如何分配字符串| 在 TypeScript 中未定义为字符串？
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How to assign string | undefined to string in TypeScript?
提问by Sohrab
I want to assign a variable, which is string | undefined, to a string variable, as you see here:
我想分配一个变量，它是字符串 | 未定义，到字符串变量，如您在此处看到的：
private selectedSerialForReplace(): string | undefined {
return this.selectedSerials.pop();
}
luminaireReplaceLuminaire(params: { "serial": string; "newserial": string; }, options?: any): FetchArgs {
............
}
luminaireReplaceLuminaire({serial: this.selectedSerialForReplace(), newserial: response.output});
I get this error:
我收到此错误：
Argument of type '{ serial: string | undefined; newserial: any; }' is not assignable to parameter of type '{ "serial": string; "newserial": string; }'
'{ 序列号：字符串 | 类型的参数 不明确的; 新闻：任何；}' 不能分配给类型 '{ "serial": string; 的参数 “新闻”：字符串；}'
I cannot change selectedSerialForReplace() function to return anything else. Could you please help me?
我无法更改 selectedSerialForReplace() 函数以返回任何其他内容。请你帮助我好吗？
回答by Acevail
The typescript compiler performs strict null checks, which means you can't pass a string | undefinedvariable into a method that expects a string.
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.
To fix this you have to perform an explicit check for undefinedbefore calling luminaireReplaceLuminaire().
要解决此问题，您必须undefined在调用luminaireReplaceLuminaire().
In your example:
在你的例子中：
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});
}
回答by sschoof
If you are sure that serial could not be undefinedyou can use the !post-fix operator
如果您确定undefined不能使用序列号，则可以使用!post-fix 运算符
luminaireReplaceLuminaire({serial: this.selectedSerialForReplace()!, newserial: response.output});
