AJAX enabled web form - php

Question

I am new to Jquery and want some help from you guys.
i want to decode json data in jquery as i am able to pass data from php to ajax but after it came back in jquery it is not parsing it says undefined. the code is bellow
JavaSript file
$.post("GetData.php", function(data) {
if(data==false)
var tpl = '<p>no record found<p>'
else
var tpl = DrawTableRowsforSection(data);
$("#result").append($(tpl));
},"json");
function DrawTableRowsforSection(p)
{
alert(p.id[0]);
var o = '<table>';
for (var i = 0; i < p.length; i++)
alert(p.id[i]);
o += '<tr><td>'+p.id[i]+'</td><td>'+p.section_name[i]+'<td></tr>';
o+='</table>'
return O;
}
PHP Script
header('Content-Type: application/json');
mysql_connect('localhost','root','') ;
mysql_select_db('news');
session_start();
$query = 'select id,section_name from section';
if ($result = mysql_query($query)) {
if (!mysql_num_rows($result)==null) {
$myArray = array();
while ($row = mysql_fetch_assoc($result)) {
$id = ToSring($row['id']);
$myArray[] = $row;
}
echo json_encode($myArray);
}
}
The database has a table named section
fields are as follows
id int(11)
section_name varchar(20)
There are total 5 records there.
What I want is to populate a table using returned data.
Can any one guide me where I am making mistake
Regards
Kashif Afzaal

Answer 1

Be sure that mysql returns results and you can take them through ajax.
Also, I've seen the following mistake:
You are using the variable tpl wrong. It is just a js variable, no need to use $ .
Use this way:
$("#result").append(tpl);

Answer 2

OK, from the top in the PHP script:
The MySQL extension is ancient and decrepit. Use MySQLi or PDO.
You never handle a failed MySQL query, so the response will quite possibly be empty. In 99% of cases, any if block should have a corresponding else block.
The result of mysql_num_rows() is never null. Do if (mysql_num_rows($result) > 0), or better yet, just if (mysql_num_rows($result)).
Never do if (!something == something) - it rarely does what you want it to. Do if (something != something)
You never handle mysql_num_rows() failing/giving no rows, so (again) the response will quite possibly be empty.
You never do anything with the $id variable you create.
I'm pretty sure ToSring should read ToString. But there's actually no such function in PHP - you would have either define it or invoke it as a method of a class/object. Indeed, this is likely to be the problem, as it will result in a fatal error.