can't get jquery output my json object

Question

I have the following code in my page. I am expecting a series of json objects to get returned from the person_output.aspx page, which it does successfully. However, when it comes to outputting the content, I receive an error.

  $.getJSON("ajax/person_output.aspx", { 'uID': 1 }, function(data) {
            $.each(data.items, function(i, item) {
            $("<span/>").html(item.first_name).appendTo("#content");
                });
            });

ajax/person_output.aspx produces the following json (this is only for one record..)

{
    "l_id": "49",
    "u_id": "1",
    "first_name": "john",
    "last_name": "doe",
    "title" : "General Manager",
    "color" : "333"
}

firebug produces the following error;

G is undefined
init()()jquery-1....2.min.js (line 12)
(?)()()URLINX5 (line 99)
I()jquery-1....2.min.js (line 19)
F()()jquery-1....2.min.js (line 19)
[Break on this error] (function(){var l=this,g,y=l.jQuery,p=l.....each(function(){o.dequeue(this,E)})}});

The error appears to be with the variable G. Is that defined anywhere? — aleemb
– aleemb, Commented May 18, 2009 at 19:39
The variable "G" is a result of jquery being compressed. you should try running it with a development version of jQuery and see if the output makes more sense in that case. — fforw
– fforw, Commented May 18, 2009 at 19:59

great_llama · Accepted Answer · 2009-05-18 19:04:36Z

2

There is a comma after the color property in your JSON result, does removing it help?

answered May 18, 2009 at 19:04

great_llama

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3 Comments

ichiban Over a year ago

Good catch. I was about to say the same thing. +1

Subliminal Hash Over a year ago

nope actually there were more lines which I did not paste for keeping things simple.. in the original one there is no comma after the last line. I am editing the question now..

great_llama Over a year ago

You say "this is only for one record",.. you're returning an array of these objects wrapped in [{square: "brackets"}] ?

Allen Rice · Accepted Answer · 2009-05-18 19:18:09Z

0

I'm new to firebug but you need to figure out what is undefined here


$.getJSON("ajax/person_output.aspx", { 'uID': 1 }, function(data) {
            $.each(data.items, function(i, item) {
            $("").html(item.first_name).appendTo("#content");
                });
            });

It is either data, data.items, item, or item.first_name. If you are returning an array, don't you need to do do something like data.items[i], rather than item.first_name? The way you have it setup now, "item" is probably 0,1,2,3...n

answered May 18, 2009 at 19:18

Allen Rice

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