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

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})`) }

其他回答

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。

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>;

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>

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。