Typescript 如何通过布尔输入guard?
我有以下资料:Typescript 如何通过布尔输入guard?,typescript,typeguards,Typescript,Typeguards,我有以下资料: exampleFunction = (param: string) => { try { const result = processString(param) as { resultExample: string }; return { isSuccess: true, ...result }; catch (error) { return { isSuccess: false }; } } 理想情况下,返回的类型为:{is
exampleFunction = (param: string) => {
try {
const result = processString(param) as { resultExample: string };
return { isSuccess: true, ...result };
catch (error) {
return { isSuccess: false };
}
}
理想情况下,返回的类型为:{issucces:true,resultExample:string}{issucces:false}
,但它的类型为{issucces:boolean,resultExample:string}{issucces:boolean}
,这令人沮丧,因为这意味着我无法利用类型保护
const data = exampleFunction("exampleParam");
if (data.isSuccess) {
functionExpectingResultExample(data.resultExample); // Type Error because "resultExample: string | {}" is not assignable to "resultExample: string"
}
以上内容不应该出错,因为true和false应该被用作类型保护,而应该被当作布尔类型
手动将它们输入为true和false是我唯一的解决方案吗?您需要将
issucess
返回类型转换为const
,以便TypeScript知道该值不是布尔值,而是一个特定的、不变的布尔值:
const exampleFunction = (param: string) => {
try {
const result = processString(param) as { resultExample: string };
return { isSuccess: true as const, ...result };
} catch (error) {
return { isSuccess: false as const };
}
}
看看这本书
如果愿意,还可以将整个返回对象强制转换为常量:
const exampleFunction = (param: string) => {
try {
const result = processString(param) as { resultExample: string };
return { isSuccess: true , ...result } as const;
} catch (error) {
return { isSuccess: false } as const;
}
}
.Whoooooaaa。魔术我可以用这个。如果我有很多这样的范例。。。。我可以这样做吗:
const TRUE=TRUE as const
,然后在不同的函数中,执行isSuccess:TRUE,它会自动为我键入一个const?只是选中了,是的,我可以:)。