在 TypeScript 中,isPossiblyUndefined 是一个类型保护,用于判断一个变量是否可能是未定义的。这种类型保护对于防止在运行时遇到未定义变量引发的问题非常有用。 当使用 isPossiblyUndefined 时 在我们的代码中使用 isPossiblyUndefined,它的值为 true,表示该变量可能为未定义。这意味着在运行时,TypeScript 编译器会尝试...
解决办法就是加一个判断 因为这里find()能够拿到正确的数据,所以else语句不会执行,只要返回非undefined且符合我们定义的泛型类型的数据最后得到的结果就不会包含undefined了
解决方案:加类型断言 (this.toolsasArray<any>).forEach(item=>{console.log(item.id);})
children) { for (let i = 0; i < example.children.length; i++) { if (example.children[i] && example.children[i].children) { console.log(example.children[i].children.length) } } } if判断example.children[i].children非空后 下一句依然提示example.children[i].children is possibly undefine...
children[i] && example.children[i].children) { console.log(example.children[i].children.length) } } } if判断example.children[i].children非空后 下一句依然提示example.children[i].children is possibly undefined请问为何会出现这种情况 如何能在不使用强制非空断言的情况下解决这个问题?
1functionprintName(value:string){ 2console.log(value) 3} 4 5letfullName="Abhi"asstring|undefined 6 7printName(fullName||"default-value") Now if thefullNameis having afalsy value, then it will pass the 'default-value'. Using nullish coalescing operator (??) ...
2. 调用函数时忽略 undefined 类型 type NumGenerator = () => number; function myFunc(numGenerator: NumGenerator | undefined) { // Object is possibly 'undefined'.(2532) // Cannot invoke an object which is possibly 'undefined'.(2722)
1.2 调用函数时忽略 undefined 类型 typeNumGenerator=() =>number;functionmyFunc(numGenerator: NumGenerator |undefined) {// Object is possibly 'undefined'.(2532)// Cannot invoke an object which is possibly 'undefined'.(2722)constnum1 =numGenerator();// Errorconstnum2 = numGenerator!();//OK}...
2533 错误 Object is possibly 'null' or 'undefined'. 2534 错误 A function returning 'never' cannot have a reachable end point. 2535 错误 Enum type '{0}' has members with initializers that are not literals. 2536 错误 Type '{0}' cannot be used to index type '{1}'. 2537 错误 Type '...
Is this a regression? Yes Description type User = { role: "guest-user" | "user" | "none" }; public user = signal<User | undefined>(undefined); public user$ = toObservable(this.user).pipe(filter((user) => user !== undefined)); ...