我得到了一个json响应并将其存储在mongodb中,但是我不需要的字段也进入了数据库,无论如何要剥离不道德的字段?
interface Test{
name:string
};
const temp :Test = JSON.parse('{ "name":"someName","age":20 }') as Test;
console.log(temp);
输出:
{ name: 'someName', age: 20 }
解决方法:
您可以使用从给定对象中选择某些属性的函数:
function pick<T, K extends keyof T>(obj: T, ...keys: K[]): Pick<T, K> {
const copy = {} as Pick<T, K>;
keys.forEach(key => copy[key] = obj[key]);
return copy;
}
然后:
let obj = { "name": "someName", "age": 20 };
let copy = pick(obj, "name") as Test;
console.log(copy); // { name: "someName" }