Wordpress - Passing javascript variable to php

Question

I'm implementing a photo tagging system.

in my php file I have:

if($_POST['type'] == "insert") {
    $pid = $post->ID;   

    $tag_name = $_POST['tag_name'];
    $tag_link = $_POST['tag_link'];
    $pic_x = $_POST['pic_x'];
    $pic_y = $_POST['pic_y'];

    $arr = array("tag_name" => $tag_name, "tag_link" => $tag_link, "pic_x" => $pic_x, "pic_y" => $pic_y);

    add_photo_tag($pid, $arr);

    wp_redirect("http://www.test.com");
}

to catch the data. in my js file i have:

$('#tagit #btnsave').live('click',function(){
    name = $('#tagname').val();
    link = $('#taglink').val();
    counter++;
    $('#taglist ol').append('<li rel="'+counter+'">'+counter+'. <a href="'+link+'#test_id" target="new">'+name+'</a> (<a class="remove">Entfernen</a>)</li>');
    $('#imgtag').append('<div class="tagview" id="view_'+counter+'">'+counter+'</div>');
    $('#view_' + counter).css({top:mouseY,left:mouseX});
    $('#tagit').fadeOut();

    $.ajax({
        type: "POST",
        url: "content.php",
        data: "tag_name=" + name + "&tag_link=" + link + "&pic_x=" + mouseX + "&pic_y=" + mouseY + "&type=insert",
        cache: true,
        success: function(data) {
            alert("success!");
        }
    });

});

Somehow the variables don't get passed to the php resulting in me not able to save the data properly to database. The problem must be somewhere in either the $.ajax part or php. Can someone help me?

please try with wordpress ajax passing method that's better one codex.wordpress.org/AJAX_in_Plugins — softsdev
– softsdev, Commented Jun 28, 2013 at 12:58

Sergio · Accepted Answer · 2013-06-28 12:01:45Z

1

Well, you're checking post variable $_POST['type'] == "insert_tag" while it's actual value is: &type=insert. Others look fine. Btw never user .live(), it's obsolete, do it with .on()

Also, if all your elements are on a form you may use $('your_form_selector').serialize() - that'l be your post data

answered Jun 28, 2013 at 12:01

Sergio

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1 Comment

Cloudzane Over a year ago

sry just edited it. that sadly wasn't the mistake i just fundled with the code too much and forgot to edit it before i posted it here

knolleary · Accepted Answer · 2013-06-28 12:03:38Z

0

Your php code is looking for 'type' to be equal to 'insert_tag'.

Your JavaScript is POSTing with type=insert - so your php code ignores it.

answered Jun 28, 2013 at 12:03

knolleary

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1 Comment

Cloudzane Over a year ago

thx that was just a copy/paste mistake. I actually had that right. Still doesn't work sadly..

montrealist · Accepted Answer · 2013-06-28 12:38:00Z

0

Why are you redirecting in your PHP code? Remove this line:

wp_redirect("http://www.test.com");

You can replace it with some sort of a simple feedback system. After all you want the server-side to pass back the status and maybe some validation messages.

$return = array('status' => 0, 'msg' => 'all is well');

if(!add_photo_tag($pid, $arr)) {
    $return['msg'] = __('Could not add photo tag.');
    $return['status'] = -1;
}

echo json_encode($return);

answered Jun 28, 2013 at 12:38

montrealist

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Comments

Code Lღver · Accepted Answer · 2013-06-28 12:40:35Z

0

Try with the following code:

$.ajax({
        type: "POST",
        url: "content.php",
        data: {"tag_name": name,"tag_link":link,"pic_x": mouseX,"pic_y":mouseY, "type":"insert"}, 
        //change here like I have done
        cache: true,
        success: function(data) {
            alert("success!");
        }
    });

And Replace the if($_POST['type'] == "insert_tag") line with following:

if($_POST['type'] == "insert")

As the value you are passing of $_POST['type'] is insert not insert_tag.

EDITED:

one more thing as I have found that is you have not used the var before initializing the variable. use the var name = ...

edited Jun 28, 2013 at 12:40

answered Jun 28, 2013 at 12:01

Code Lღver

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2 Comments

Cloudzane Over a year ago

thx I edited the code adding "," after the mouseY. Still doesn't trigger the alert proving it has successed nor does anything get transmitted to the php :S

Code Lღver Over a year ago

@Cloudzane use the firebug to debug your code and then you will surely find the proper error in your code.

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