React-Native setState不是函数

Question

我知道之前也曾问过类似的问题。但是这次的例子有点不同，所以请忍受我。

我正在尝试在访问键盘侦听器时在react-native组件内部设置状态-

componentDidMount() {
    this.keyboardDidShowListener = Keyboard.addListener(
      'keyboardDidShow',
      this._keyboardDidShow,
    );
    this._keyboardDidShow.bind(this);
    this.keyboardDidHideListener = Keyboard.addListener(
      'keyboardDidHide',
      this._keyboardDidHide,
    );
    this._keyboardDidHide.bind(this);
  }

  componentWillUnmount() {
    this.keyboardDidShowListener.remove();
    this.keyboardDidHideListener.remove();
  }

  _keyboardDidShow() {
    alert('Keyboard Shown');
    this.setState({keyboardOpen: true});
  }

  _keyboardDidHide() {
    alert('Keyboard Hidden');
    this.setState({keyboardOpen: false});
  }

我遇到了错误-

我在这里做什么错了？

Answer 1

您需要授予这些功能访问this的权限。

您有2种方法：

1）使用箭头功能

_keyboardDidShow = () => {
    alert('Keyboard Shown');
    this.setState({keyboardOpen: true});
  }

  _keyboardDidHide = () => {
    alert('Keyboard Hidden');
    this.setState({keyboardOpen: false});
  }

2）在构造函数中使用bind绑定它们：

constructor(props){
    super(props)
    this._keyboardDidShow = this._keyboardDidShow.bind(this)  
    this._keyboardDidHide = this._keyboardDidHide.bind(this)  
}

您没有将它们正确绑定到componentDidMount

中