我有一个React组件,在组件的渲染方法中,我有这样的东西:

render() {
    return (
        <div>
            <div>
                // removed for brevity
            </div>

           { switch(...) {} }

            <div>
                // removed for brevity
            </div>
        </div>
    );
}

Now the point is that I have two div elements, one at the top and one at the bottom, that are fixed. In the middle I want to have a switch statement, and according to a value in my state I want to render a different component. So basically, I want the two div elements to be fixed always, and just in the middle to render a different component each time. I'm using this to implement a multi-step payment procedure). Though, as is the code currently it doesn't work, as it gives me an error saying that switch is unexpected. Any ideas how to achieve what I want?


当前回答

我将接受的答案转换为箭头功能组件解决方案,看到James提供了类似的答案,可以得到未定义的错误。这就是解决方案:

  const renderSwitch = (param) => {
    switch (param) {
      case "foo":
        return "bar";
      default:
        return "foo";
    }
  };

  return (
    <div>
      <div></div>

      {renderSwitch(param)}

      <div></div>
    </div>
  );

其他回答

试试这个,它也更干净:在一个函数中获得渲染的开关,并调用它传递你想要的参数。例如:

renderSwitch(param) {
  switch(param) {
    case 'foo':
      return 'bar';
    default:
      return 'foo';
  }
}

render() {
  return (
    <div>
      <div>
          // removed for brevity
      </div>
      {this.renderSwitch(param)}
      <div>
          // removed for brevity
      </div>
    </div>
  );
}

一种在渲染块中使用条件操作符表示一种开关的方法:

{(someVar === 1 &&
    <SomeContent/>)
|| (someVar === 2 &&
    <SomeOtherContent />)
|| (this.props.someProp === "something" &&
    <YetSomeOtherContent />)
|| (this.props.someProp === "foo" && this.props.someOtherProp === "bar" &&
    <OtherContentAgain />)
||
    <SomeDefaultContent />
}

应该确保条件严格返回布尔值。

我不太喜欢当前的任何答案,因为它们要么太啰嗦,要么需要您在代码中跳跃才能理解发生了什么。

我更喜欢用一个更以react组件为中心的方式来做这件事,通过创建一个<Switch/>。这个组件的任务是获取一个道具,并且只呈现子道具与该道具匹配的子元素。所以在下面的例子中,我在开关上创建了一个测试道具,并将其与子节点上的值道具进行比较,只渲染匹配的值道具。

例子:

const Switch = props => { const { test, children } = props // filter out only children with a matching prop return children.find(child => { return child.props.value === test }) } const Sample = props => { const someTest = true return ( <Switch test={someTest}> <div value={false}>Will display if someTest is false</div> <div value={true}>Will display if someTest is true</div> </Switch> ) } ReactDOM.render( <Sample/>, document.getElementById("react") ); <script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.6.3/umd/react.production.min.js"></script> <script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.6.3/umd/react-dom.production.min.js"></script> <div id="react"></div>

您可以根据自己的需要进行简单或复杂的切换。不要忘记对子节点及其值道具执行更健壮的检查。

Lenkan的回答是一个很好的解决方案。

<div>
  {{ beep: <div>Beep</div>,
     boop: <div>Boop</div>
  }[greeting]}
</div>

如果需要一个默认值,那么您甚至可以这样做

<div>
  {{ beep: <div>Beep</div>,
     boop: <div>Boop</div>
  }[greeting] || <div>Hello world</div>}
</div>

或者,如果这对你来说不太好,那么你可以做一些

<div>
  { 
    rswitch(greeting, {
      beep: <div>Beep</div>,
      boop: <div>Boop</div>,
      default: <div>Hello world</div>
    }) 
  }
</div>

with

function rswitch (param, cases) {
  if (cases[param]) {
    return cases[param]
  } else {
    return cases.default
  }
}

React组件中的Switch-Case语句可以如下使用:

<div  id="time-list">
{   
    (() => {
        switch (groupByFilterId) {
            case 0:/*Case 0 */
                return (
                    <div>Case 0</div>
                )
               break;
           case 1: /*Case 1 */
           return ( 
            <div>Case 1</div>
            )
            break;
           case 2:/*Case 2 */
           return ( 
            <div>Case 2</div>
            )
            break;
        }
     })()}

      
       
    
    </div>