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

改进了一点 马特·韦的回答。

export const Switch = ({ test, children }) => { const defaultResult = children.find((child) => child.props.default) || null; const result = children.find((child) => child.props.value === test); return result || defaultResult; }; export const Case = ({ children }) => children; const color = getColorFromTheMostComplexFnEver(); <Switch test={color}> <Case value="Green">Forest</Case> <Case value="Red">Blood</Case> <Case default>Predator</Case> </Switch> <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>

其他回答

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>;

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。

你可以这样做。

 <div>
          { object.map((item, index) => this.getComponent(item, index)) }
 </div>

getComponent(item, index) {
    switch (item.type) {
      case '1':
        return <Comp1/>
      case '2':
        return <Comp2/>
      case '3':
        return <Comp3 />
    }
  }

下面是一个使用按钮在组件之间切换的完整工作示例

可以按如下方式设置构造函数

constructor(props)
{
    super(props);
    this.state={
        currentView: ''
    }
}

然后您就可以像下面这样渲染组件了

  render() 
{
    const switchView = () => {

    switch(this.state.currentView) 
    {

      case "settings":   return <h2>settings</h2>;
      case "dashboard":   return <h2>dashboard</h2>;

      default:      return <h2>dashboard</h2>
    }
  }

    return (

       <div>

            <button onClick={(e) => this.setState({currentView: "settings"})}>settings</button>
            <button onClick={(e) => this.setState({currentView: "dashboard"})}>dashboard</button>

            <div className="container">
                { switchView() }
            </div>


        </div>
    );
}

}

正如你所看到的,我正在使用一个按钮来切换状态。

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>
  );