CODEWITHSUNDEEP
javascript
1future
1future

Reputation: 255

ReferenceError message in javascript?

Why am i getting the following error:ReferenceError: result is not defined"

function returnData() {        
  _myService.getData().then(function(data) {
    var result = data;
  });
  return result;
}

Upvotes: 0

Views: 93

Answers (3)

Yosvel Quintero
Yosvel Quintero

Reputation: 19070

I suppose that calling _myService.getData() you are making an Asynchronous call to a web service..

In that case you can return your service response data like this:

function returnData() {
  return _myService
    .getData()
    .then(function(data) {
      // Here you can work with the response data
      return data;
    });
}

Notice that you can work with the response in the callback, as you correctly passed into then.

Upvotes: 0

Rohan Kanjani
Rohan Kanjani

Reputation: 19

@Yosvel's answer is correct, the reason is because you are making an Asynchronous call to your service.

In doing so, you must wait for a result to be returned to you, which can then be processed in your then function.

By having two return calls, your return value is undefined because it is not waiting for the callback function, but simply returning a null value.

Upvotes: 0

T.J. Crowder
T.J. Crowder

Reputation: 1074495

Because result is declared within the callback you passed into then. It doesn't exist outside.

You can declare it outside, but note that it won't have the data as of your return result; line later:

function returnData(){
    var result;           // <==== Note

    _myService.getData().then(function(data){

      result = data;      // <==== No `var` here
    })

    // WON'T HAVE THE VALUE HERE, it's too soon
    return result;
}

See How do I return the response from an asynchronous call? for why, and what to do about it. Basically, you don't need the returnData function at all, it doesn't add any value (other than perhaps encapsulation around _myService). It cannot return the value. So either use _myService directly, or if you're trying to hide that, just

function returnData() {
    return _myService.getData();
}

...and use then in the calling code to receive it. Since again, returnData can't return the data.

Upvotes: 2

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