对象解析（{x，y，... rest}）用于列出对象的属性 [英] Object destructuring ({ x, y, ...rest }) for whitelisting properties of an object

问题描述

使用对象休息解构非常简单黑名单对象的属性，如下例所示：

Using Object rest destructuring is straightforward to blacklist properties of an object, like in the following example:

const original = {
  a: 1,
  b: 2,
  c: 3,
  evil: "evil",
  ugly: "ugly",
};

const { evil, ugly, ...sanitized } = original;

console.log(sanitized);   // prints { a: 1, b: 2, c: 3 }

我想知道是否存在一个类似的简单的方法来做同样的事情，但是使用白名单的属性（例如： {a，b，c} ）。很多时候，我必须将可用属性的一部分转换为JSON，而这样的功能会使代码更易于阅读和更安全。

I wonder if there exists a similar terse way to do the same, but using a white list of properties (in the example: { a, b, c }). Very often, I have to convert a subset of the available properties as JSON, and such a functionality would make the code much more readable and safer.

我发现了一个类似的问题，但这并不完全相同：
在ES6 / ES2015中有一个更简单的方法将一个对象的属性映射到另一个对象？

I found a similar question, but it is not exactly the same problem: Is there a more terse way to map properties of one object to another in ES6/ES2015?

编辑：遗憾的是，下一个代码不起作用，因为它返回原始对象而不是过滤的对象。

It is a pity that the next code doesn't work, as it is returning the original object instead of the filtered one.

const sanitized = {a, b, c} = original;     
// sanitized === original

推荐答案

为此，我使用2帮助函数

For this purpose I use 2 helper functions

export const pickProps = (object, ...props) => (
  props.reduce((a, x) => {
    if (object.hasOwnProperty(x)) a[x] = object[x];
    return a;
  }, {})
);

export const omitProps = (object, ...props) => {
  const no = {...object};
  props.forEach(p => delete no[p]);
  return no;
};

您还可以执行

const original = {
  a: 1,
  b: 2,
  c: 3,
  evil: "evil",
  ugly: "ugly",
};

const { a, b, c } = original;
const filtered = { a, b, c };

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