How to access data of [object object] AngularJS

Question

I am new to AngularJS. I am sending a JsonObject to another state.

Ex:

viewScope.edit=function(d) {

   var viewData = {
      'name' : d.name,
   };

   $state.go('edit', {'viewD': viewData}, {reload: true});
};

My State is-

viewApp.config(function($stateProvider){
    $stateProvider
        .state('edit', {
            name: '#/edit',
            url: '/register/{viewD}',
            templateUrl: function(){    
                 return path+'/register.jsp';
             },
             controller:'registerCtrl',
             resolve : {
                 loadPlugin: function($ocLazyLoad) {
                     return $ocLazyLoad.load([{
                         name : 'registerApp',
                         files: [path+'/resources/js/register.js'],
                     }])
                }
            }
        })
});

In register Controller getting data-

regApp.controller('registerCtrl',function($stateParams){                 
        if($stateParams != undefined){
            console.log($stateParams.viewD);
        }
});

On console output is- [object object]

How can i access the name key from this [object object].

console.log($StateParams.viewD.name); // Not Working

JSON.parse, JSON.stringify not working.

What is the content of $stateParams ? What is the type of $stateParams.viewD ? — Weedoze
– Weedoze, Commented Apr 20, 2018 at 5:42
@Weedoze The content of $stateParams is viewD : "[object Object]" — Deepak Verma
– Deepak Verma, Commented Apr 20, 2018 at 5:50
Where do you see this [object Object] ? In the console ? Can you expand it ? What is the result of console.log(typeof $stateParamas.viewD) — Weedoze
– Weedoze, Commented Apr 20, 2018 at 6:22
@Weedoze yes in the console. No i can not expand it. Its type is String — Deepak Verma
– Deepak Verma, Commented Apr 20, 2018 at 6:30
Try to replace your url in the state config by url: '/register/{viewD:json}', — Weedoze
– Weedoze, Commented Apr 20, 2018 at 6:40

Weedoze · Accepted Answer · 2018-05-15 07:07:52Z

1

You have to change your state config URL from

url: '/register/{viewD}',

to

url: '/register/{viewD:json}',

The JSON parameter type has been added in version 0.2.13

answered May 15, 2018 at 7:07

Weedoze

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Comments

Senal · Accepted Answer · 2018-04-20 05:56:32Z

1

Change your config method as following,

viewApp.config(function($stateProvider){
$stateProvider
    .state('edit', {
        name: '#/edit',
        url: '/register',
        params: {
           viewD: null
        }
        templateUrl: function(){    
             return path+'/register.jsp';
         },
         controller:'registerCtrl',
         resolve : {
             loadPlugin: function($ocLazyLoad) {
                 return $ocLazyLoad.load([{
                     name : 'registerApp',
                     files: [path+'/resources/js/register.js'],
                 }])
            }
        }
    })
});

And then you can access your object from $state like this in the controller, $state.params.viewD or from $stateParams like this $stateParams.viewD

Now try console.log($state.params.viewD.name)

edited Apr 20, 2018 at 5:56

answered Apr 20, 2018 at 5:50

Senal

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6 Comments

Deepak Verma Over a year ago

console.log($stateParams.viewD); output= {name : "abc"} but console.log($stateParams.viewD.name); gives error during app load undefined name.

Senal Over a year ago

Try this and see the type of your viewD console.log(typeof $stateParams.viewD), if it's a string JSON.parse($stateParams.viewD).name would return the name

Deepak Verma Over a year ago

Give error --- SyntaxError: Unexpected token u in JSON at position 0 on JSON.parse($stateParams.viewD).name

Senal Over a year ago

Seems like your $stateParams.viewD object is undefined. if($stateParams.viewD) JSON.parse($stateParams.viewD).name try this. If it doesn't go inside the if check your source of the viewD

Deepak Verma Over a year ago

Its not undefined it came under if($StateParams.viewD){ and still error. What i do?

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