javascript - 更改屏幕后 react native 保存按钮状态

Question

我的应用程序中有 5 个按钮 [“运行”、“骑行”、“阅读”、“编码”、“牛儿”]，当我点击它时，按钮会改变颜色并在屏幕上显示标题。我正在使用这个库:react-native-selectmultiple-button .

假设我点击了“运行”和“骑行”按钮，这些按钮将突出显示并且文本将显示在屏幕上，但是当我将屏幕切换到另一页并返回到上一个屏幕时，按钮状态将设置回默认值.

下面是我的代码:

const multipleData = ["running", "riding", "reading", "coding", "Niuer"];

export default class SimpleButton extends Component {
  constructor(props) {
    super(props);
    this.state = {
      multipleSelectedDataLimited: []
    };
  }

  render() {
    return (
      <View style={{paddingTop:200}}>
      <Text style={styles.welcome}>
        implement the multiple-select buttons demo by SelectMultipleButton
      </Text>
      <Text style={{ color: 'blue', marginLeft: 10 }}>
      I like {_.join(this.state.multipleSelectedDataLimited, ", ")}
      </Text>
        <View
          style={{
            flexWrap: "wrap",
            flexDirection: "row",
            justifyContent: "center"
          }}
        >
          {multipleData.map(interest => (
            <SelectMultipleButton
              key={interest}
              buttonViewStyle={{
                borderRadius: 10,
                height: 40
              }}
              textStyle={{
                fontSize: 15
              }}
              highLightStyle={{
                borderColor: "gray",
                backgroundColor: "transparent",
                textColor: "gray",
                borderTintColor: 'blue',
                backgroundTintColor: 'blue',
                textTintColor: "white"
              }}
              value={interest}
              selected={this.state.multipleSelectedDataLimited.includes(
                interest
              )}
              singleTap={valueTap =>
                this._singleTapMultipleSelectedButtons_limited(interest)
              }
            />
          ))}
        </View>
      </View>
    );
  }

  _singleTapMultipleSelectedButtons_limited(interest) {
    if (this.state.multipleSelectedDataLimited.includes(interest)) {
      _.remove(this.state.multipleSelectedDataLimited, ele => {
        return ele === interest;
      });
    } else {
      if (this.state.multipleSelectedDataLimited.length < 3)
        this.state.multipleSelectedDataLimited.push(interest);
    }
    this.setState({
      multipleSelectedDataLimited: this.state.multipleSelectedDataLimited
    });
  }
}

const styles = StyleSheet.create({
  welcome: {
    margin: 10,
    marginTop: 30,
    color: "gray"
  }
});

有没有办法在更换屏幕后保持按钮的状态？

如有任何建议或意见，我们将不胜感激。提前致谢!

Answer 1

你能做的最好的事情就是将你的状态保存在 Redux 中并使用 redux-persist 。您还可以使用 AsyncStorage。我有类似的情况，我必须保持两个组件来回导航之间的状态，我使用了这样的导航参数:

屏幕 A:

this.props.navigation.navigate('ScreenB', {
              onPressScreenAFun: (params) => {
                this.screenAFun(params),
                ButtonState1: true // you can send multiple params
              },
            })

screenAFun = (ButtonState1) => {
 this.setState({buttonState1: ButtonState1})
}

屏幕 B:

    screenBFun = (params) => {
       const { onPressScreenAFun, ButtonState1 } = this.props.navigation.navigate.state.params

      onPressScreenAFun(ButtonState1)
      this.props.navigation.goBack()
    }