How to directly return JSON object from $http?

Question

I want to create a factory that always returns the json object retrieved from a webservice:

angular.module('test').factory('myService', myService);
myService.$inject = ['$http'];

function myService($http) {
    var urlBase;

    return {
        getContent: function(id) {
            return $http.get(urlBase).then(function(response) {
                return response.data;
            });
        }
    };
}

When I call MyService.getContent();, I'm not getting the JSON object, but an object with $$state and __proto__.

Why? How can I fore the factory to directly return the content only?

Note: I know I could write

MyService.getContent().then(function(response) { console.log(response); });

but this way I'd always have to repeat the then... function statement when I access the factory.

This is why promises are there for. How would you ensure the sequence if you won't use promsie returned by $http in controller. In case you wan't to setup some model with the data returned from service. — Diljohn5741, Feb 09 '16 at 13:40

score 2 · Accepted Answer · answered Feb 09 '16 at 13:38

2

You can't return the result of an asynchronous function. Your return response.data; statement is just exiting the promise .then() callback. You should modify your function like so:

getContent: function(id) {
  return $http.get(urlBase);
}

And then call it like this:

MyService.getContent().then(function (response) {
  // do something with the response
});

answered Feb 09 '16 at 13:38

Joe Clay

33,401
4
85
85

I posted this before your edit, but it still applies - there's no way to synchronously return a piece of data that's retrieved asynchronously! – Joe Clay Feb 09 '16 at 13:41

score 0 · Answer 2 · answered Feb 09 '16 at 13:42

0

That is because you are returning a promise that has already been resolve and not the promise instance. Just returning

$http.get(urlBase)

from your getContent function should do the trick :)

answered Feb 09 '16 at 13:42

shanks

912
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23

How to directly return JSON object from $http?

2 Answers2