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


function Notification({ text, status }) {
  return (
    <div>
      {(() => {
        switch (status) {
          case 'info':
            return <Info text={text} />;
          case 'warning':
            return <Warning text={text} />;
          case 'error':
            return <Error text={text} />;
          default:
            return null;
        }
      })()}
    </div>
  );
}

其他回答

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 />
    }
  }

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>;

这是另一种方法。

render() {
   return {this[`renderStep${this.state.step}`]()}

renderStep0() { return 'step 0' }
renderStep1() { return 'step 1' }

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})`) }