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up vote 0 down vote favorite

A json object has a key lastLogin. Value of it is a string.

I am trying to print firstName John and Blake

$scope._users = [{
        "User": {
            "userid": "dummy",
            "lastlogin": "{\"employees\":[{\"firstName\":\"John\"},   {\"firstName\":\"Blake\"}]}",
        }
    }];

https://jsfiddle.net/xu0vqmff/

Any help would be appreciated.

share|improve this question
2  
Pet peeve: There are no JSON objects in JavaScript. There are objects, and there are JSON strings. $scope._users[0] is an example of an object (not JSON object). $scope._users[0].User.lastlogin is an example of a JSON string. – Amadan Jul 9 '15 at 2:33
    
JavaScript is case-sensitive. lastlogin !== lastLogin. – tmack Jul 9 '15 at 2:42
up vote 1 down vote accepted

Try like this

View

<div ng-controller="MyCtrl">
    <div ng-repeat="user in _users" ng-init="myInfo=parJson(user.User.lastlogin)">
        <div ng-repeat="emp in myInfo.employees">{{emp.firstName}}</div>
    </div>
</div>

Controller

var myApp = angular.module('myApp', []);

function MyCtrl($scope) {
    $scope.getName = function (user) {
        return "Names";
    };

    $scope._users = [{
        "User": {
            "userid": "dummy",
                "lastlogin": "{\"employees\":[{\"firstName\":\"John\"},                   {\"firstName\":\"Blake\"}]}",
        }
    }];
    $scope.parJson = function (json) {
        return JSON.parse(json);
    }
    //console.log(JSON.parse($scope._users[0].User.lastlogin));
}

JSFIDDLE

share|improve this answer
2  
Anik, why would you use JSON.parse instead of the built-in angular.fromJson? – Daniel Nalbach Jul 9 '15 at 2:56
    
@DanielNalbach , does it make any big difference between JSON.parse and angular.fromJson ? – Anik Islam Abhi Jul 9 '15 at 4:25
    
Not really, angular.fromJson is just a wrapper over JSON.parse with a string check to prevent double parsing. I thought it did more, but checked the source, and that's it. – Daniel Nalbach Jul 9 '15 at 15:49

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