如何分配字符串在TypeScript中未定义为字符串? [英] How to assign string | undefined to string in TypeScript?

问题描述

我想分配一个变量，它是字符串|未定义，则为字符串变量，如您在此处看到的:

I want to assign a variable, which is string | undefined, to a string variable, as you see here:

private selectedSerialForReplace(): string | undefined {
    return this.selectedSerials.pop();
  }

luminaireReplaceLuminaire(params: {  "serial": string; "newserial": string; }, options?: any): FetchArgs {
............
}

luminaireReplaceLuminaire({serial: this.selectedSerialForReplace(), newserial: response.output});

我收到此错误:

'{类型的参数:serial:字符串|不明确的;新闻:任何； }' 是 不可分配给类型'{"serial"的参数:字符串； 新闻": 细绳; }'

Argument of type '{ serial: string | undefined; newserial: any; }' is not assignable to parameter of type '{ "serial": string; "newserial": string; }'

我无法将selectedSerialForReplace()函数更改为返回其他任何内容.你能帮我吗?

I cannot change selectedSerialForReplace() function to return anything else. Could you please help me?

推荐答案

打字稿编译器执行严格的空检查，这意味着您不能将string | undefined变量传递给需要string的方法.

The typescript compiler performs strict null checks, which means you can't pass a string | undefined variable into a method that expects a string.

要解决此问题，您必须在调用luminaireReplaceLuminaire()之前对undefined进行显式检查.

To fix this you have to perform an explicit check for undefined before calling luminaireReplaceLuminaire().

在您的示例中:

private selectedSerialForReplace(): string | undefined {
    return this.selectedSerials.pop();
}

luminaireReplaceLuminaire(params: {  "serial": string; "newserial": string; }, options?: any): FetchArgs {
    ............
}

const serial = this.selectedSerialForReplace();
if(serial !== undefined) {
    luminaireReplaceLuminaire({serial, newserial: response.output});
}

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