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。

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 />
    }
  }

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>;

这个答案专门用来解决@tonyfat提出的“重复”问题，关于如何使用条件表达式来处理相同的任务。

---

Avoiding statements here seems like more trouble than it's worth, but this script does the job as the snippet demonstrates:

// Runs tests let id = 0, flag = 0; renderByFlag(id, flag); // jobId out of range id = 1; // jobId in range while(++flag < 5){ // active flag ranges from 1 to 4 renderByFlag(id, flag); } // Defines a function that chooses what to render based on two provided values function renderByFlag(jobId, activeFlag){ jobId === 1 ? ( activeFlag === 1 ? render("A (flag = 1)") : activeFlag === 2 ? render("B (flag = 2)") : activeFlag === 3 ? render("C (flag = 3)") : pass(`flag ${activeFlag} out of range`) ) : pass(`jobId ${jobId} out of range`) } // Defines logging functions for demo purposes function render(val){ console.log(`Rendering ${val}`); } function pass(reason){ console.log(`Doing nothing (${reason})`) }

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。

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
  }
}