2018-10-10 22:03:03 +00:00
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---
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id: 5a24c314108439a4d4036176
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2020-12-16 07:37:30 +00:00
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title: 使用 State 切换元素
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2018-10-10 22:03:03 +00:00
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challengeType: 6
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2020-09-17 16:13:42 +00:00
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forumTopicId: 301421
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2018-10-10 22:03:03 +00:00
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---
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2020-12-16 07:37:30 +00:00
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# --description--
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2021-01-12 16:22:19 +00:00
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有时可能在更新状态的时候想知道上一个状态是什么。但是状态更新是异步的,这意味着 React 可能会把多个 `setState()` 集中在一起批量更新。所以设置 `this.state` 或者 `this.props` 后值后可能没有立即更新。所以最好不要写如下的代码:
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2020-09-17 16:13:42 +00:00
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2021-01-12 16:22:19 +00:00
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```jsx
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2020-09-17 16:13:42 +00:00
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this.setState({
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counter: this.state.counter + this.props.increment
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});
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```
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2020-12-16 07:37:30 +00:00
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正确的做法是,给 `setState` 传入一个函数,这个函数可以访问 state 和 props。给 `setState` 传入函数可以返回赋值后的 state 和 props。代码可以重写为这样:
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2020-09-17 16:13:42 +00:00
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2021-01-12 16:22:19 +00:00
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```jsx
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2020-09-17 16:13:42 +00:00
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this.setState((state, props) => ({
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counter: state.counter + props.increment
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}));
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```
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如果只需要 `state`,那么用下面的格式也是可以的:
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2021-01-12 16:22:19 +00:00
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```jsx
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2020-09-17 16:13:42 +00:00
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this.setState(state => ({
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counter: state.counter + 1
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}));
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```
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注意一定要把 object 放在括号里,否则 JavaScript 会认为这只是代码片段。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --instructions--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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`MyComponent`有一个初始值为`false`的`visibility`属性。如果`visibility`的值为 true,render 方法返回一个视图,如果为 false,返回另一个视图。
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目前,无法更新组件的`state`中的`visibility`属性,该值应在 true 和 false 之间来回切换。按钮上有一个单击处理程序,它触发一个名为`toggleVisibility()`的类方法。定义此方法,以便`visibility`的`state`在调用方法时切换到相反的值。如果`visibility`是`false`,则该方法将其设置为`true`,反之亦然。
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最后,单击按钮以查看基于其`state`的组件的条件渲染。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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**提示:** 不要忘记将`this`关键字绑定到构造函数中的方法上!
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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# --hints--
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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`MyComponent`应该返回一个`div`元素,其中包含一个`button`元素。
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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assert.strictEqual(
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Enzyme.mount(React.createElement(MyComponent)).find('div').find('button')
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.length,
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1
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);
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2018-10-10 22:03:03 +00:00
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```
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2020-12-16 07:37:30 +00:00
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`MyComponent`应该使用设置为`false`的`visibility`属性来初始化其 state。
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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```js
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assert.strictEqual(
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Enzyme.mount(React.createElement(MyComponent)).state('visibility'),
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false
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);
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```
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2018-10-10 22:03:03 +00:00
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2020-12-16 07:37:30 +00:00
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单击按钮元素应在`true`和`false`之间切换`visibility`属性的状态。
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2020-09-17 16:13:42 +00:00
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2018-10-10 22:03:03 +00:00
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```js
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2020-12-16 07:37:30 +00:00
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async () => {
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const waitForIt = (fn) =>
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new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250));
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const mockedComponent = Enzyme.mount(React.createElement(MyComponent));
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const first = () => {
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mockedComponent.setState({ visibility: false });
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return waitForIt(() => mockedComponent.state('visibility'));
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};
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const second = () => {
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mockedComponent.find('button').simulate('click');
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return waitForIt(() => mockedComponent.state('visibility'));
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};
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const third = () => {
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mockedComponent.find('button').simulate('click');
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return waitForIt(() => mockedComponent.state('visibility'));
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};
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const firstValue = await first();
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const secondValue = await second();
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const thirdValue = await third();
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assert(!firstValue && secondValue && !thirdValue);
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2020-09-17 16:13:42 +00:00
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};
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2018-10-10 22:03:03 +00:00
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```
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2020-08-13 15:24:35 +00:00
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2020-12-16 07:37:30 +00:00
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应该传入`setState` 一个匿名函数。
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```js
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const paramRegex = '[a-zA-Z$_]\\w*(,[a-zA-Z$_]\\w*)?';
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const noSpaces = code.replace(/\s/g, '');
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assert(
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new RegExp(
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'this\\.setState\\((function\\(' +
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paramRegex +
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'\\){|([a-zA-Z$_]\\w*|\\(' +
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paramRegex +
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'\\))=>)'
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).test(noSpaces)
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);
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```
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不要在 `setState` 里面使用 `this`。
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```js
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assert(!/this\.setState\([^}]*this/.test(code));
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```
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# --solutions--
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