR.fail() and jqXHR.always() instead.
Please look at the jQuery documentation.

you can use jquery load like this:
$(".targeted").load('/api/something');
if you want to wait untill after the page is loaded, wrap it with window load like so:
$(window).load(function () {
$(".targeted").load('/api/something');
});
P.S. $(window).load(..) and $(".class").load(url) are two different functions

You can do:
$(function() {
$.ajax({
type: "GET",
url: "/api/something",
})
.done(function(data) {
$('.targeted').text(data);
});
});

(this).parent().find not working for getting response in ajax call

$(".content-short").click(function() {
$(".content-full").empty();
var contentid=$(this).parent().find(".content-full").attr('data-id');
var content=$(this).parent().find(".content-full");
alert(contentid);
var collegename = $(this).attr('data-id');
$.ajax({
type: "post",
url: "contenthome.php",
data: 'collegename=' + collegename,
dataType: "text",
success: function(response) {
$content.html(response);
}
});
});
here the alert displays the specific data-id but
content=$(this).parent().find(".content-full");
this didn't displays data in content-full div with that specific data-id
anything wrong in the code or something else?
the query displays data if i use(."content-full"); instead of
$(this).parent().find(".content-full");

Inside the ajax callback you are using $content, but you declare your variable as content. May that be the problem?
Your question is not clear. What are you trying to achieve?

Ajax request with on method of jquery

I have a doubt, I use Jquery to load dynamic wall posts of people who follow me. Now since Jquery dosent work on dynamic content in the tradition click method I use on method for the response.If i dont trigger a Jquery method but do other things it works. I want to know how to launch Ajax method.
My Code:
$('body').on('click','.likebutton', $.ajax({
method:"POST",
url:"../assets/backend/likepost/index.php",
data:"postid="+$('.likebutton').attr('id'),
success:function(response) {
$(this).find(".liketext").html("<p class='liketext'><span class='glyphicon glyphicon-ok'></span> You like this</p>");
}
Any improvment to the code would be greatly appreciated. Thnx

Use following format:
$('body').on('click','.likebutton', function(){
var that = this;
/*
place code of other processing
*/
$.ajax({
method:"POST",
url:"../assets/backend/likepost/index.php",
data:"postid="+$(that).attr('id'),
success:function(response) {
$(that).find(".liketext").html("<p class='liketext'><span class='glyphicon glyphicon-ok'></span> You like this</p>");
}
});
}

You need to wrap ajax function call in an anonymous functions.
Use
$('body').on('click', '.likebutton', function() {
//Store the reference in a variable
var _this = $(this);
$.ajax({
method: "POST",
url: "../assets/backend/likepost/index.php",
data: "postid=" + this.id, //fetch id using this
success: function(response) {
//Use the refrence to set html
_this.find(".liketext").html("<p class='liketext'><span class='glyphicon glyphicon-ok'></span> You like this</p>");
}
});
});

How to redirect response from ajax request

I need to redirect to a page from response. I made a ajax call and can handle success. There is html page in response, but how to redirect it to that page.
Here's my code.
$("#launchId").live('click',function(){
var id= $("#id").val();
var data = 'id='+id;
$.ajax({
url: "xyz.json",
type: "post",
data: data,
dataType: 'json',
complete : function(response) {
window.location.href = response;
}
});
});

Not using ajax would make this easier:
<form type="POST" action="xyz.json">
<label for="id">Enter ID:</label><input id="id" name="id">
<button type="submit" id="launchId">Send</button>
</form>
If you really want to use ajax, you should generate a distinct server response, containing only the HTML parts you want to update in your page or actual JSON.
If you insist on using the response which you currently get, the appropriate way of dealing with it would be document.write:
$.ajax({
url: "xyz.json",
type: "post",
data: data,
dataType: 'html', // it's no JSON response!
success: function(response) {
document.write(response); // overwrite current document
},
error: function(err) {
alert(err+" did happen, please retry");
}
});

Please try this.
var newDoc = document.open("text/html", "replace");
newDoc.write(response.responseText);
newDoc.close();

Your response is an object containing the full HTML for a page in the responseText property.
You can probably do $(body).html(response.responseText); instead of window.location.href = ...; to overwrite the current page content with what you got a response.
...
complete : function(response) {
$(body).html(response.responseText);
}
But i suggest you don't and there could be style and other conflicts with whats already there on the page.

In your HTML add a div with id as 'content', something like this
<div id='content'/>
Since your response is html in your complete function append the content into the div like this -
complete : function(response) {
$('#content').append(response.responseText);
}
Let me know if you still face issues.

try this
$("#launchId").live('click',function(){
var id= $("#id").val();
var data = 'id='+id;
$.ajax({
url: "xyz.json",
type: "post",
data: data,
dataType: 'json',
complete : function(response) {
window.location.href = '/yourlocation?'+response;
}
});
});

Search box in a jQuery ajax success page - issues in loop

Firstly, there have some tag links in my main page. click each one, post value to b.php with jquery.ajax and turn back value in div#result.
b.php have a search box. when search something in it. the result data will still show in the div#result.
my problem is: I know if I will do jQuery ajax in the b.php, I shall write the jQuery code in the first success part. but this only can control one time, when I continue search in the search box, the jQuery not work. I think I met a loop problem. How to solve it?
a.php
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('.click').click(function(){
var value1 = $(this).text();
$.ajax({
url: "b.php",
dataType: "html",
type: 'POST',
data: "data=" + value1,
success: function(data){
$("#result").html(data);
$('#search').click(function(){
var value = $('#search1').val();
$.ajax({
url: "b.php",
dataType: "html",
type: 'POST',
data: "data=" + value,
success: function(data){
$("#result").html(data);
}
});
});
}
});
});
});
</script>
<a rel="aa" class="click">aa</a>
<a rel="aa" class="click">bb</a>
<div id="result"></div>
b.php
<?php
echo $_POST['data'];
?>
<form name="form">
<input type="text" value="" id="search1">
<a name="nfSearch" id="search">search</a>
</form>

When a new element is introduced to the page the jQuery .click() method becomes useless because it can only see elements that were part of the original DOM. What you need to use instead is the jQuery .live() method which allows you to bind events to elements that were created after the DOM was loaded. You can read more about how to use it at the below link.
.live() – jQuery API
$('#search').live('click', function(e) {
// Prevent the default action
e.preventDefault();
// Your code here....
});

First of all i think you should attach the ajax call to the click on the link: the way you are doing right now just execute an ajax call as soon as the page is loaded.
$(document).ready(function(){
//when you click a link call b.php
$('a.yourclass').click(function(){
$.ajax({
url: "b.php",
dataType: "html",
type: 'POST',
data: "data = something",
success: function(data){
$("#result").html(data);
var value = $('#search').val();
$.ajax({
url: "b.php",
dataType: "html",
type: 'POST',
data: "data =" + value,
success: function(data){
$("#result").html(data);
}
});
}
});
});
});
In this way, each time a link with the class of "yourclass" is clicked an ajax call to b.php is sent and if it succed, another call is made (always to b.php). I don't understand if this is what you are looking fo, if you post your html my answer can be better.
In b.php of course you need to echo some html that can be used in the callback

It's strange how your attempting to do two ajax requests like that, surely one is enough. If you need to support multiple text boxes then you just adjust your selectors.
Your whole code can be shortended down to something like this:
$(document).ready(function() {
$('#result').load('b.php', { data: $('#search').val() });
});
So if you wanted to search for the value when clicking on a link (for links within #container):
$('#container').delegate('a', 'click', function() {
// .text() will get what's inside the <a> tag
$('#result').load('b.php', { data: $(this).text() });
});