4.4 KiB
4.4 KiB
id | title | challengeType | isRequired | videoUrl | localeTitle |
---|---|---|---|---|---|
5a24c314108439a4d4036185 | Use && for a More Concise Conditional | 6 | false | 使用&&获得更简洁的条件 |
Description
else if
语句来返回略有不同的UI,你可能会重复代码,这会留下错误的余地。相反,您可以使用&&
logical运算符以更简洁的方式执行条件逻辑。这是可能的,因为您要检查条件是否为true
,如果是,则返回一些标记。下面是一个示例: {condition && <p>markup</p>}
如果condition
为true
,则返回标记。如果条件为false
,则在评估condition
后操作将立即返回false
并且不返回任何内容。您可以直接在JSX中包含这些语句,并在每个语句之后写入&&
多个条件串在一起。这允许您在render()
方法中处理更复杂的条件逻辑,而无需重复大量代码。 Instructions
h1
仅在display
为true
呈现,但使用&&
logical运算符而不是if/else
语句。 Tests
tests:
- text: <code>MyComponent</code>应该存在并呈现。
testString: 'assert((function() { const mockedComponent = Enzyme.mount(React.createElement(MyComponent)); return mockedComponent.find("MyComponent").length; })(), "<code>MyComponent</code> should exist and render.");'
- text: 当<code>display</code>设置为<code>true</code> ,应该渲染<code>div</code> , <code>button</code>和<code>h1</code> 。
testString: 'async () => { const waitForIt = (fn) => new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250)); const mockedComponent = Enzyme.mount(React.createElement(MyComponent)); const state_1 = () => { mockedComponent.setState({display: true}); return waitForIt(() => mockedComponent )}; const updated = await state_1(); assert(updated.find("div").length === 1 && updated.find("div").children().length === 2 && updated.find("button").length === 1 && updated.find("h1").length === 1, "When <code>display</code> is set to <code>true</code>, a <code>div</code>, <code>button</code>, and <code>h1</code> should render."); }; '
- text: 当<code>display</code>设置为<code>false</code> ,只应呈现<code>div</code>和<code>button</code> 。
testString: 'async () => { const waitForIt = (fn) => new Promise((resolve, reject) => setTimeout(() => resolve(fn()), 250)); const mockedComponent = Enzyme.mount(React.createElement(MyComponent)); const state_1 = () => { mockedComponent.setState({display: false}); return waitForIt(() => mockedComponent )}; const updated = await state_1(); assert(updated.find("div").length === 1 && updated.find("div").children().length === 1 && updated.find("button").length === 1 && updated.find("h1").length === 0, "When <code>display</code> is set to <code>false</code>, only a <code>div</code> and <code>button</code> should render."); }; '
- text: render方法应该使用&& logical运算符来检查this.state.display的条件。
testString: 'getUserInput => assert(getUserInput("index").includes("&&"), "The render method should use the && logical operator to check the condition of this.state.display.");'
Challenge Seed
class MyComponent extends React.Component {
constructor(props) {
super(props);
this.state = {
display: true
}
this.toggleDisplay = this.toggleDisplay.bind(this);
}
toggleDisplay() {
this.setState({
display: !this.state.display
});
}
render() {
// change code below this line
return (
<div>
<button onClick={this.toggleDisplay}>Toggle Display</button>
<h1>Displayed!</h1>
</div>
);
}
};
After Test
console.info('after the test');
Solution
// solution required