1

假设我的动画被迭代 20 次，但我不希望所有动画同时播放。我将如何触发一个特定的动画，其值为1. 如果我点击了，1那么其他 19 不应该触发除此之外1。

    export default class AnimateScreen extends React.PureComponent {
    constructor(props){

      super(props);

      this.forceUpdateHandler = this.forceUpdateHandler.bind(this);

      this.state = {
      dataSource: '',
      progress: new Animated.Value(0),
    };


    animate = (id) => {

          Animated.timing(this.state.progress, {
            toValue: 1,
            duration: 5000,
            easing: Easing.linear,
          }).start([id]); <!-- Here is my attempt in trying to animate that one specific animation. 


render(){
return(
    <FlatList
               data={this.state.dataSource}
               renderItem={({item}) => 
              <View>
               <View>
                 <Text>Work in progress</Text>
                 <View>
                 <TouchableHighlight
                 onPress={this.animate.bind(this, item.id)}>
                  <Animation
                  progress={this.state.progress}
                  source={require('../tools/animations/heart_icon.json')}
                  />
        </TouchableHighlight>

                   <Text> Hello</Text>

                 </View>

               </View>


               </View>
             }
             keyExtractor={(item, index) => index.toString()}
          />
        );
        }
    }

我试过了，仍然触发了所有的动画。有没有办法专门触发它们？

数据源：

"dataSource":[{"id":"10","images":"Emerson live in the sunshine swim in the sea drink the wild air.jpg","note":"Hello","tag":"sunshine"}


    componentDidUpdate(prevProps, prevState) {
  if (!prevState.dataSource) {

      return fetch(`https://www.website.com/React/json-data.php` , {
       method: 'POST',
       headers: {
         'Accept': 'application/json',
         'Content-Type': 'application/json',
       }

      })
        .then((response) => response.json())
        .then((responseJson) => {
          this.setState({
            dataSource: responseJson,
            },function() {
              // In this block you can do something with new state.
            });
        })
        .catch((error) => {
          console.error(error);
        });
    }
  }
4

1 回答 1

1

基本上，您可以将进度键添加到数据源中的每个项目，并将动画值设置为 0，然后单击您将为该项目的进度设置动画。它应该大致如下所示：

export default class AnimateScreen extends React.PureComponent {
  constructor(props){

    super(props);

    this.forceUpdateHandler = this.forceUpdateHandler.bind(this);

    this.state = {
      dataSource: [],
      progress: new Animated.Value(0),
    };

    componentDidUpdate(prevProps, prevState) {
      if (!prevState.dataSource) {

        return fetch(`https://www.website.com/React/json-data.php` , {
          method: 'POST',
          headers: {
            'Accept': 'application/json',
            'Content-Type': 'application/json',
          }

        })
        .then((response) => response.json())
        .then((responseJson) => {
          this.setState({
            dataSource: responseJson.map(item => ({ ...item, progress: new Animated.Value(0) })), // Add progress key
          },function() {
            // In this block you can do something with new state.
          });
        })
        .catch((error) => {
          console.error(error);
        });
      }
    }


    animate = (item) => {
      Animated.timing(item.progress, {
        toValue: 1,
        duration: 5000,
        easing: Easing.linear,
      }).start(); <!-- Here is my attempt in trying to animate that one specific animation. 


      render(){
        return(
          <FlatList
            data={this.state.dataSource}
            renderItem={({item}) => 
            <View>
              <View>
                <Text>Work in progress</Text>
                <View>
                  <TouchableHighlight
                    onPress={() => this.animate(item)}>
                    <Animation
                      progress={item.progress}
                      source={require('../tools/animations/heart_icon.json')}
                    />
                  </TouchableHighlight>

                  <Text> Hello</Text>

                </View>

              </View>


            </View>
          }
          keyExtractor={(item, index) => index.toString()}
        />
      );
    }
  }
于 2018-05-17T22:27:16.117 回答