typescript 打字稿 - 对象可能是“空”
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typescript - object is possibly 'null'
提问by paan
I get this error for the following but the program running perfectly
我收到以下错误，但程序运行完美
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);
}
);
}
I tried the solution in this questionsbut didn't work for me.
What is the proper fix for this error?
此错误的正确修复方法是什么？
采纳答案by Simon
Try casting:
尝试铸造：
var video = document.querySelector('#camera-stream')
to:
到：
var video = <HTMLVideoElement>document.querySelector('#camera-stream')
回答by Aluan Haddad
TypeScript has a special syntax for handling this scenario.
TypeScript 有一种特殊的语法来处理这种情况。
When you knowthe value is actually neither nullnor undefinedbut the compiler does not, you can use the non-null assertion operator !to communicate this. This works on an expression by expression basis.
当您知道该值实际上既不是null也不是undefined但编译器没有时，您可以使用非空断言运算符!来传达这一点。这适用于一个表达式一个表达式。
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 nullnor undefinedand, 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
然而，更有可能的是，我们不能明确地知道所讨论的对象既不是也不null是undefined，毕竟，这种可能性是故意编写的类型来传达的。在这种情况下，使用该!语法将抑制编译时错误，但可能导致运行时失败。在这种情况下，我们应该通过在取消引用之前确保对象为真来处理这种可能性。编写此代码的常用习惯用法是
if (video) {
video.member
}
In fact, TypeScript uses a contextual type checking technique known as control flow analysis and thereby determines that videocan safely be dereferenced in the ifstatement block. Therefore, the above code does not result in any errors because TypeScript knows that it is safe.
事实上，TypeScript 使用一种称为控制流分析的上下文类型检查技术，从而确定video可以安全地在if语句块中取消引用。因此，上面的代码不会导致任何错误，因为 TypeScript 知道它是安全的。
It is best to use the !syntax very sparingly.
最好!非常谨慎地使用该语法。
回答by Alex Trn
Generally, if you want to disable the strict null checks functionin TypeScript, you can use the character !where the error is shown, as below:
一般情况下，如果要禁用TypeScript中的严格空检查功能，可以使用!显示错误的字符，如下所示：
this.myRef.current!.value = ''
Note: do this if you're sure about the object
注意：如果您确定对象，请执行此操作
回答by wfreude
Simon's answer can also be written using as(preferred by some linters like airbnb):
西蒙的答案也可以使用as（首选一些像airbnb这样的短绒）：
var video = document.querySelector('#camera-stream') as HTMLVideoElement;
