37

通常，有了道具，我们可以写

componentDidUpdate(oldProps) {
  if (oldProps.foo !== this.props.foo) {
    console.log('foo prop changed')
  }
}

为了检测道具变化。

但是如果我们使用React.createRef()，我们如何检测 ref 何时更改为新的组件或 DOM 元素？React 文档并没有真正提及任何内容。

铁，

class Foo extends React.Component {
  someRef = React.createRef()

  componentDidUpdate(oldProps) {
    const refChanged = /* What do we put here? */

    if (refChanged) {
      console.log('new ref value:', this.someRef.current)
    }
  }

  render() {
    // ...
  }
}

我们是否应该自己实现某种旧值的东西？

铁，

class Foo extends React.Component {
  someRef = React.createRef()
  oldRef = {}

  componentDidMount() {
    this.oldRef.current = this.someRef.current
  }

  componentDidUpdate(oldProps) {
    const refChanged = this.oldRef.current !== this.someRef.current

    if (refChanged) {
      console.log('new ref value:', this.someRef.current)

      this.oldRef.current = this.someRef.current
    }
  }

  render() {
    // ...
  }
}

这是我们应该做的吗？我原以为 React 会为此添加一些简单的功能。

4

2 回答 2

63

React 文档推荐使用回调 refs来检测ref值的变化。

挂钩

export function Comp() {
  const onRefChange = useCallback(node => {
    if (node === null) { 
      // DOM node referenced by ref has been unmounted
    } else {
      // DOM node referenced by ref has changed and exists
    }
  }, []); // adjust deps

  return <h1 ref={onRefChange}>Hey</h1>;
}

useCallback用于防止ref 回调与元素的双重调用。null

您可以通过使用以下方式存储当前 DOM 节点来触发更改时的重新渲染useState：

const [domNode, setDomNode] = useState(null);
const onRefChange = useCallback(node => {
  setDomNode(node); // trigger re-render on changes
  // ...
}, []);

类组件

export class FooClass extends React.Component {
  state = { ref: null, ... };

  onRefChange = node => {
    // same as Hooks example, re-render on changes
    this.setState({ ref: node });
  };

  render() {
    return <h1 ref={this.onRefChange}>Hey</h1>;
  }
}

注意：useRef不通知ref更改。/ object refs也没有运气。React.createRef()

这是一个测试用例，它在触发onRefChange回调时删除并重新添加一个节点：

const Foo = () => {
  const [ref, setRef] = useState(null);
  const [removed, remove] = useState(false);

  useEffect(() => {
    setTimeout(() => remove(true), 3000); // drop after 3 sec
    setTimeout(() => remove(false), 5000); // ... and mount it again
  }, []);

  const onRefChange = useCallback(node => {
    console.log("ref changed to:", node);
    setRef(node); // or change other state to re-render
  }, []);

  return !removed && <h3 ref={onRefChange}>Hello, world</h3>;
}

ReactDOM.render(<Foo />, document.getElementById("root"));
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.10.1/umd/react.production.min.js" integrity="sha256-vMEjoeSlzpWvres5mDlxmSKxx6jAmDNY4zCt712YCI0=" crossorigin="anonymous"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.10.1/umd/react-dom.production.min.js" integrity="sha256-QQt6MpTdAD0DiPLhqhzVyPs1flIdstR4/R7x4GqCvZ4=" crossorigin="anonymous"></script>

<script> var {useState, useEffect, useCallback} = React</script>

<div id="root"></div>

于 2020-02-04T22:12:53.827 回答
3

componentDidUpdate当组件状态或道具改变时被调用，所以它不一定会在 aref改变时被调用，因为它可以在你认为合适的时候被改变。

但是，如果您想检查一个 ref 是否与之前的渲染相比发生了变化，您可以保留另一个 ref 来检查真实的 ref。

例子

class App extends React.Component {
  prevRef = null;
  ref = React.createRef();
  state = {
    isVisible: true
  };

  componentDidMount() {
    this.prevRef = this.ref.current;

    setTimeout(() => {
      this.setState({ isVisible: false });
    }, 1000);
  }

  componentDidUpdate() {
    if (this.prevRef !== this.ref.current) {
      console.log("ref changed!");
    }

    this.prevRef = this.ref.current;
  }

  render() {
    return this.state.isVisible ? <div ref={this.ref}>Foo</div> : null;
  }
}

ReactDOM.render(<App />, document.getElementById("root"));
<script src="https://unpkg.com/react@16/umd/react.development.js"></script>
<script src="https://unpkg.com/react-dom@16/umd/react-dom.development.js"></script>

<div id="root"></div>

于 2019-04-24T21:15:49.973 回答