ES6语法参考:使用扩展和布尔短路在声明期间有条件地向对象添加字段 [英] ES6 syntax reference: use spread and boolean short circuiting to conditionally add fields to an object during declaration

问题描述

我想构造一个这样的对象:

I want to construct an object like this:

const obj = {
    a: 'a',  // only add this if "someCondition" is true
    b: 'b',  // only add this if "someCondition" is false
    always: 'present', // add this in any case
}

这有效:

const obj = { always: 'present' }
if (someCondition) { obj.a = 'a' }
if (!someCondition) { obj.b = 'b' }

但是，我正在寻找一种使用ES6语法的更简洁的方法.

However, I'm looking for a more concise way using ES6 syntax.

推荐答案

可以使用ES6语法在对象声明期间有条件地添加字段.

It is possible using ES6 syntax to conditionally add fields during declaration of an object.

如果对象的使用者不能容忍具有null/undefined/任何值的字段，并且您不想编写多个语句来正确声明该对象，这将很有用:

This is useful if the consumer of the object will not tolerate fields with null / undefined / whatever values, and you do not want to have to write multiple statements to correctly declare the object:

const obj = {
  ...(someCondition && {a: 'a'}),
  ...(!someCondition && {b: 'b'}),
  always: 'present'
}

那如何运作?让我们看一下...(true && {a: 'a'}). ES6 扩展运算符"..." 将迭代{"a":"a"}中的每个field-> value对，并将它们应用于x.

So how does that work ? Lets look at ...(true && {a: 'a'}). The ES6 spread operator "..." will iterate each of the field->value pairs in { "a": "a" } applying them to x.

true && x表达式将返回x，而false && x表达式将返回false.这称为短路评估

The true && x expression will return x, whereas false && x will return false. This is known as short circuit evaluation

因此，如果逻辑表达式为true，则散布运算符将添加字段，如果逻辑表达式为true，则将不添加任何内容.

So if the logical expression is true then the spread operator will add the fields, and if it is not true it will not add anything.

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