类型'string | number | boolean‘不能赋值给类型'undefined’。类型'string‘不能赋值给类型’undefined‘。to (2322) 类型'string | null‘不能赋值给类型'SetStateAction<string>’的参数。类型'null‘不能赋值给类型’SetStateAction<string>‘ Typescript类型字符串不能赋值给类型keyof ...
The error message “Type ‘string’ is not assignable to type ‘string & …'” is indicating that the variable being assigned is of type ‘string’, but the type it is being assigned to is expecting a more complex type that includes a string but also has additional properties o...
问类型'string‘不能赋值给类型keyof [ Type ]ENpublic static void main(String[] args) { ...
function useKey<T, K extends keyof T>(o: T, k: K) { var name: string = k; // Type 'string | number | symbol' is not assignable to type 'string'.} 如果你确定只使用字符串类型的属性名，你可以这样写：function useKey<T, K extends Extract<keyof T, string>>(o: T, k: K)...
Type 'T[Extract<keyof T, string>]' is not assignable to type '(T & U)[Extract<keyof T, string>]'. Type 'T' is not assignable to type 'T & U'. Type 'object' is not assignable to type 'T & U'. Type 'object' is not assignable to type 'T'. 'object' is assignable to the...
// Type 'string' is not assignable to type 'number' let numList_2: data = [1, "", 3]; 1. 2. 3. 4. 5. 6. 7. 8. 对象的定义 // 方式1 interface Data_1 { name: string, age: number, } let data_1: Data_1 = {
Type 'undefined' is not assignable to type 'string'. 如何解决此错误? 这是我发现的唯一解决方案,用于检查是否未定义不生成警告的属性 type NotUndefined<T, K extends keyof T> = T & Record<K, Exclude<T[K], undefined>>; function checkIfKeyIsDefined<T, K extends keyof T>(item: T, key: ...
TypeScript也具备 Number、String、Boolean、Symbol 等类型(注意不是number、string等)。 在TS中,number和Number是两种数据类型,不可以等价。测试如下: let a: String = "S"; let b: string = "s"; a = b; // success b = a; // error TS2322: Type 'String' is not assignable to type 'string'...
interface Person { name: string age: number address: string } 在Person 类型上使用 keyof,将会得到一个新的类型,如下: type SomeNewType = keyof Person SomeNewType 是一个联合字面量类型("name" | "age" | address),它是由 Person 的属性组成的类型。 现在,你可以创建 SomeNewType 类型的对象了:...
function useKey<T, K extends keyof T>(o: T, k: K) { var name: string = k; // Type 'string | number | symbol' is not assignable to type 'string'. } 如果你确定只使用字符串类型的属性名,你可以这样写: function useKey<T, K extends Extract<keyof T, string>>(o: T, k: K) {...