angularjs function returned array is an object, not a string

Question

I am having some issues getting the function below to return as a usable string?

my code is this:

app.factory('Urls', ['$http', function($http) {
    var urls = {};
    urls.getUrls = function () {
        return  $http.get('json/dataUrls.json'); 
    }
    return urls
}]);

app.factory('Emails', ['$http', 'Urls', function($http, Urls) {

    var dataUrl = Urls.getUrls().then(function(response) {
        return response.data.emails;
    });
    console.log(dataUrl);

    var query = {};
    query.getItems = function() {
        return  $http.get('json/emails.json');  
    };
    return query;
}]);

below is the result of console.log(dataUrl); what i am trying to get is the value string...

f {$$state: {…}}
$$state
:
status
:
1
value
:
"json/emails.json"
__proto__
:
Object
__proto__
:
Object

could you please put your code in plunker/fiddler/codepen and make a runnable code snippet? — Vikasdeep Singh
– Vikasdeep Singh, Commented Nov 6, 2017 at 4:09

Derek Brown · Accepted Answer · 2017-11-06 04:11:32Z

2

Urls.getUrls().then(...) returns a promise (read more about this here), a datatype used to save a "reference" to a value which is fetched asynchronously. Therefore, running console.log on dataUrl will return an object, not a string.

In order to get this value, you need to access it asynchronously. The best way of doing this would be via promise chaining:

app.factory('Urls', ['$http', function($http) {
    var urls = {};
    urls.getUrls = function () {
        return  $http.get('json/dataUrls.json'); 
    }
    return urls
}]);

app.factory('Emails', ['$http', 'Urls', function($http, Urls) {

    Urls.getUrls().then(function(response) {
        return response.data.emails;
    }).then(function(emails){
        console.log(emails);
    });

    var query = {};
    query.getItems = function() {
        return  $http.get('json/emails.json');  
    };
    return query;
}]);

answered Nov 6, 2017 at 4:11

Derek Brown

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1 Comment

Sandra Willford Over a year ago

thank you for the explanation and links. I have some reading to do to wrap my head around the concept i guess!

Khiem Tran · Accepted Answer · 2017-11-06 04:31:39Z

1

DataUrls you are getting from var dataUrl = Urls.getUrls().then(... is actually a Promise, not a String. I guess your intention is something like bellow:

app.factory('Urls', ['$http', function($http) {
    var urls = {};
    urls.getUrls = function () {
        return  $http.get('json/dataUrls.json'); 
    }
    return urls
}]);

app.factory('Emails', ['$http', 'Urls', function($http, Urls) {

    var query = {};
    query.getItems = function() {
        return Urls.getUrls().then(function(response) {
            return response.data.emails;
        }).then(dataUrls => { // dataUrls here is a string, you can use it.
            return  $http.get(dataUrls); 
        });
    };
    return query;
}]);

edited Nov 6, 2017 at 4:31

answered Nov 6, 2017 at 4:17

Khiem Tran

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1 Comment

Khiem Tran Over a year ago

Look at the code, you can pass the dataUrls because it's in the same scope. then(dataUrls => { return $http.get(dataUrls + '/emails.json'); });

Shree Wankhede · Accepted Answer · 2017-11-07 05:29:55Z

0

app.factory('Urls', ['$http', function($http) {
    var urls = {};
    urls.getUrls = function () {
        return  $http.get('json/dataUrls.json'); 
    }
    return urls
}]);

app.factory('Emails', ['$http', 'Urls', function($http, Urls) {

    Urls.getUrls().then(function(response) {
        return response.data.emails;
    }).then(function(emails){
        console.log(emails);
    });

    var query = {};
    query.getItems = function() {
        return  $http.get('json/emails.json');  
    };
    return query;
}]);

answered Nov 7, 2017 at 5:29

Shree Wankhede

856 bronze badges

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