Question

我得到一个json响应并将其存储在mongodb中，但是我不需要的字段也会进入数据库，无论如何都要删除那些无用的字段？

interface Test{
    name:string
};
const temp :Test = JSON.parse('{ "name":"someName","age":20 }') as Test;
console.log(temp);

输出：

{ name: 'someName', age: 20 }

Answer 1

您可以使用从给定对象中选择某些属性的函数：

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然后：

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;
}

Answer 2

假设您要删除年龄

temp = {...temp, age: undefined}

这将从您的对象中删除age。

Answer 3

由于接口在运行时不存在，因此接口不会影响运行时行为。您必须使用delete

从对象中删除属性

let obj = JSON.parse('{ "name":"someName","age":20 }')
delete obj.age;

Answer 4

如果要以强类型方式执行此操作，则可以定义一个满足您的界面（const dummy: IMyInterface = {someField: "someVal"};）的虚拟/理想对象，然后根据Object.keys(dummy)过滤传入对象的字段。这样，如果您在不更新此“过滤”代码的情况下更新接口，则编译器将抱怨。

从typescript接口对象中删除字段

4 个答案: