jQuery Ajax POST example with PHP

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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" />
</form>

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 (in example, form.php)?

phpjavascriptjqueryajaxpost

Answers

answered 7 years ago mekwall #1

Basic usage of .ajax would look something like this:

HTML:

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

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

JQuery:

// Variable to hold request
var request;

// Bind to the submit event of our form
$("#foo").submit(function(event){

    // Prevent default posting of form - put here to work in case of errors
    event.preventDefault();

    // Abort any pending request
    if (request) {
        request.abort();
    }
    // 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
    request.fail(function (jqXHR, textStatus, errorThrown){
        // Log the error to the console
        console.error(
            "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.

answered 5 years ago NullPoiиteя #2

To make ajax request using jQuery you can do this by following code

HTML:

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

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

JavaScript:

Method 1

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

 $.ajax({
        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 */
    event.preventDefault();

    /* Clear result div*/
    $("#result").html('');

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

    /* Send the data using post and put the results in a div */
    /* I am not aborting 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 cab be abort 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  */
     ajaxRequest.fail(function (){

       // 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 ajax request now its time to grab data to server.

PHP

As we make a POST request in 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 POST request by simply either, Btw make sure that $_POST is set other wise you will get error.

var_dump($_POST);
// or
print_r($_POST);

And you are inserting value to database make sure you are sensitizing or escaping All request ( weather you made GET or POST) properly before making query, Best would be using prepared statements.

and if you want to return any data back to 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 than you can get it like

 ajaxRequest.done(function (response){  
    alert(response);
 });

There are Couple of Shorthand Methods you can use below code it do the same work.

var ajaxRequest= $.post( "test.php",values, function(data) {
  alert( data );
})
  .fail(function() {
    alert( "error" );
  })
  .always(function() {
    alert( "finished" );
});

answered 5 years ago user2610222 #3

You can use serialize. Below is an example.

$("#submit_btn").click(function(){
    $('.error_status').html();
        if($("form#frm_message_board").valid())
        {
            $.ajax({
                type: "POST",
                url: "<?php echo site_url('message_board/add');?>",
                data: $('#frm_message_board').serialize(),
                success: function(msg) {
                    var msg = $.parseJSON(msg);
                    if(msg.success=='yes')
                    {
                        return true;
                    }
                    else
                    {
                        alert('Server error');
                        return false;
                    }
                }
            });
        }
        return false;
    });

answered 4 years ago Mr. Alien #4

I would like to share a detailed way of how to post with PHP + Ajax along with errors thrown back on failure.

First of all, create two files, for example form.php and process.php.

We will first create a form which will be then submitted using the jQuery .ajax() method. The rest will be explained in the comments.


form.php

<form method="post" name="postForm">
    <ul>
        <li>
            <label>Name</label>
            <input type="text" name="name" id="name" placeholder="Bruce Wayne">
            <span class="throw_error"></span>
            <span id="success"></span>
       </li>
   </ul>
   <input type="submit" value="Send" />
</form>


Validate the form using jQuery client-side validation and pass the data to process.php.

$(document).ready(function() {
    $('form').submit(function(event) { //Trigger on form submit
        $('#name + .throw_error').empty(); //Clear the messages first
        $('#success').empty();

        //Validate fields if required using jQuery

        var postForm = { //Fetch form data
            'name'     : $('input[name=name]').val() //Store name fields value
        };

        $.ajax({ //Process the form using $.ajax()
            type      : 'POST', //Method type
            url       : 'process.php', //Your form processing file URL
            data      : postForm, //Forms name
            dataType  : 'json',
            success   : function(data) {
                            if (!data.success) { //If fails
                                if (data.errors.name) { //Returned if any error from process.php
                                    $('.throw_error').fadeIn(1000).html(data.errors.name); //Throw relevant error
                                }
                            }
                            else {
                                    $('#success').fadeIn(1000).append('<p>' + data.posted + '</p>'); //If successful, than throw a success message
                                }
                            }
        });
        event.preventDefault(); //Prevent the default submit
    });
});

Now we will take a look at process.php

$errors = array(); //To store errors
$form_data = array(); //Pass back the data to `form.php`

/* Validate the form on the server side */
if (empty($_POST['name'])) { //Name cannot be empty
    $errors['name'] = 'Name cannot be blank';
}

if (!empty($errors)) { //If errors in validation
    $form_data['success'] = false;
    $form_data['errors']  = $errors;
}
else { //If not, process the form, and return true on success
    $form_data['success'] = true;
    $form_data['posted'] = 'Data Was Posted Successfully';
}

//Return the data back to form.php
echo json_encode($form_data);

The project files can be downloaded from http://projects.decodingweb.com/simple_ajax_form.zip.

answered 4 years ago john #5

<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<form method="post" id="form_content" action="Javascript:void(0);">
    <button id="desc" name="desc" value="desc" style="display:none;">desc</button>
    <button id="asc" name="asc"  value="asc">asc</button>
    <input type='hidden' id='check' value=''/>
</form>

<div id="demoajax"></div>

<script>
    numbers = '';
    $('#form_content button').click(function(){
        $('#form_content button').toggle();
        numbers = this.id;
        function_two(numbers);
    });

    function function_two(numbers){
        if (numbers === '')
        {
            $('#check').val("asc");
        }
        else
        {
            $('#check').val(numbers);
        }
        //alert(sort_var);

        $.ajax({
            url: 'test.php',
            type: 'POST',
            data: $('#form_content').serialize(),
            success: function(data){
                $('#demoajax').show();
                $('#demoajax').html(data);
                }
        });

        return false;
    }
    $(document).ready(function_two());
</script>

answered 3 years ago DDeme #6

HTML:

    <form name="foo" action="form.php" method="POST" id="foo">
        <label for="bar">A bar</label>
        <input id="bar" class="inputs" name="bar" type="text" value="" />
        <input type="submit" value="Send" onclick="submitform(); return false;" />
    </form>

JavaScript :

   function submitform()
   {
       var inputs = document.getElementsByClassName("inputs");
       var formdata = new FormData();
       for(var i=0; i<inputs.length; i++)
       {
           formdata.append(inputs[i].name, inputs[i].value);
       }
       var xmlhttp;
       if(window.XMLHttpRequest)
       {
           xmlhttp = new XMLHttpRequest;
       }
       else
       {
           xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
       }
       xmlhttp.onreadystatechange = function()
       {
          if(xmlhttp.readyState == 4 && xmlhttp.status == 200)
          {

          }
       }
       xmlhttp.open("POST", "insert.php");
       xmlhttp.send(formdata);
   }

answered 3 years ago Shaiful Islam #7

I use this way.It submit everything like files

$(document).on("submit", "form", function(event)
{
    event.preventDefault();

    var url=$(this).attr("action");
    $.ajax({
        url: url,
        type: 'POST',
        dataType: "JSON",
        data: new FormData(this),
        processData: false,
        contentType: false,
        success: function (data, status)
        {

        },
        error: function (xhr, desc, err)
        {
            console.log("error");

        }
    });        

});

answered 2 years ago Uchiha Itachi #8

I am using this simple one line code for years without problem. (It requires jquery)

<script type="text/javascript">
function ap(x,y) {$("#" + y).load(x);};
function af(x,y) {$("#" + x ).ajaxSubmit({target: '#' + y});return false;};
</script>

Here ap() means ajax page and af() means ajax form. In a form just simply calling af() function will post form to the url and load response on desired html element.

<form>
...
<input type="button" onclick="af('http://example.com','load_response')"/>
</form>
<div id="load_response">this is where response will be loaded</div>

answered 2 years ago pawan sen #9

handling ajax error and loader before submit and after submitting success show alert boot box with an example:

var formData = formData;

$.ajax({
    type: "POST",
    url: url,
    async: false,
    data: formData, //only input
    processData: false,
    contentType: false,
    xhr: function ()
    {
        $("#load_consulting").show();
        var xhr = new window.XMLHttpRequest();
        //Upload progress
        xhr.upload.addEventListener("progress", function (evt) {
            if (evt.lengthComputable) {
                var percentComplete = (evt.loaded / evt.total) * 100;
                $('#addLoad .progress-bar').css('width', percentComplete + '%');
            }
        }, false);
        //Download progress
        xhr.addEventListener("progress", function (evt) {
            if (evt.lengthComputable) {
                var percentComplete = evt.loaded / evt.total;
            }
        }, false);
        return xhr;
    },
    beforeSend: function (xhr) {
        qyuraLoader.startLoader();
    },
    success: function (response, textStatus, jqXHR) {
        qyuraLoader.stopLoader();
        try {
            $("#load_consulting").hide();

            var data = $.parseJSON(response);
            if (data.status == 0)
            {
                if (data.isAlive)
                {
                    $('#addLoad .progress-bar').css('width', '00%');
                    console.log(data.errors);
                    $.each(data.errors, function (index, value) {
                        if (typeof data.custom == 'undefined') {
                            $('#err_' + index).html(value);
                        }
                        else
                        {
                            $('#err_' + index).addClass('error');

                            if (index == 'TopError')
                            {
                                $('#er_' + index).html(value);
                            }
                            else {
                                $('#er_TopError').append('<p>' + value + '</p>');
                            }
                        }

                    });
                    if (data.errors.TopError) {
                        $('#er_TopError').show();
                        $('#er_TopError').html(data.errors.TopError);
                        setTimeout(function () {
                            $('#er_TopError').hide(5000);
                            $('#er_TopError').html('');
                        }, 5000);
                    }
                }
                else
                {
                    $('#headLogin').html(data.loginMod);
                }
            } else {
                //document.getElementById("setData").reset();
                $('#myModal').modal('hide');
                $('#successTop').show();
                $('#successTop').html(data.msg);
                if (data.msg != '' && data.msg != "undefined") {

                    bootbox.alert({closeButton: false, message: data.msg, callback: function () {
                            if (data.url) {
                                window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
                            } else {
                                location.reload(true);
                            }
                        }});
                } else {

                    bootbox.alert({closeButton: false, message: "Success", callback: function () {
                            if (data.url) {
                                window.location.href = '<?php echo site_url() ?>' + '/' + data.url;
                            } else {
                                location.reload(true);
                            }
                        }});
                }

            }
        } catch (e) {
            if (e) {
                $('#er_TopError').show();
                $('#er_TopError').html(e);
                setTimeout(function () {
                    $('#er_TopError').hide(5000);
                    $('#er_TopError').html('');
                }, 5000);
            }
        }
    }
});

answered 1 year ago Juned Ansari #10

If you want to send data using jquery Ajax then there is no need of form tag and submit button

Example:

<script>
    $(document).ready(function () {
        $("#btnSend").click(function () {
            $.ajax({
                url: 'process.php',
                type: 'POST',
                data: {bar: $("#bar").val()},
                success: function (result) {
                    alert('success');
                }
            });
        });
    });
</script>
<label for="bar">A bar</label>
<input id="bar" name="bar" type="text" value="" />
<input id="btnSend" type="button" value="Send" />

answered 8 months ago Waseem Bashir #11

Hi please check this is the complete ajax request code.

        $('#foo').submit(function(event) {
        // get the form data
        // there are many ways to get this data using jQuery (you can use the 
    class or id also)
    var formData = $('#foo').serialize();
    var url ='url of the request';
    // process the form.

    $.ajax({
        type        : 'POST', // define the type of HTTP verb we want to use
        url         : 'url/', // the url where we want to POST
        data        : formData, // our data object
        dataType    : 'json', // what type of data do we expect back.
        beforeSend : function() {
        //this will run before sending an ajax request do what ever activity 
         you want like show loaded 
         },
        success:function(response){
            var obj = eval(response);
            if(obj)
            {  
                if(obj.error==0){
                alert('success');
                }
            else{  
                alert('error');
                }   
            }
        },
        complete : function() {
           //this will run after sending an ajax complete                   
                    },
        error:function (xhr, ajaxOptions, thrownError){ 
          alert('error occured');
        // if any error occurs in request 
        } 
    });
    // stop the form from submitting the normal way and refreshing the page
    event.preventDefault();
});

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