CODEWITHSUNDEEP
angulartypescript
paan
paan

Reputation: 355

typescript - object is possibly 'null'

I get this error for the following but the program running perfectly

error

var video = document.querySelector('#camera-stream'),

if(!navigator.getMedia){
        displayErrorMessage("Your browser doesn't have support for the navigator.getUserMedia interface.");
    }
    else{
        // Request the camera.
        navigator.getMedia(
            {
                video: true
            },
            // Success Callback
            function(stream:any){

                // Create an object URL for the video stream and
                // set it as src of our HTLM video element.
                video.src = window.URL.createObjectURL(stream);

                // Play the video element to start the stream.
                video.play();
                video.onplay = function() {
                    showVideo();
                };

            },
            // Error Callback
            function(err:any){
                displayErrorMessage("There was an error with accessing the camera stream: " + err.name, err);
            }
        );

    }

example error

I tried the solution in this questions but didn't work for me.

What is the proper fix for this error?

Upvotes: 20

Views: 94408

Answers (4)

Aluan Haddad
Aluan Haddad

Reputation: 31873

TypeScript has a special syntax for handling this scenario, the non-null assertion operator.

When you know the value is actually neither null nor undefined but the compiler does not, you can use the non-null assertion operator, !, to communicate this. This works on an expression by expression basis.

declare let video: HTMLVideoElement | null | undefined;

video.src = window.URL.createObjectURL(stream); // error

video!.src = window.URL.createObjectURL(stream); // OK

video.autoplay = true; // error as the `!` does not percolate forward

video!.autoplay = true; // OK

However, it is far more likely that we do not definitively know that the object in question is neither null nor undefined and, after all, that possibility is what the type was deliberately written to convey. In such a case, using the ! syntax would suppress a compile time error but could result in a runtime failure. In this case we should rather handle the possibility by ensuring that the object is truthy before dereferencing it. A common idiom for writing this code is

if (video) {
  video.member
}

In fact, TypeScript uses a contextual type checking technique known as control flow based type analysis and thereby determines that video can safely be dereferenced in the if statement block because the null and undefined types have be removed from the union by truthy check. Therefore, the above code does not result in any errors because TypeScript knows that it is safe.

It is best to use the ! syntax very sparingly.

Upvotes: 37

wfreude
wfreude

Reputation: 508

Simon's answer can also be written using as (preferred by some linters like airbnb):

var video = document.querySelector('#camera-stream') as HTMLVideoElement;

Upvotes: 13

Ali Tourani
Ali Tourani

Reputation: 1247

Generally, if you want to disable the strict null checks function in TypeScript, you can use the character ! where the error is shown, as below:

this.myRef.current!.value = ''

Note: do this if you're sure about the object

Upvotes: 6

Simon
Simon

Reputation: 2643

Try casting:

var video = document.querySelector('#camera-stream')

to:

var video = <HTMLVideoElement>document.querySelector('#camera-stream')

Upvotes: 18

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