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...
报错信息如下 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 assi...
private _name: string = ""; private _age: number = 0; // 构造函数 constructor(name:string); constructor(name: string, age:number); constructor(name: string, age?:number) { this._name = name; if (age) { this._age = age; } } } // 实例2: class Calculator { add(a: number, ...
问TypeScript w/ React类型'string‘的参数不能分配给'keyof T’类型的参数EN原因:因为当您请求object_...
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 '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: ...
{ [key: string]: any } 如果您希望N仅被约束为字符串,那么您可以将该需求与N的约束相交。 const foo = < T extends Foo, N extends keyof T & string >(dict: T, key: N) => { return bar(key); } foo({ a: 123 }, 'a') // fine foo({ a: 123 }, 'b') // error See Play...
TypeScript Version: Version 2.9.0-dev.20180426 Search Terms: keyof not assignable string Code //Just a minimal example, not what I am actually doing function isKey<E> (str : string) : str is keyof E { return true; } Expected behavior: Sh...
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) {...