Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Reading JSON data in jQuery AJAX post output [duplicate]

I have the code below which i am passing through json to jquery.

$string['msg']['1']['Name'] = "John Doe";
$string['msg']['1']['Said'] = "Hello there";

$string['msg']['2']['Name'] = "Jane King";
$string['msg']['2']['Said'] = "Hello there";

$string['errors']['1']['Person'] = "Jane King";
$string['errors']['1']['type'] = "Wrong Screwdriver";


$string['errors']['2']['Person'] = "John Doe";
$string['errors']['2']['type'] = "Wrong Spanner";

The JSON output is below:

{"msg":{"1":{"Name":"John Doe","Said":"Hello there"},"2":{"Name":"Jane King","Said":"Hello there"}},"errors":{"1":{"Person":"Jane King","type":"Wrong Screwdriver"},"2":{"Person":"John Doe","type":"Wrong Spanner"}}}

This is the ajax code that is retreiving the json.

$.ajax({
    dataType : 'json',
    url: 'polling.php',
    type: 'post',
    data: 'id=1',
    success: function(data) {
        //read json     

});

I am trying to read all 'msg' in order of number.. 1, 2.. and get 'Name' and 'Said' values for each.

Then do the same for 'errors'

but i cant correctly retrieve any values

like image 903
sam fisher Avatar asked May 16 '13 05:05

sam fisher


2 Answers

As per your given output:

Try this:

$(function(){

    var  data = {"msg":{"1":{"Name":"John Doe","Said":"Hello there"},"2":{"Name":"Jane King","Said":"Hello there"}},"errors":{"1":{"Person":"Jane King","type":"Wrong Screwdriver"},"2":{"Person":"John Doe","type":"Wrong Spanner"}}};

    $.each(data.msg, function(index, value) {
      alert(index + ': ' + value.Name);
    });

});

JSFiddle Example

like image 136
Satpal Avatar answered Nov 07 '22 21:11

Satpal


Demo Link

var data = {"msg":{"1":{"Name":"John Doe","Said":"Hello there"},"2":{"Name":"Jane King","Said":"Hello there"}},"errors":{"1":{"Person":"Jane King","type":"Wrong Screwdriver"},"2":{"Person":"John Doe","type":"Wrong Spanner"}}};


$.each(data['msg'], function(key, element){
    console.log(key+' : '+element['Name']);
});

$.each(data['errors'], function(key, element){
    console.log(key+' : '+element['Person']);
});
like image 24
Jayendra Avatar answered Nov 07 '22 21:11

Jayendra