包含所需键和字符串索引的记录

Question

这就是我想要的

type Report = {
branches: number; // should be required
functions: number; // should be required
lines: number; // should be required
statements: number; // should be required
};
const report: Report = {
branches: 10,
functions: 19,
lines: 54,
statements: 67,
}

但是我不能这样做

const items = Object.keys(report).map(key => report[key])

typescript(版本4.3.5)显示report[key]上的以下错误:

元素隐式具有'any'类型，因为'string'类型的表达式不能用于索引'Report'类型。在类型"报告"上找不到带有"字符串"类型参数的索引签名。ts(7053)

所以我试试这个

export type Report = {
[key: string]: number;
branches: number;
functions: number;
lines: number;
statements: number;
};

但现在它允许像

这样的字符

const withFoo: Report = {
branches: 25.3,
imABug: 45, // should not be allowed!
functions: 20.1,
lines: 70.01, 
statements: 45,
},

Answer 1

那是因为你正在创建一个Object.keys()字符串数组，并试图映射到Report对象，我觉得这只是糟糕的实现。

为什么不试试这样来遍历属性呢:

type Report = {
branches: number; // should be required
functions: number; // should be required
lines: number; // should be required
statements: number; // should be required
};
const foo: Report = {
branches: 10,
functions: 19,
lines: 54,
statements: 67,
}

Object.entries(foo).map(([key, value]) => console.log(key, value));