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

我真的很喜欢https://stackoverflow.com/a/60313570/770134中的建议,所以我把它改成了Typescript

import React, { FunctionComponent } from 'react'
import { Optional } from "typescript-optional";
const { ofNullable } = Optional

interface SwitchProps {
  test: string
  defaultComponent: JSX.Element
}

export const Switch: FunctionComponent<SwitchProps> = (props) => {
  return ofNullable(props.children)
    .map((children) => {
      return ofNullable((children as JSX.Element[]).find((child) => child.props['value'] === props.test))
        .orElse(props.defaultComponent)
    })
    .orElseThrow(() => new Error('Children are required for a switch component'))
}

const Foo = ({ value = "foo" }) => <div>foo</div>;
const Bar = ({ value = "bar" }) => <div>bar</div>;
const value = "foo";
const SwitchExample = <Switch test={value} defaultComponent={<div />}>
  <Foo />
  <Bar />
</Switch>;

其他回答

让它变得简单,只需使用许多if语句。

例如:

<Grid>
   {yourVar==="val1"&&(<> your code for val1 </>)}
   {yourVar==="val2"&&(<> your code for val2 </>)}
   .... other statments
</Grid>

我在render()方法中做了这个:

  render() {
    const project = () => {
      switch(this.projectName) {

        case "one":   return <ComponentA />;
        case "two":   return <ComponentB />;
        case "three": return <ComponentC />;
        case "four":  return <ComponentD />;

        default:      return <h1>No project match</h1>
      }
    }

    return (
      <div>{ project() }</div>
    )
  }

我试图保持render()返回干净,所以我把我的逻辑放在一个'const'函数上面。这样我也可以缩进我的开关盒整齐。

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。

与其他答案相比,我更倾向于在渲染函数中内联“开关”。它使什么组件可以在那个位置呈现更清楚。你可以使用一个简单的javascript对象来实现一个类似switch的表达式:

render () {
  return (
    <div>
      <div>
        {/* removed for brevity */}
      </div>
      {
        {
          'foo': <Foo />,
          'bar': <Bar />
        }[param]
      }
      <div>
        {/* removed for brevity */}
      </div>
    </div>
  )
}

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>