How to get json key and value in javascript?

Question

I am returning a json as shown below

{"name": "", "skills": "", "jobtitel": "Entwickler", "res_linkedin": "GwebSearch"}

I am trying to get each element key and value:

..
}).done(function(data){
    alert(data['jobtitel']);
});

I am getting undefined in alert. WHY? I tried data.jobtitel, i tried loop but no success..

first u need to convert that json to array then have to try data.jobtitel — sawan
– sawan, Commented Sep 20, 2013 at 7:15
JSON is string and u want to access there that then u must have to convert... — sawan
– sawan, Commented Sep 20, 2013 at 7:18
@swan, $.parseJSON is not always available in every browser.. what to do ? — doniyor
– doniyor, Commented Sep 20, 2013 at 7:19
if u add jquery then $.parseJSON will work...otherwise u have to use pure javascript JSON converter JSON.eval(jsonstring) — sawan
– sawan, Commented Sep 20, 2013 at 7:22

sawan · Accepted Answer · 2020-07-17 10:14:49Z

169

//By using jquery json parser    
var obj = $.parseJSON('{"name": "", "skills": "", "jobtitel": "Entwickler", "res_linkedin": "GwebSearch"}');
alert(obj['jobtitel']);

//By using javasript json parser
var t = JSON.parse('{"name": "", "skills": "", "jobtitel": "Entwickler", "res_linkedin": "GwebSearch"}');
alert(t['jobtitel'])

Check this jsfiddle

As of jQuery 3.0, $.parseJSON is deprecated. To parse JSON strings use the native JSON.parse method instead.

Source: http://api.jquery.com/jquery.parsejson/

edited Jul 17, 2020 at 10:14

answered Sep 20, 2013 at 7:16

sawan

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2 Comments

acrawly Over a year ago

This is not documented enough. The ability to use json['KeyAsString']. Really great. Thank you.

dGRAMOP Over a year ago

How would I resolve Entwickler to jobtitel? Is there anythind native or do I have to write my own code?

NaYaN · Accepted Answer · 2013-09-20 07:19:41Z

24

you have parse that Json string using JSON.parse()

..
}).done(function(data){
    obj = JSON.parse(data);
    alert(obj.jobtitel);
});

answered Sep 20, 2013 at 7:19

NaYaN

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1 Comment

Jonas Over a year ago

Simplified: JSON.parse(data).jobtitle

onihime · Accepted Answer · 2013-09-20 07:26:59Z

13

var data = {"name": "", "skills": "", "jobtitel": "Entwickler", "res_linkedin": "GwebSearch"}

var parsedData = JSON.parse(data);
alert(parsedData.name);
alert(parsedData.skills);
alert(parsedData.jobtitel);
alert(parsedData.res_linkedin);

answered Sep 20, 2013 at 7:26

onihime

1294 bronze badges

1 Comment

Ishara Amarasekera Over a year ago

This is how you do it in javascript.

Chandan Kumar · Accepted Answer · 2019-03-17 18:39:44Z

6

For getting key

var a = {"a":"1","b":"2"};
var keys = []
for(var k in a){
  keys.push(k)
}

For getting value.

var a = {"a":"1","b":"2"};
var values = []
for(var k in a){
  values.push(a[k]);
}

answered Mar 17, 2019 at 18:39

Chandan Kumar

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Comments

user2672373 · Accepted Answer · 2013-09-20 07:21:10Z

4

http://jsfiddle.net/v8aWF/

Worked out a fiddle. Do check it out

(function() {
    var oJson = {
        "name": "", 
        "skills": "", 
        "jobtitle": "Entwickler", 
        "res_linkedin": "GwebSearch"
    }
    alert(oJson.jobtitle);
})();

answered Sep 20, 2013 at 7:21

user2672373

Comments

knoxgon · Accepted Answer · 2017-03-01 09:16:15Z

3

A simple approach instead of using JSON.parse

 success: function(response){
     var resdata = response;
     alert(resdata['name']);
}

edited Mar 1, 2017 at 9:16

knoxgon

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answered Dec 30, 2016 at 4:58

ricky

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Comments

Kamil Kiełczewski · Accepted Answer · 2019-04-25 10:41:32Z

3

It looks like data not contains what you think it contains - check it.

let data={"name": "", "skills": "", "jobtitel": "Entwickler", "res_linkedin": "GwebSearch"};

console.log( data["jobtitel"] );
console.log( data.jobtitel );

answered Apr 25, 2019 at 10:41

Kamil Kiełczewski

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Comments

Grant Miller · Accepted Answer · 2018-08-27 16:05:19Z

0

You can use the following solution to get a JSON key and value in JavaScript:

var dt = JSON.stringify(data).replace('[', '').replace(']', '');
if (dt) {
  var result = jQuery.parseJSON(dt);
  var val = result.YOUR_OBJECT_NAME;
}

edited Aug 27, 2018 at 16:05

Grant Miller

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answered Aug 27, 2018 at 12:44

Monzur

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1 Comment

Narendra Jadhav Over a year ago

While this code snippet may be the solution, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion.

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