如何在React中将事件处理程序传递给子组件 [英] How to pass an event handler to a child component in React

问题描述

我在React中创建了一个< Button /> 组件，它抽象出了我的应用程序中的一些样式。我在两个不同的上下文中使用它 - 一个用于提交登录表单，另一个用于导航到注册页面（以及将来可能的其他上下文）。

I have a <Button /> component I've created in React that abstracts out some of the styling in my application. I am using it in two different contexts - one to submit a login form, and the other to navigate to the registration page (and probably other contexts in the future).

我试图弄清楚如何将事件处理程序从父组件传递到< Button /> 。我想为登录表单调用 onSubmit 处理程序，但是导航按钮的 onClick 处理程序。这可能吗？

I am trying to figure out how to pass the event handlers from the parent component to the <Button />. I want to call an onSubmit handler for the login form, but an onClick handler for the navigation button. Is this possible?

我试过这样调用这个组件：

I have tried calling the component like this:

<Button text={callToAction} style={styles.callToActionButton} onClick={() => FlowRouter.go("Auth")}/>

<Button text="Go!" style={styles.registerButton} onSubmit={() => this.register(this.state.user, this.state.password)}/>

我也尝试删除箭头功能，这只会导致函数在组件运行时执行已加载：

I've also tried removing the arrow function, which just causes the functions to execute when the component is loaded:

// executes event handlers on page load
<Button text={callToAction} style={styles.callToActionButton} onClick={FlowRouter.go("Auth")}/>

<Button text="Go!" style={styles.registerButton} onSubmit={this.register(this.state.user, this.state.password)}/>

推荐答案

通常，您可以将onClick处理程序转发给您按钮类通过将其作为属性传递。你可以通过简单地定义你的按钮组件的propTypes来制作一个必需的道具。

In general, you can forward the onClick handler to your button class by passing it as a property. You could this make a required prop by simply defining the propTypes for your button component.

作为一个例子，我添加了一个小片段，展示了它是如何工作的

As an example, I added a small snippet that shows how it works

var StyledButton = React.createClass({
  propTypes: {
    // the StyledButton requires a clickHandler
    clickHandler: React.PropTypes.func.Required,
    // and I guess the text can be seen as required as well
    text: React.PropTypes.string.required
  },
  render: function() {
    // as you are sure you have a clickHandler, you can just reference it directly from the props
    return <button type="button" onClick={this.props.clickHandler}>{this.props.text}</button>;
  }
});

var MyForm = React.createClass({
  getInitialState() {
    return {
      clicked: 0
    };
  },
  click() {
    this.setState({clicked: this.state.clicked+1});
  	alert('ouch');
  },
  secondClickHandler() {
    this.setState({clicked: 0});
    alert(':(');
  },
  render() {
    // your parent component simply sets which button  
    return <fieldset>
        <div>
    	  <StyledButton clickHandler={this.click} text="Click me" /> 
          { (this.state.clicked > 0) && <StyledButton clickHandler={this.secondClickHandler} text="Not again" /> }
        </div>
    </fieldset>;
  }
});

ReactDOM.render(
  <MyForm />,
  document.getElementById('container')
);

<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.0.2/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.0.2/react-dom.min.js"></script>
<div id="container">
    <!-- This element's contents will be replaced with your component. -->
</div>

另外，你不会通常使用按钮的提交方法，您宁愿将收到的数据发送到Web服务，并在收到结果时处理任何更改。提交会杀死当前网站并需要重新加载所有内容，而使用ajax调用或商店，它可以等待结果，然后根据响应重定向用户

Also, you wouldn't in general use the submit method of a button, you would rather send the data received to a webservice, and handle any changes when the result is received. The submit kills the current website and needs to load everything anew, while with an ajax call, or a store, it can just wait for the result and then redirect the user based on the response

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