4

我有一个如下的reduce函数:

let el = scopes.reduce ((tot, {actions}) => tot + actions.length, 0);

我试图这样改造它,但似乎它不是正确的方法:

let el = scopes.reduce ((tot, {actions.length: len}) => tot + len, 0);

有没有办法做到这一点,或者这是不可能的。

4

1 回答 1

6

你很接近,但你使用嵌套而不是点表示法:

// Outer −−−−−−−−−−−−−−−−−−−−v−−−−−−−−−−−−−−−−−−−−−−v
let el = scopes.reduce((tot, {actions: {length: len}}) => tot + len, 0);
// Inner −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^−−−−−−−−−−−−^

现场示例:

const scopes = [
    {actions: [1, 2, 3]},
    {actions: []},
    {actions: [4, 5]}
];
let el = scopes.reduce((tot, {actions: {length: len}}) => tot + len, 0);
console.log(el); // 5

解构要记住的关键是语法与对象和数组字面量相同,包括嵌套;只是信息流是另一个方向。例如,在对象字面量中,数据从右向左流动,从源 ( source) 到目标 ( prop):

let source = 42;
let obj = {prop: source};
//           <−−−−−*

在解构中,数据从左到右流动,从源 ( prop) 到目标 ( target):

let {prop: target};
//     *−−−−−&gt;
console.log(target); // 42

目标可以是变量、对象属性,甚至是另一种解构模式。这就是我们上面使用的:actions属性的目标是解构模式{length: len},它将值length放入变量len中。这是我的新书中的图 7-1(请参阅我的个人资料以获取链接):

图 7-1 来自 JavaScript:新玩具


您还可以使用速记符号并length在回调中使用:

let el = scopes.reduce((tot, {actions: {length}}) => tot + length, 0);

现场示例:

const scopes = [
    {actions: [1, 2, 3]},
    {actions: []},
    {actions: [4, 5]}
];
let el = scopes.reduce((tot, {actions: {length}}) => tot + length, 0);
console.log(el); // 5

于 2020-01-23T09:31:44.580 回答