Jquery Ajax and json - php

Question

I am sending a value to mysql query throught ajax. I get output from query as array. I used echo json_encode($var);
If my array size is more than 1, my success function is not getting called. I could see the response and status code 200 in fire bug.
How should I retrieve the value in my success function ?
success: function(data){
ob = jQuery.parseJSON(data);
}
Update
my Json response from firebug
{"uid":"4",
"name":"ram\u00fcrmeg\u00f6zl\u00fcer",
"pic_big":"http:\/\/profile.ak.fbcdn.net\/hprofile-ak-prn1\/4149__3333_n.jpg"}
$.ajax({
type: "POST",
url: "x.php",
dataType: 'json',
data: y,
success: function(data){
}
})

Answer 1

Try the below and see if you get any alert.
if(!$.isEmptyObject(data))
{
alert (data);
var len = data.length;
for (var i = 0; i< len; i++) {
var var1 = [data[i].uid];
var var2 = [data[i].name];
var var3 = [data[i].pic_big];
}
else{alert('No data')}

Answer 2

Your response is already an Object, no need to parseJSON it. Just do this for getting the values:
console.log(data.uid);
console.log(data.name);
console.log(data.pic_big);
Or whatever you wanna do with those values.

Answer 3

You can try:
$.ajax({
type: "POST",
url: "x.php",
dataType: 'json',
data: y,
success: function(data){
$.each(data, function() {
var uid = this.uid; // etc
}
// this works alone if your response is wrapped in [ ], if not:
if(!data.length) {
var uid = data.uid;
} else { // above $.each code
}
}
jsFiddle of this here: http://jsfiddle.net/Y73xe/