pass jquery value to php variable - php

Question

Am a noob so hope you can help!
I've got the following jquery code:
$("#single-home-container").html(post_id);
It then displays the value I want within the HTML on the page:
<div id="single-home-container"></div>
What I would like is to pass the value into a PHP variable to I can use the info in a MySQL query.
How do I do so? It's in WordPress so no separate files
Thanks

Answer 1

You can use jQuery.post() or jQuery.ajax().
Here some examples:
<script>
$.post("test.php", { "func": "getNameAndTime" },
function(data){
console.log(data.name); // John
console.log(data.time); // 2pm
}, "json");
</script>
<script>
$.ajax({
url: "test.html",
context: document.body,
success: function(){
$(this).addClass("done");
}
});
</script>

Answer 2

You'd need to use ajax. Something like this:
JQuery:
$('.locationID').click(function(){
var post_id = $(this).attr('rel'),
$container = $("#single-home-container");
old_html = $container.html();
$container.html('loading...');
$.ajax({
url : '/path/to/php/file.php',
data:{"post_id":post_id},
type: 'post',
dataType: 'json',
success: function(data){
$container.html(data.postBody);
},
error: function(data){
// possibly notify the user of the error
$container.html(old_html);
}
});
});
That assumes you have a field in your posts table called postBody
PHP
<?php
header('Content-type: application/json');
// $query_result = do mysql query with $_POST['post_id']
echo json_encode($query_result);
exit();
?>
That assumes that you want all the fields and you're returning all the fields, including postBody - and of course you have PHP 5.2.6+ or whatever version they added json_encode().