Jquery loop data

Question

How can i loop the following JSON data string into jquery? Should I use the function $.getJSON ?

jQuery17209521235961001366_1380903443191({"type":"result","rid":"hopkinsa","data":
[[{"artist":"NAME1","title":"SONG1","album":null,"royaltytrackid":null,"url":null,"image":"test1.jpg","time":1380910069,"localtime":"11:07 AM"},
{"artist":"NAME2","title":"SONG2","album":null,"royaltytrackid":null,"url":null,"image":"test2.jpg","time":1380909866,"localtime":"11:04 AM"},  
{"artist":"NAME2","title":"SONG3","album":null,"royaltytrackid":null,"url":null,"image":"test3.jpg","time":1380909864,"localtime":"11:04 AM"}],false,0,10,0]})

I trying with this code:

$(document).ready(function() {
    $.getJSON("json.php", function(data){
        $.each(data.data[0], function(i, item){
            $("#data").append(item.artist); // Name1, Name2
        });      
    })
});

That would be JSONP, so yes, if you're trying to get that from another site you can use $.getJSON — adeneo
– adeneo, Commented Oct 4, 2013 at 18:21

Ruan Mendes · Accepted Answer · 2013-10-04 18:28:02Z

1

That's not JSON, it's a literal JS object because jQuery will already have decoded it. Just use brackets [], dot notation ., and loops.

In your callback, if the param name is data, you could do the following

$.getJSON("url.json", function(data){
    console.log(data.type, data.data[0][0].artist) //"result", "Name1"
    $.each(data.data[0], function(i, item){
        console.log(item.artist); // Name1, Name2
    });        
})

edited Oct 4, 2013 at 18:28

answered Oct 4, 2013 at 18:22

Ruan Mendes

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4 Comments

Stefan Over a year ago

I have the literal JS object in php file json.php. I don't get any results if I use your code...

Ruan Mendes Over a year ago

Stefan, please provide actual error messages, the page is blank is not enough information to know what is going on. Did you check the network tab? Is the request going out? Did you debug it? Step through the code? Please do some debugging on your own and post it in your answer.

Stefan Over a year ago

Sorry but I don't know how to debug it :/

Stefan Over a year ago

Can I shoe the result whit this: $("#data").html(data.type, data.data[0][0].artist) ?

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