我有一个函数应该返回某种类型,但使用扩展运算符导致它分配一个拼写错误的键。
interface State {
fieldA: number,
fieldB: string
}
const reducer: (state: State, action: {payload: string}) => State = (state, action) => {
// please note that those variables are not desired
const tmp = { ...state, feildB: action.payload }; // No compile time error, :(
// This is too verbose... but works
const tmp2 = Object.assign<State, Partial<State>>(state, {feildB: action.payload}) // ERROR - this is what I need
return tmp
}
const t = reducer({fieldA: 1, fieldB: 'OK'}, {payload: 'Misspelled'}) // Misspelled
console.log("feildB", (t as any).feildB) // Misspelled
console.log("fieldB", (t as any).fieldB) // OK
有没有办法使其类型安全,同时将样板代码保持在最低限度?
最佳答案
TypeScript 正在做它应该做的事情。在您的例子中,您正在创建一个新对象 tmp
,其新类型具有 3 个字段,即:
interface State {
fieldA: number;
fieldB: string;
}
interface Tmp {
fieldA: string;
fieldB: string;
payload: string;
}
换句话说,传播运算符执行以下操作:
interface Obj {
[key: string]: any;
}
const spread = (...objects: Obj[]) => {
const merged: Obj = {};
objects.forEach(obj => {
Object.keys(obj).forEach(k => merged[k] = obj[k]);
});
return merged;
}
传播运算符正在为您创建一种新类型的对象;如果你想推断类型,那么你应该这样做:
// this now throws an error
const tmp: State = { ...state, feildB: action.payload };
关于typescript - 展开运算符类型不安全,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59129738/