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

import React from 'react';

import ListView from './ListView';
import TableView from './TableView';

function DataView({
    currView,
    data,
    onSelect,
    onChangeStatus,
    viewTodo,
    editTodo,
    deleteTodo,
}) {
    return (
        <div>
            {(function () {
                switch (currView) {
                    case 'table':
                        return (
                            <TableView
                                todos={data}
                                onSelect={onSelect}
                                onChangeStatus={onChangeStatus}
                                viewTodo={viewTodo}
                                editTodo={editTodo}
                                deleteTodo={deleteTodo}
                            />
                        );

                    case 'list':
                        return (
                            <ListView
                                todos={data}
                                onSelect={onSelect}
                                onChangeStatus={onChangeStatus}
                                viewTodo={viewTodo}
                                editTodo={editTodo}
                                deleteTodo={deleteTodo}
                            />
                        );

                    default:
                        break;
                }
            })()}
        </div>
    );
}

export default DataView;

其他回答


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>
  );
}

试试这个,它也更干净:在一个函数中获得渲染的开关,并调用它传递你想要的参数。例如:

renderSwitch(param) {
  switch(param) {
    case 'foo':
      return 'bar';
    default:
      return 'foo';
  }
}

render() {
  return (
    <div>
      <div>
          // removed for brevity
      </div>
      {this.renderSwitch(param)}
      <div>
          // removed for brevity
      </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>

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。

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。