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

我不太喜欢当前的任何答案,因为它们要么太啰嗦,要么需要您在代码中跳跃才能理解发生了什么。

我更喜欢用一个更以react组件为中心的方式来做这件事,通过创建一个<Switch/>。这个组件的任务是获取一个道具,并且只呈现子道具与该道具匹配的子元素。所以在下面的例子中,我在开关上创建了一个测试道具,并将其与子节点上的值道具进行比较,只渲染匹配的值道具。

例子:

const Switch = props => { const { test, children } = props // filter out only children with a matching prop return children.find(child => { return child.props.value === test }) } const Sample = props => { const someTest = true return ( <Switch test={someTest}> <div value={false}>Will display if someTest is false</div> <div value={true}>Will display if someTest is true</div> </Switch> ) } 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>

您可以根据自己的需要进行简单或复杂的切换。不要忘记对子节点及其值道具执行更健壮的检查。

其他回答

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>
  );

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>

这是因为switch语句是一个语句,但这里javascript需要一个表达式。

虽然,不建议在呈现方法中使用switch语句,但你可以使用自调用函数来实现这一点:

render() {
    // Don't forget to return a value in a switch statement
    return (
        <div>
            {(() => {
                switch(...) {}
            })()}
        </div>
    );
}

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。