Most votes on ajax questions 2

Most votes on ajax questions 2. #11 Is Safari on iOS 6 caching $.ajax results? #12 jQuery AJAX submit form #13 Scroll to bottom of div? #14 Ajax request returns 200 OK, but an error event is fired instead of success #15 How to make an AJAX call without jQuery? #16 jQuery Ajax File Upload #17 jQuery Ajax error handling, show custom exception messages #18 Wait until all jQuery Ajax requests are done? #19 jQuery Ajax POST example with PHP #20 Updating address bar with new URL without hash or reloading the page

Read all the top votes questions and answers in a single page.

#11: Is Safari on iOS 6 caching $.ajax results? (Score: 1089)

Created: 2012-09-20 Last updated: 2021-03-12

Tags: javascript, jquery, ajax, caching, mobile-safari

Since the upgrade to iOS 6, we are seeing Safari’s web view take the liberty of caching $.ajax calls. This is in the context of a PhoneGap application so it is using the Safari WebView. Our $.ajax calls are POST methods and we have cache set to false {cache:false}, but still this is happening. We tried manually adding a TimeStamp to the headers but it did not help.

We did more research and found that Safari is only returning cached results for web services that have a function signature that is static and does not change from call to call. For instance, imagine a function called something like:


This function receives the same input parameters over and over again, but the data it returns should be different every time.

Must be in Apple’s haste to make iOS 6 zip along impressively they got too happy with the cache settings. Has anyone else seen this behavior on iOS 6? If so, what exactly is causing it?

The workaround that we found was to modify the function signature to be something like this:

getNewRecordID(intRecordType, strTimestamp)

and then always pass in a TimeStamp parameter as well, and just discard that value on the server side. This works around the issue.

#11 Best answer 1 of Is Safari on iOS 6 caching $.ajax results? (Score: 448)

Created: 2012-09-20 Last updated: 2012-09-24

After a bit of investigation, turns out that Safari on iOS6 will cache POSTs that have either no Cache-Control headers or even “Cache-Control: max-age=0”.

The only way I’ve found of preventing this caching from happening at a global level rather than having to hack random querystrings onto the end of service calls is to set “Cache-Control: no-cache”.


  • No Cache-Control or Expires headers = iOS6 Safari will cache
  • Cache-Control max-age=0 and an immediate Expires = iOS6 Safari will cache
  • Cache-Control: no-cache = iOS6 Safari will NOT cache

I suspect that Apple is taking advantage of this from the HTTP spec in section 9.5 about POST:

Responses to this method are not cacheable, unless the response includes appropriate Cache-Control or Expires header fields. However, the 303 (See Other) response can be used to direct the user agent to retrieve a cacheable resource.

So in theory you can cache POST responses…who knew. But no other browser maker has ever thought it would be a good idea until now. But that does NOT account for the caching when no Cache-Control or Expires headers are set, only when there are some set. So it must be a bug.

Below is what I use in the right bit of my Apache config to target the whole of my API because as it happens I don’t actually want to cache anything, even gets. What I don’t know is how to set this just for POSTs.

Header set Cache-Control "no-cache"

Update: Just noticed that I didn’t point out that it is only when the POST is the same, so change any of the POST data or URL and you’re fine. So you can as mentioned elsewhere just add some random data to the URL or a bit of POST data.

Update: You can limit the “no-cache” just to POSTs if you wish like this in Apache:

SetEnvIf Request_Method "POST" IS_POST
Header set Cache-Control "no-cache" env=IS_POST

#11 Best answer 2 of Is Safari on iOS 6 caching $.ajax results?(Score: 147)

Created: 2012-10-12

I hope this can be of use to other developers banging their head against the wall on this one. I found that any of the following prevents Safari on iOS 6 from caching the POST response:

  • adding [cache-control: no-cache] in the request headers
  • adding a variable URL parameter such as the current time
  • adding [pragma: no-cache] in the response headers
  • adding [cache-control: no-cache] in the response headers

My solution was the following in my Javascript (all my AJAX requests are POST).

	type: 'POST',
	headers: { "cache-control": "no-cache" }

I also add the [pragma: no-cache] header to many of my server responses.

If you use the above solution be aware that any $.ajax() calls you make that are set to global: false will NOT use the settings specified in $.ajaxSetup(), so you will need to add the headers in again.

See also original question in stackoverflow

#12: jQuery AJAX submit form (Score: 1012)

Created: 2009-12-25 Last updated: 2020-01-21

Tags: javascript, jquery, ajax, submit, forms

I have a form with name orderproductForm and an undefined number of inputs.

I want to do some kind of jQuery.get or ajax or anything like that that would call a page through Ajax, and send along all the inputs of the form orderproductForm.

I suppose one way would be to do something like

          {action : document.orderproductForm.action.value,
           cartproductid : document.orderproductForm.cartproductid.value,
           productid : document.orderproductForm.productid.value,

However I do not know exactly all the form inputs. Is there a feature, function or something that would just send ALL the form inputs?

#12 Best answer 1 of jQuery AJAX submit form (Score: 1529)

Created: 2011-08-05 Last updated: 2020-07-11

This is a simple reference:

// this is the id of the form
$("#idForm").submit(function(e) {

    e.preventDefault(); // avoid to execute the actual submit of the form.

	var form = $(this);
	var url = form.attr('action');
           type: "POST",
           url: url,
           data: form.serialize(), // serializes the form's elements.
           success: function(data)
               alert(data); // show response from the php script.


#12 Best answer 2 of jQuery AJAX submit form(Score: 845)

Created: 2009-12-25 Last updated: 2017-03-18

You can use the ajaxForm/ajaxSubmit functions from Ajax Form Plugin or the jQuery serialize function.


$("#theForm").ajaxForm({url: 'server.php', type: 'post'})


$("#theForm").ajaxSubmit({url: 'server.php', type: 'post'})

ajaxForm will send when the submit button is pressed. ajaxSubmit sends immediately.


$.get('server.php?' + $('#theForm').serialize())

$.post('server.php', $('#theForm').serialize())

AJAX serialization documentation is here.

See also original question in stackoverflow

#13: Scroll to bottom of div? (Score: 906)

Created: 2008-11-06 Last updated: 2020-06-21

Tags: javascript, html, ajax, chat

I am creating a chat using Ajax requests and I’m trying to get messages div to scroll to the bottom without much luck.

I am wrapping everything in this div:

#scroll {

Is there a way to keep it scrolled to the bottom by default using JS?

Is there a way to keep it scrolled to the bottom after an ajax request?

#13 Best answer 1 of Scroll to bottom of div? (Score: 1496)

Created: 2008-11-06 Last updated: 2018-07-19

Here’s what I use on my site:

var objDiv = document.getElementById("your_div");
objDiv.scrollTop = objDiv.scrollHeight;

#13 Best answer 2 of Scroll to bottom of div?(Score: 370)

Created: 2010-04-19 Last updated: 2019-04-29

This is much easier if you’re using jQuery scrollTop:


See also original question in stackoverflow

#14: Ajax request returns 200 OK, but an error event is fired instead of success (Score: 851)

Created: 2011-05-31 Last updated: 2020-04-17

Tags: javascript, jquery,, ajax, json

I have implemented an Ajax request on my website, and I am calling the endpoint from a webpage. It always returns 200 OK, but jQuery executes the error event.
I tried a lot of things, but I could not figure out the problem. I am adding my code below:

###jQuery Code

var row = "1";
var json = "{'TwitterId':'" + row + "'}";
    type: 'POST',
    url: 'Jqueryoperation.aspx?Operation=DeleteRow',
    contentType: 'application/json; charset=utf-8',
    data: json,
    dataType: 'json',
    cache: false,
    success: AjaxSucceeded,
    error: AjaxFailed
function AjaxSucceeded(result) {
function AjaxFailed(result) {
    alert(result.status + ' ' + result.statusText);

###C# code for JqueryOpeartion.aspx

protected void Page_Load(object sender, EventArgs e) {
private void test() {
    Response.Write("<script language='javascript'>alert('Record Deleted');</script>");

I need the ("Record deleted") string after successful deletion. I am able to delete the content, but I am not getting this message. Is this correct or am I doing anything wrong? What is the correct way to solve this issue?

#14 Best answer 1 of Ajax request returns 200 OK, but an error event is fired instead of success (Score: 1171)

Created: 2011-05-31 Last updated: 2018-07-18

jQuery.ajax attempts to convert the response body depending on the specified dataType parameter or the Content-Type header sent by the server. If the conversion fails (e.g. if the JSON/XML is invalid), the error callback is fired.

Your AJAX code contains:

dataType: "json"

In this case jQuery:

Evaluates the response as JSON and returns a JavaScript object. […] The JSON data is parsed in a strict manner; any malformed JSON is rejected and a parse error is thrown. […] an empty response is also rejected; the server should return a response of null or {} instead.

Your server-side code returns HTML snippet with 200 OK status. jQuery was expecting valid JSON and therefore fires the error callback complaining about parseerror.

The solution is to remove the dataType parameter from your jQuery code and make the server-side code return:

Content-Type: application/javascript

alert("Record Deleted");

But I would rather suggest returning a JSON response and display the message inside the success callback:

Content-Type: application/json

{"message": "Record deleted"}

#14 Best answer 2 of Ajax request returns 200 OK, but an error event is fired instead of success(Score: 33)

Created: 2011-06-15 Last updated: 2017-02-25

I’ve had some good luck with using multiple, space-separated dataTypes (jQuery 1.5+). As in:

    type: 'POST',
    url: 'Jqueryoperation.aspx?Operation=DeleteRow',
    contentType: 'application/json; charset=utf-8',
    data: json,
    dataType: 'text json',
    cache: false,
    success: AjaxSucceeded,
    error: AjaxFailed

See also original question in stackoverflow

#15: How to make an AJAX call without jQuery? (Score: 833)

Created: 2011-12-19 Last updated: 2015-12-18

Tags: javascript, ajax

How to make an AJAX call using JavaScript, without using jQuery?

#15 Best answer 1 of How to make an AJAX call without jQuery? (Score: 618)

Created: 2011-12-19 Last updated: 2017-10-12

With “vanilla” JavaScript:

<script type="text/javascript">
function loadXMLDoc() {
    var xmlhttp = new XMLHttpRequest();

    xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == XMLHttpRequest.DONE) {   // XMLHttpRequest.DONE == 4
           if (xmlhttp.status == 200) {
               document.getElementById("myDiv").innerHTML = xmlhttp.responseText;
           else if (xmlhttp.status == 400) {
              alert('There was an error 400');
           else {
               alert('something else other than 200 was returned');
    };"GET", "ajax_info.txt", true);

With jQuery:

    url: "test.html",
    context: document.body,
    success: function(){

#15 Best answer 2 of How to make an AJAX call without jQuery?(Score: 226)

Created: 2013-08-06 Last updated: 2018-11-27

Using the following snippet you can do similar things pretty easily, like this:

ajax.get('/test.php', {foo: 'bar'}, function() {});

Here is the snippet:

var ajax = {};
ajax.x = function () {
    if (typeof XMLHttpRequest !== 'undefined') {
        return new XMLHttpRequest();
    var versions = [

    var xhr;
    for (var i = 0; i < versions.length; i++) {
        try {
            xhr = new ActiveXObject(versions[i]);
        } catch (e) {
    return xhr;

ajax.send = function (url, callback, method, data, async) {
    if (async === undefined) {
        async = true;
    var x = ajax.x();, url, async);
    x.onreadystatechange = function () {
        if (x.readyState == 4) {
    if (method == 'POST') {
        x.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');

ajax.get = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    ajax.send(url + (query.length ? '?' + query.join('&') : ''), callback, 'GET', null, async)
}; = function (url, data, callback, async) {
    var query = [];
    for (var key in data) {
        query.push(encodeURIComponent(key) + '=' + encodeURIComponent(data[key]));
    ajax.send(url, callback, 'POST', query.join('&'), async)

See also original question in stackoverflow

#16: jQuery Ajax File Upload (Score: 801)

Created: 2010-02-23 Last updated: 2019-07-05

Tags: javascript, jquery, ajax, post, file-upload

Can I use the following jQuery code to perform file upload using POST method of an ajax request ?

    type: "POST",
    timeout: 50000,
    url: url,
    data: dataString,
    success: function (data) {
        return false;

If it is possible, do I need to fill data part? Is it the correct way? I only POST the file to the server side.

I have been googling around, but what I found was a plugin while in my plan I do not want to use it. At least for the moment.

#16 Best answer 1 of jQuery Ajax File Upload (Score: 616)

Created: 2010-02-23 Last updated: 2020-03-23

File upload is not possible through AJAX.
You can upload file, without refreshing page by using IFrame.
You can check further details here.


With XHR2, File upload through AJAX is supported. E.g. through FormData object, but unfortunately it is not supported by all/old browsers.

FormData support starts from following desktop browsers versions.

  • IE 10+
  • Firefox 4.0+
  • Chrome 7+
  • Safari 5+
  • Opera 12+

For more detail, see MDN link.

#16 Best answer 2 of jQuery Ajax File Upload(Score: 343)

Created: 2012-05-30 Last updated: 2018-04-23

Iframes is no longer needed for uploading files through ajax. I’ve recently done it by myself. Check out these pages:

Updated the answer and cleaned it up. Use the getSize function to check size or use getType function to check types. Added progressbar html and css code.

var Upload = function (file) {
    this.file = file;

Upload.prototype.getType = function() {
    return this.file.type;
Upload.prototype.getSize = function() {
    return this.file.size;
Upload.prototype.getName = function() {
Upload.prototype.doUpload = function () {
    var that = this;
    var formData = new FormData();
    // add assoc key values, this will be posts values
    formData.append("file", this.file, this.getName());
    formData.append("upload_file", true);

        type: "POST",
        url: "script",
        xhr: function () {
            var myXhr = $.ajaxSettings.xhr();
            if (myXhr.upload) {
                myXhr.upload.addEventListener('progress', that.progressHandling, false);
            return myXhr;
        success: function (data) {
            // your callback here
        error: function (error) {
            // handle error
        async: true,
        data: formData,
        cache: false,
        contentType: false,
        processData: false,
        timeout: 60000

Upload.prototype.progressHandling = function (event) {
    var percent = 0;
    var position = event.loaded || event.position;
    var total =;
    var progress_bar_id = "#progress-wrp";
    if (event.lengthComputable) {
        percent = Math.ceil(position / total * 100);
    // update progressbars classes so it fits your code
    $(progress_bar_id + " .progress-bar").css("width", +percent + "%");
    $(progress_bar_id + " .status").text(percent + "%");

How to use the Upload class

//Change id to your id
$("#ingredient_file").on("change", function (e) {
    var file = $(this)[0].files[0];
    var upload = new Upload(file);
    // maby check size or type here with upload.getSize() and upload.getType()

    // execute upload

Progressbar html code

<div id="progress-wrp">
    <div class="progress-bar"></div>
    <div class="status">0%</div>

Progressbar css code

#progress-wrp {
  border: 1px solid #0099CC;
  padding: 1px;
  position: relative;
  height: 30px;
  border-radius: 3px;
  margin: 10px;
  text-align: left;
  background: #fff;
  box-shadow: inset 1px 3px 6px rgba(0, 0, 0, 0.12);

#progress-wrp .progress-bar {
  height: 100%;
  border-radius: 3px;
  background-color: #f39ac7;
  width: 0;
  box-shadow: inset 1px 1px 10px rgba(0, 0, 0, 0.11);

#progress-wrp .status {
  top: 3px;
  left: 50%;
  position: absolute;
  display: inline-block;
  color: #000000;

See also original question in stackoverflow

#17: jQuery Ajax error handling, show custom exception messages (Score: 767)

Created: 2008-12-18 Last updated: 2016-12-19

Tags: jquery, ajax, custom-exceptions

Is there some way I can show custom exception messages as an alert in my jQuery AJAX error message?

For example, if I want to throw an exception on the server side via Struts by throw new ApplicationException("User name already exists");, I want to catch this message (‘user name already exists’) in the jQuery AJAX error message.

jQuery("#save").click(function () {
  if (jQuery('#form').jVal()) {
      type: "POST",
      url: "",
      dataType: "html",
      data: "userId=" + encodeURIComponent(trim(document.forms[0].userId.value)),
      success: function (response) {
        alert("Details saved successfully!!!");
      error: function (xhr, ajaxOptions, thrownError) {

On the second alert, where I alert the thrown error, I am getting undefined and the status code is 500.

I am not sure where I am going wrong. What can I do to fix this problem?

#17 Best answer 1 of jQuery Ajax error handling, show custom exception messages (Score: 371)

Created: 2009-01-16

Make sure you’re setting Response.StatusCode to something other than 200. Write your exception’s message using Response.Write, then use…

xhr.responseText your javascript.

#17 Best answer 2 of jQuery Ajax error handling, show custom exception messages(Score: 229)

Created: 2010-05-10 Last updated: 2011-12-20


public class ClientErrorHandler : FilterAttribute, IExceptionFilter
    public void OnException(ExceptionContext filterContext)
        var response = filterContext.RequestContext.HttpContext.Response;
        response.ContentType = MediaTypeNames.Text.Plain;
        filterContext.ExceptionHandled = true;

public class SomeController : Controller
    public ActionResult SomeAction()
        throw new Exception("Error message");

View script:

    type: "post", url: "/SomeController/SomeAction",
    success: function (data, text) {
    error: function (request, status, error) {

See also original question in stackoverflow

#18: Wait until all jQuery Ajax requests are done? (Score: 717)

Created: 2010-09-14 Last updated: 2017-09-27

Tags: javascript, jquery, ajax

How do I make a function wait until all jQuery Ajax requests are done inside another function?

In short, I need to wait for all Ajax requests to be done before I execute the next. But how?

#18 Best answer 1 of Wait until all jQuery Ajax requests are done? (Score: 966)

Created: 2012-03-25 Last updated: 2018-04-13

jQuery now defines a when function for this purpose.

It accepts any number of Deferred objects as arguments, and executes a function when all of them resolve.

That means, if you want to initiate (for example) four ajax requests, then perform an action when they are done, you could do something like this:

$.when(ajax1(), ajax2(), ajax3(), ajax4()).done(function(a1, a2, a3, a4){
    // the code here will be executed when all four ajax requests resolve.
    // a1, a2, a3 and a4 are lists of length 3 containing the response text,
    // status, and jqXHR object for each of the four ajax calls respectively.

function ajax1() {
    // NOTE:  This function must return the value 
    //        from calling the $.ajax() method.
    return $.ajax({
        url: "someUrl",
        dataType: "json",
        data:  yourJsonData,            

In my opinion, it makes for a clean and clear syntax, and avoids involving any global variables such as ajaxStart and ajaxStop, which could have unwanted side effects as your page develops.

If you don’t know in advance how many ajax arguments you need to wait for (i.e. you want to use a variable number of arguments), it can still be done but is just a little bit trickier. See (and maybe

If you need deeper control over the failure modes of the ajax scripts etc., you can save the object returned by .when() - it’s a jQuery Promise object encompassing all of the original ajax queries. You can call .then() or .fail() on it to add detailed success/failure handlers.

#18 Best answer 2 of Wait until all jQuery Ajax requests are done?(Score: 311)

Created: 2012-12-26 Last updated: 2020-03-25

If you want to know when all ajax requests are finished in your document, no matter how many of them exists, just use $.ajaxStop event this way:

$(document).ajaxStop(function () {
  // 0 === $.active

In this case, neither you need to guess how many requests are happening in the application, that might finish in the future, nor dig into functions complex logic or find which functions are doing HTTP(S) requests.

$.ajaxStop here can also be bound to any HTML node that you think might be modified by requst.

If you want to stick with ES syntax, then you can use Promise.all for known ajax methods:

Promise.all([ajax1(), ajax2()]).then(() => {
  // all requests finished successfully
}).catch(() => {
  // all requests finished but one or more failed

An interesting point here is that it works both with Promises and $.ajax requests.

Here is the jsFiddle demonstration.

Update 2:
Yet more recent version using async/await syntax:

try {
  const results = await Promise.all([ajax1(), ajax2()])
  // do other actions
} catch(ex) { }

See also original question in stackoverflow

#19: jQuery Ajax POST example with PHP (Score: 712)

Created: 2011-02-15 Last updated: 2020-06-22

Tags: javascript, jquery, ajax, post

I am trying to send data from a form to a database. Here is the form I am using:

<form name="foo" action="form.php" method="POST" id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />
    <input type="submit" value="Send" />

The typical approach would be to submit the form, but this causes the browser to redirect. Using jQuery and Ajax, is it possible to capture all of the form’s data and submit it to a PHP script (an example, form.php)?

#19 Best answer 1 of jQuery Ajax POST example with PHP (Score: 970)

Created: 2011-02-15 Last updated: 2019-07-12

Basic usage of .ajax would look something like this:


<form id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />

    <input type="submit" value="Send" />


// Variable to hold request
var request;

// Bind to the submit event of our form

    // Prevent default posting of form - put here to work in case of errors

    // Abort any pending request
    if (request) {
    // setup some local variables
    var $form = $(this);

    // Let's select and cache all the fields
    var $inputs = $form.find("input, select, button, textarea");

    // Serialize the data in the form
    var serializedData = $form.serialize();

    // Let's disable the inputs for the duration of the Ajax request.
    // Note: we disable elements AFTER the form data has been serialized.
    // Disabled form elements will not be serialized.
    $inputs.prop("disabled", true);

    // Fire off the request to /form.php
    request = $.ajax({
        url: "/form.php",
        type: "post",
        data: serializedData

    // Callback handler that will be called on success
    request.done(function (response, textStatus, jqXHR){
        // Log a message to the console
        console.log("Hooray, it worked!");

    // Callback handler that will be called on failure (jqXHR, textStatus, errorThrown){
        // Log the error to the console
            "The following error occurred: "+
            textStatus, errorThrown

    // Callback handler that will be called regardless
    // if the request failed or succeeded
    request.always(function () {
        // Reenable the inputs
        $inputs.prop("disabled", false);


Note: Since jQuery 1.8, .success(), .error() and .complete() are deprecated in favor of .done(), .fail() and .always().

Note: Remember that the above snippet has to be done after DOM ready, so you should put it inside a $(document).ready() handler (or use the $() shorthand).

Tip: You can chain the callback handlers like this: $.ajax().done().fail().always();

PHP (that is, form.php):

// You can access the values posted by jQuery.ajax
// through the global variable $_POST, like this:
$bar = isset($_POST['bar']) ? $_POST['bar'] : null;

Note: Always sanitize posted data, to prevent injections and other malicious code.

You could also use the shorthand .post in place of .ajax in the above JavaScript code:

$.post('/form.php', serializedData, function(response) {
    // Log the response to the console
    console.log("Response: "+response);

Note: The above JavaScript code is made to work with jQuery 1.8 and later, but it should work with previous versions down to jQuery 1.5.

#19 Best answer 2 of jQuery Ajax POST example with PHP(Score: 227)

Created: 2013-01-08 Last updated: 2019-07-12

To make an Ajax request using jQuery you can do this by the following code.


<form id="foo">
    <label for="bar">A bar</label>
    <input id="bar" name="bar" type="text" value="" />
    <input type="submit" value="Send" />

<!-- The result of the search will be rendered inside this div -->
<div id="result"></div>


Method 1

 /* Get from elements values */
 var values = $(this).serialize();

        url: "test.php",
        type: "post",
        data: values ,
        success: function (response) {

           // You will get response from your PHP page (what you echo or print)
        error: function(jqXHR, textStatus, errorThrown) {
           console.log(textStatus, errorThrown);

Method 2

/* Attach a submit handler to the form */
$("#foo").submit(function(event) {
    var ajaxRequest;

    /* Stop form from submitting normally */

    /* Clear result div*/

    /* Get from elements values */
    var values = $(this).serialize();

    /* Send the data using post and put the results in a div. */
    /* I am not aborting the previous request, because it's an
       asynchronous request, meaning once it's sent it's out
       there. But in case you want to abort it you can do it
       by abort(). jQuery Ajax methods return an XMLHttpRequest
       object, so you can just use abort(). */
       ajaxRequest= $.ajax({
            url: "test.php",
            type: "post",
            data: values

    /*  Request can be aborted by ajaxRequest.abort() */

    ajaxRequest.done(function (response, textStatus, jqXHR){

         // Show successfully for submit message
         $("#result").html('Submitted successfully');

    /* On failure of request this function will be called  */ (){

        // Show error
        $("#result").html('There is error while submit');

The .success(), .error(), and .complete() callbacks are deprecated as of jQuery 1.8. To prepare your code for their eventual removal, use .done(), .fail(), and .always() instead.

MDN: abort() . If the request has been sent already, this method will abort the request.

So we have successfully send an Ajax request, and now it’s time to grab data to server.


As we make a POST request in an Ajax call (type: "post"), we can now grab data using either $_REQUEST or $_POST:

  $bar = $_POST['bar']

You can also see what you get in the POST request by simply either. BTW, make sure that $_POST is set. Otherwise you will get an error.

// Or

And you are inserting a value into the database. Make sure you are sensitizing or escaping All requests (whether you made a GET or POST) properly before making the query. The best would be using prepared statements.

And if you want to return any data back to the page, you can do it by just echoing that data like below.

// 1. Without JSON
   echo "Hello, this is one"

// 2. By JSON. Then here is where I want to send a value back to the success of the Ajax below
echo json_encode(array('returned_val' => 'yoho'));

And then you can get it like:

 ajaxRequest.done(function (response){

There are a couple of shorthand methods. You can use the below code. It does the same work.

var ajaxRequest= $.post("test.php", values, function(data) {
  .fail(function() {
  .always(function() {

See also original question in stackoverflow

#20: Updating address bar with new URL without hash or reloading the page (Score: 688)

Created: 2010-07-26 Last updated: 2017-05-23

Tags: javascript, ajax, google-chrome

I either dreamt about chrome (dev channel) implementing a way to update the address bar via javascript (the path, not domain) without reloading the page or they really have done this.

However, I can’t find the article I think I read.

Am I crazy or is there a way to do this (in Chrome)?

p.s. I’m not talking about window.location.hash, et al. If the above exists the answer to this question will be untrue.

#20 Best answer 1 of Updating address bar with new URL without hash or reloading the page (Score: 920)

Created: 2010-07-27 Last updated: 2019-01-23

You can now do this in most “modern” browsers!

Here is the original article I read (posted July 10, 2010): HTML5: Changing the browser-URL without refreshing page.

For a more in-depth look into pushState/replaceState/popstate (aka the HTML5 History API) see the MDN docs.

TL;DR, you can do this:

window.history.pushState("object or string", "Title", "/new-url");

See my answer to Modify the URL without reloading the page for a basic how-to.

#20 Best answer 2 of Updating address bar with new URL without hash or reloading the page(Score: 175)

Created: 2016-09-01 Last updated: 2018-06-17

Changing only what’s after hash - old browsers

document.location.hash = 'lookAtMeNow';

Changing full URL. Chrome, Firefox, IE10+

history.pushState('data to be passed', 'Title of the page', '/test');

The above will add a new entry to the history so you can press Back button to go to the previous state. To change the URL in place without adding a new entry to history use

history.replaceState('data to be passed', 'Title of the page', '/test');

Try running these in the console now!

See also original question in stackoverflow

  1. This page use API to get the relevant data from stackoverflow community.
  2. Content license on this page is CC BY-SA 3.0.
  3. score = up votes - down votes.