Reading a JSON object elements

Question

I'm pretty much new to jquery, i'm trying to read an JSON object.

My Jquery ajax call

$(document).ready(function () {

        var resource = "v1/projects";

        var url = '@Url.Action("Proxy")?resource=' + resource;

        var method = 'GET';


        var settings = {
            dataType: "text",
            type: method,
            success: function (data, textStatus, jqXHR) {

               var res= JSON.stringify(data);


               for (var k in res) {
                   alert(res)
               }

            },
            error: function (jqXHR, textStatus) {

                alert("error");
            }

        };



        $.ajax(url, settings);
    });

Ajax call works fine and i receive the following result.

enter image description here

I need to extract name and description from the result.

I tried using the for loop but it prints all the elements in the result

bipen · Accepted Answer · 2013-09-11 06:37:02Z

1

use . operator

var res=JSON.parse(data);
$.each(res,function(i,v){
    alert(v.name);
    alert(v.description);
});

answered Sep 11, 2013 at 6:37

bipen

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1 Comment

bipen Over a year ago

i would recommned you to use dataType: "json", and return as JSON from server ... that you you don't have to parse or stringify your json

Herman Tran · Accepted Answer · 2013-09-11 06:39:54Z

1

Instead of JSON.stringify, you would want to use var res= JSON.parse(data) to convert it into a JS object and then use alert(res[k]) in the for loop

answered Sep 11, 2013 at 6:39

Herman Tran

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Comments

RahulG · Accepted Answer · 2013-09-11 06:42:53Z

1

try this

for (var k in res) {
    alert(' name=' +k+ ' description=' +k+');
}

answered Sep 11, 2013 at 6:42

RahulG

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Comments

chamara · Accepted Answer · 2013-09-11 08:48:24Z

0

got it working with the following.

 success: function (data, textStatus, jqXHR) {

                     var obj = eval("(" + jqXHR.responseText + ")");

                      for (var k in obj.data) {
                        alert(obj.data[k].name);

                      }

                },

answered Sep 11, 2013 at 8:48

chamara

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