CODEWITHSUNDEEP
jqueryajax
Byzmo PRopagands
Byzmo PRopagands

Reputation: 11

pass a variable with JQuery and ajax to PHP and fetch it with $_POST

i am trying to return a variable and I really don't understand how to do it. This is the 

jquery part $('a[href="if_blog"]').click( function( e ) {
        e.preventDefault(); 

        var bblog = $( this ).attr('href');
        var dataString = 'bblog=' + bblog;

        $.ajax({
            type: 'POST',
            url: '', 
            data: dataString, 
            success: function(data){

            $('#php').html(data);
        }
        });
    });

this is the php part and it is part of a big wordpress file:
$byzmo_blog = $_POST['bblog'];
switch ($byzmo_blog) {
    case 'if_blog':
        $byzmo_postype = 'if_posts';
        break;
    case 'food_mood_blog':
        $byzmo_postype = 'food_mood';
        break;
    case 'look_good_blog':
        $byzmo_postype = 'look_good';
        break;
    case 'feel_good_blog':
        $byzmo_postype = 'feel_good';
        break;
}
?>

<div id="php"></div>
<?php var_dump($_POST);

$query = array(
    'post_type' => $byzmo_postype,
    'posts_per_page' => get_theme_option('blog_posts'),
    'orderby' => get_theme_option('blog_orderby'),
    'order' => get_theme_option('blog_order'),
    'paged' => ( get_query_var('paged') ? get_query_var('paged') : true )
);
?>

So when I click the link (a[href="if_blog"]') I want to load some content according to what $_POST shows, but bcs I am in the same page data loads the whole index.php file, so it is loaded twise in the page(because I use .html()) and then it loads in $_POST too.. how can use only the variable in data with $_POST for this part $byzmo_blog = $_POST['bblog']; ?

I hope you can understand me......

EDIT: Ok, i am afraid I don't get it: so I have the index.file which is sth like this:

<a href='if_blog'>The Link</a>

<?php
$post_type = $_POST['posttype'];

$query = array(
    'post_type' => $post_type,
    'posts_per_page' => get_theme_option('blog_posts')
);
?>

<?php query_posts( $query ); if ( have_posts() ) :  ?>

<?php while ( have_posts() ) : the_post(); endwhile; endif; ?>

So, clicking the link is fetched by the jquery file:

jQuery.noConflict();
jQuery( document ).ready( function( $ ) {
$('a[href="if_blog"]').click( function( e ) {


        var bblog = $( this ).attr('href'); 
        var url = 'test.php'; 

        $.post(url, { byzmo_blog: bblog },
            function(data) {
                    alert(data);
                    //console.log(data); //
        }); //, "json");
        e.preventDefault();
});

} );

returns either a string from text.php, or an object if I apply json_encode.

Now from this point I want to return this data to index.html and use it $post_type = $_POST['posttype']; further.. when I use the console.log nothing happens, do I have to use json_decode here? i searched a lot but just can't get it at this moment... :/

Upvotes: 0

Views: 761

Answers (1)

cybertextron
cybertextron

Reputation: 10961

use $.post('url', '{data}', function(answer)); instead. You can gather more info from postJquery. Basically, you have:

$.post("test.php", { name: "John", time: "2pm" },
   function(data) {
     alert("Data Loaded: " + data);
   });

in the client site. So, in you server side, retrieve the value as

$name = $_POST['name'];
$time = $_POST['time'];

You can use json_decode($_POST['name']) to the JavaScript literal object to a PHP array, and to send it back to the page use json_encode($array); In your example, you did not give POST to where send the data to. You definitely need the url to where to POST to.

Upvotes: 1

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