Using Jquery AJAX to POST variable in PHP

Question

I believe I have the right syntax but am missing something important. Been searching on here for a while but can't figure out why the POST variable is not being detected. Basically my .ajax is firing because my test statement has been changing due to the value but some reason can't receive variable via $_POST (i.e. my echo in php echo that it is not firing) Also the native file and php that I am sending it to are the same file blankFormTemplate.php but don't think that should be an issue.

$(document).ready(function()
{
    var $selectedContexts = [];

    $('.allContextField').change(function(){
        //alert($(this).val());
        hideField = $(this).val();
        $('#'+hideField).remove();

        $.ajax
        ({
            type: "POST",
            url: "blankFormTemplate.php",
            data: addedContext=hideField,
            cache: false,
            success: function(addedContext)
            {
                $('#test').html($('#test').html()+hideField);
            } 
        });
    });
});

in my PHP blankFormTemplate.php:

<?php

if(isset($_POST['addedContext']))
{
    echo 'hello';
}
else
{
    echo 'why';
}

?>

Any help would be greatly appreciated.

Thanks,

Are you getting why everytime? Have you looked at the request being sent in the network tab of your developer console/firebug? — chris85
– chris85, Commented Apr 28, 2015 at 23:55
change data: addedContext=hideField, to data: 'addedContext='+hideField,. you need to set data to a sting, but you are just doing an assignment. — Sean
– Sean, Commented Apr 28, 2015 at 23:57
See the @Sean comment. Also print_r($_POST); might be useful in the future to see what is being sent, or inspecting the request in the developer console, stackoverflow.com/questions/1820927/…. — chris85
– chris85, Commented Apr 29, 2015 at 0:03
@sean tried your fix but didn't seem to affect things one way or the other — Norman Wei
– Norman Wei, Commented Apr 29, 2015 at 0:03

user2560539 · Accepted Answer · 2015-04-29 00:29:35Z

0

Your javascript

$(document).ready(function()
{

    $('.allContextField').change(function(){
        hideField = $(this).val();
        $('#'+hideField).remove();
});
if (hideField.trim()) {
 $.ajax
        ({
            type: "POST",
            url: "blankFormTemplate.php",
            data: (addedContext:hideField},
            cache: false,
            success: function(msg)
            {
                $('#test').html(msg);
            } 
        });
}  
});

Your PHP

<?php

if(isset($_POST['addedContext']))
{
    echo 'hello';
}
else
{
    echo 'why';
}

?>

answered Apr 29, 2015 at 0:29

user2560539

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