Codeigniter & Jquery Ajax, Post data not going through

Question

I am having problem getting the post data in the controller when I send the post request via Jquery Ajax, I have checked with firebug and the form post data is being submitted, but in the controller when I do print_r($_POST); it returns an empty array. What could be wrong ?

Here is the relevant code :

FORM HTML

<form id="contact-form" class="form-horizontal subscribe" accept-charset="utf-8" action="<?= base_url ( 'Contact/' ); ?>" method="post">
    <!--Name-->
    <div class="form-group">
        <label class="sr-only" for="name">Name</label>
        <div class="input-group">
            <div class="input-group-addon"><span class="glyphicon glyphicon-user" aria-hidden="true"></span></div>
            <input id="name" type="text" class="form-control validate" name="name" placeholder="Your full name" value="">
        </div>
    </div>
    <!--Email-->
    <div class="form-group">
        <label class="sr-only" for="email">Email</label>
        <div class="input-group">
            <div class="input-group-addon"><span class="glyphicon glyphicon-envelope" aria-hidden="true"></span></div>
            <input id="email" type="text" class="form-control validate" name="email" placeholder="Your email address" value="">
        </div>
    </div>
    <!--Message-->
    <div class="form-group">
        <label class="sr-only" for="message">Message</label>
        <div class="input-group">
            <div class="input-group-addon"><span class="glyphicon glyphicon-envelope" aria-hidden="true"></span></div>
            <textarea id="message" name="message" class="form-control" rows="3" placeholder="Message"></textarea>
        </div>
    </div>
    <!--Submit Button-->
    <div class="form-group text-right">
        <input id="contact_us" type="hidden" name="contact_us" value="Submit" />

        <button type="submit" id="contact_us" class="btn btn-warning" name="contact_us" value="Submit">
            Send &nbsp;<span class="glyphicon glyphicon-send" aria-hidden="true"></span>
        </button>
    </div>
</form>

JAVASCRIPT

<script>
$( '#contact-form' ).submit ( function ( event ) {

    event.preventDefault (  );
    event.stopPropagation (  );

    var $scriptUrl = $( '#contact-form' ).attr ( 'action' );

    $.ajax ( {
        method : 'POST',
        url : $scriptUrl,
        data    : $( this ).serialize ( ),
        cache : false,
        processData: false,
        contentType: false,
        dataType : 'json',
        success : function ( data, textStatus, jqXHR ) {
            if ( data.success === true ) { alert ('success'); }
            else { alert ('failure'); }
        },
        error : function ( jqXHR, textStatus, errorThrown ) {
            alert (  jqXHR.responseText  );/*This returns the empty array*/
        }
    } );

} );
</script>

Controller (index function) ( http://mysite/Contact - Localhost-wamp)

public function index (  )
{
    print_r($_POST);
}

Answer 1

If you are using jquery then you should use the syntax as follows:

$.post('url',$("#contact-form").serialize(),function(data){
 //here take action on returned data
});

Answer 2

You just change the
data : $(this).serialize(),
to
data : $('#contact-form').serialize(),

In ajax $this not working because when you call $this in ajax then $this call always parent object ajax.

Answer 3

将data : $( this ).serialize ( ),更改为data : new FormData($('#contact-form')[0]) remove dataType : 'json'如果这对您不起作用，请告诉我？

Answer 4

Student x. This may not be your answer, but this is all you need to pass a form

$(function() {
"use strict";
$("#form1").submit(function() {
    var data = $("#form1").serialize();
    //alert(data); return false;
    $.ajax({
        url: "/forms/form1",
        data: data,
        type: "POST",
        success: function(msg) {
            if (msg) {
                $("#display").html(msg);
            } else {
                $("#display").text("nothing came back For some reason");
            }
        }
    });
    return false;
});

});

You can also use this. I use this for all my forms, then I only need 1 script. Of course you would change the success. Just name all your forms with an ID of ajax

(function() {
"use strict";
$('form#ajax').on('submit', function() {
    var that = $(this),
        url  = that.attr('action'),
        type = that.attr('method'),
        data = {};
    that.find('[name]').each(function(index, value) {
        var that   = $(this),
            name   = that.attr('name'),
            value  = that.val();
        data[name] = value;
    });
$.ajax({
    url: url,
    type: type,
    data: data,
    success: function(response) {
        $('#display').html(response).delay(8000).fadeOut(1000);
    }
  });
return false;

}); })(jQuery);