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leviding merged 9 commits into from
Feb 20, 2020
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Update translation of 1-js/05-data-types/10-destructuring-assignment #613

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2 changes: 1 addition & 1 deletion 1-js/05-data-types/10-destructuring-assignment/1-destruct-user/solution.md
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Expand Up @@ -10,4 +10,4 @@ let {name, years: age, isAdmin = false} = user;
alert( name ); // John
alert( age ); // 30
alert( isAdmin ); // false
```
```
8 changes: 4 additions & 4 deletions 1-js/05-data-types/10-destructuring-assignment/1-destruct-user/task.md
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Expand Up @@ -4,7 +4,7 @@ importance: 5

# 解构赋值

有以下对象：
我们有一个对象：

```js
let user = {
Expand All @@ -16,10 +16,10 @@ let user = {
写一个解构赋值语句使得：

- `name` 属性赋值给变量 `name`。
- `years` 属性赋值给 `age`。
- `isAdmin` 属性赋值给变量 `isAdmin`（如果属性缺失则赋值为 false）。
- `years` 属性赋值给变量 `age`。
- `isAdmin` 属性赋值给变量 `isAdmin`（如果属性缺失则取默认值 false）。

赋值语句后的值必须是：
下面是赋值完成后的值的情况：

```js
let user = { name: "John", years: 30 };
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8 changes: 4 additions & 4 deletions 1-js/05-data-types/10-destructuring-assignment/6-max-salary/task.md
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Expand Up @@ -4,7 +4,7 @@ importance: 5

# 最高薪资

以下是一个 `salaries` 对象：
这儿有一个 `salaries` 对象：

```js
let salaries = {
Expand All @@ -14,9 +14,9 @@ let salaries = {
};
```

新建一个函数 `topSalary(salaries)`，返回拥有最高薪资的人。
新建一个函数 `topSalary(salaries)`，返回收入最高的人的姓名。

- 如果 `salaries` 是空的，函数应该返回 `null`。
- 如果有多个最高薪资的人，返回其中任意一个。
- 如果有多个收入最高的人，返回其中任意一个即可。

提示：使用 `Object.entries` 和解构语法来遍历键/值对。
P.S. 使用 `Object.entries` 和解构语法来遍历键/值对。
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