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

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>

其他回答

你不能在渲染中有开关。放置访问一个元素的对象文字的伪切换方法并不理想,因为它会导致所有视图都要处理,并且可能导致在该状态下不存在的道具的依赖错误。

这里有一个很好的干净的方法来做到这一点,不需要每个视图提前渲染:

render () {
  const viewState = this.getViewState();

  return (
    <div>
      {viewState === ViewState.NO_RESULTS && this.renderNoResults()}
      {viewState === ViewState.LIST_RESULTS && this.renderResults()}
      {viewState === ViewState.SUCCESS_DONE && this.renderCompleted()}
    </div>
  )

如果视图状态的条件不只基于一个简单的属性——比如每行有多个条件,那么枚举和getViewState函数封装条件是分离条件逻辑和清理呈现的好方法。

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
  }
}

我正在使用这个帮助器,它允许我在JSX中有开关语句

// in helpers folder const switchTrue = (object) => { const { default: defaultValue, ...rest } = object; const obj = { default: defaultValue, ...rest }; const result = Object.keys(obj).reduce((acc, cur) => { return { ...acc, [cur === 'default' ? 'true' : cur]: obj[cur], }; }, {}); return result['true']; }; const Sample = () => { const isDataLoading = false; return ( <div> { switchTrue({ [`${isDataLoading}`]: <div>Loading</div>, [`${!isDataLoading}`]: <div>Data Ready</div>, default: <div>Default</div>, }) } </div> ) } ReactDOM.render( <Sample/>, document.getElementById("react") ); <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> <div id="react"></div>

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})`) }

这个助手应该可以做到这一点。 使用示例:

{componentSwitch(3, (switcher => switcher
    .case(1, () =>
        <p>It is one</p>
    )
    .case(2, () =>
        <p>It is two</p>
    )
    .default(() =>
        <p>It is something different</p>
    )
))}

助手:

interface SwitchCases<T> {
    case: (value: T, result: () => React.ReactNode) => SwitchCases<T>;
    default: (result: () => React.ReactNode) => SwitchCases<T>;
}

export function componentSwitch<T>(value: T, cases: (cases: SwitchCases<T>) => void) {

    var possibleCases: { value: T, result: () => React.ReactNode }[] = [];
    var defaultResult: (() => React.ReactNode) | null = null;

    var getSwitchCases: () => SwitchCases<T> = () => ({
        case: (value: T, result: () => React.ReactNode) => {
            possibleCases.push({ value: value, result });

            return getSwitchCases();
        },
        default: (result: () => React.ReactNode) => {
            defaultResult = result;

            return getSwitchCases();
        },
    })
    
    // getSwitchCases is recursive and will add all possible cases to the possibleCases array and sets defaultResult.
    cases(getSwitchCases());

    // Check if one of the cases is met
    for(const possibleCase of possibleCases) {
        if (possibleCase.value === value) {
            return possibleCase.result();
        }
    }

    // Check if the default case is defined
    if (defaultResult) {
        // Typescript wrongly assumes that defaultResult is always null.
        var fixedDefaultResult = defaultResult as (() => React.ReactNode);

        return fixedDefaultResult();
    }

    // None of the cases were met and default was not defined.
    return undefined;
}