how to access this complex json object with angular js?

Question

$scope.data = {
    "statusCode": 200,
    "message": "success",
    "result": {
        "data2": [
            {
                "viewTitle": "one",
                "viewType": 1,
                "viewData": "{\"data3\":{\"_id\":\"one\"," +
                "\"channelId\":\"0\",\"channelName\":\"animals\"," +
                "\"rate\":3.5,\"visited\":21,\"meta\":{\"currency\":\"t\"" +
                ",\"size\":\"5 mb\",\"updatedAt\":\"2 2 2016\"" +
                ",\"createdAt\":\"2 2 2016\",\"duration\":\"3 minutes \"}" +
                ",\"thumbnail\":\"127.0.0.1\"" +
                ",\"banner\":\"127.0.0.1\"" +
                ",\"trailerUrl\":\"\",\"url\":\"127.0.0.1\"" +
                ",\"desc\":\"my informations \"" +
                ",\"title\":\"my videos number 4\",\"packages\"" +
                ":[{\"_id\":\"1\",\"price\":500" +
                ",\"type\":\"purchase\"}],\"packageId\":\"2\"" +
                ",\"packagePrice\":500,\"userFavorite\":0,\"userRate\":0,\"locked\":1}}"
            }
        ]
    }
};`

I try to access for example chanelName or size please help to resolve.

docs.angularjs.org/api/ng/function/angular.fromJson Make the Json string in result.data2.viewData an object, and treat it as any other object in your js code. — Max Sorin
– Max Sorin, Commented Mar 23, 2016 at 13:36

hellol11 · Accepted Answer · 2016-03-23 13:56:34Z

0

first, turn the JSON string into an object via JSON.parse();
var data = JSON.parse(); //data is now the result of JSON.parse
and then access the property you want to use in $scope.data

edited Mar 23, 2016 at 13:56

answered Mar 23, 2016 at 13:44

hellol11

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Comments

Johannes Ferner · Accepted Answer · 2016-03-23 13:58:36Z

0

I have created a CodePen for your Problem: http://codepen.io/JohannesFerner/pen/PNmmBd/

You have JSON.parse(<escaped JSON property>) or angular.fromJson(<escaped JSON property>) (see the comment above) your escaped JSON.

Without further knowledge of your backend I would suggest that you do not deliver escaped JSON to the client.

edited Mar 23, 2016 at 13:58

answered Mar 23, 2016 at 13:52

Johannes Ferner

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Community · Accepted Answer · 2017-05-23 12:15:51Z

0

First level objects can be accessed like this.

person.result.data2[0].viewTitle

JSON.parse is the easiest way here to disassemble the complex object. I have given my example below

http://jsfiddle.net/MisterFantastic/u8bhc1ds/1/

JSON.parse(person.result.data2[0].viewData)

For complex objects please refer this link. AngularJS format JSON string output

edited May 23, 2017 at 12:15

CommunityBot

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answered Mar 23, 2016 at 14:33

Thanigainathan

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