Question

我需要的是当我触摸并按住一个按钮时，我也应该能够触摸按钮1.

<View>
  
  <View 
  onStartShouldSetResponder={()=>this.console("Button 2 Clicked")}>
    <Text>BUTTON 2</Text>
  </View>
  
  <TouchableOpacity 
  onPressIn={()=>this.console('Button 1 pressed')}
  onPressOut={()=>this.console('Button 1 released')}>
    <View>
      <Text>BUTTON 1</Text>
    </View>
  </TouchableOpacity>

</View>

基本上，我有一个屏幕，我可以通过点击并按住录制按钮（按钮1）来录制视频。在同一个屏幕上，我有一个翻盖相机按钮（按钮2）。我希望在录制视频时能够点击翻盖相机按钮。

Answer 1

这是我的多个按钮解决方案

import React, { Component } from 'react';
import {
    View,
    PanResponder,
} from 'react-native';
import ReactNativeComponentTree from'react-native/Libraries/Renderer/shims/ReactNativeComponentTree';

export default class MultiTouch extends Component{
    constructor(props) {
        super(props);

        this.onTouchStart = this.onTouchStart.bind(this);
        this.onTouchEnd = this.onTouchEnd.bind(this);
        this.onTouchCancel = this.onTouchCancel.bind(this);

        this.triggerEvent = this.triggerEvent.bind(this);
    }
    onTouchStart(event){
        const element = ReactNativeComponentTree.getInstanceFromNode(event.target)._currentElement;
        this.triggerEvent(element._owner, 'onPressIn');
    }
    onTouchEnd(event){
        const element = ReactNativeComponentTree.getInstanceFromNode(event.target)._currentElement;
        this.triggerEvent(element._owner, 'onPressOut');
    }
    onTouchCancel(event){
        const element = ReactNativeComponentTree.getInstanceFromNode(event.target)._currentElement;
        this.triggerEvent(element._owner, 'onPressOut');
    }
    onTouchMove(event){
       // console.log(event);
    }
    triggerEvent(owner, event){ // Searching down the 
        if(!owner || !owner.hasOwnProperty('_instance')){
            return;
        }
        if(owner._instance.hasOwnProperty(event)){
            owner._instance[event]();
        }else{
            this.triggerEvent(owner._currentElement._owner, event);
        }
    }
    render(){
        return (
            <View
                onTouchStart={this.onTouchStart}
                onTouchEnd={this.onTouchEnd}
                onTouchCancel={this.onTouchCancel}
                onTouchMove={this.onTouchMove}>
                {this.props.children}
            </View>
        );
    }
}

然后我只需用组件

包裹需要同时按下的按钮

<MultiTouch style={this.style.view}>
    <UpDownButton />
    <UpDownButton />
</MultiTouch>

干杯！

<强>更新

由于本机react v.0.51中的更改中断，我以前的解决方案不再起作用。但我设法创造一个新的。我没有使用TouchableWithoutFeedback和onPress，而是在每个应该有多点触控的按钮上使用View和onTouch。

import React, { Component } from 'react';
import {
    View,
} from 'react-native';
export default class RoundButtonPart extends Component{
    constructor(props) {
        super(props);

        this.state = { active: false };

        this.onTouchStart = this.onTouchStart.bind(this);
        this.onTouchEnd = this.onTouchEnd.bind(this);
        this.onTouchCancel = this.onTouchCancel.bind(this);
    }

    onTouchStart(event){
        this.setState({ active: true });
        this.props.onPressIn && this.props.onPressIn();
    }
    onTouchEnd(event){
        this.setState({ active: false });
        this.props.onPressOut && this.props.onPressOut();
    }
    onTouchCancel(event){
        this.setState({ active: false });
        this.props.onPressOut && this.props.onPressOut();
    }
    onTouchMove(event){

    }
    render(){
        return (
            <View
                onTouchStart={this.onTouchStart}
                onTouchEnd={this.onTouchEnd}
                onTouchCancel={this.onTouchCancel}
                onTouchMove={this.onTouchMove}>

                 {this.props.children}
            </View>
        );
    }
}

Answer 2

使用onTouchStart，onTouchEnd视图组件的道具可以轻松解决此问题，而无需使用手势响应方法。

因此修改后的代码看起来像

<View>

  <View onTouchStart={()=>this.console("Button 2 Clicked")}>
    <Text>BUTTON 2</Text>
  </View>

  <View 
    onTouchStart={()=>this.console('Button 1 pressed')}
    onTouchEnd={()=>this.console('Button 1 released')}>
      <Text>BUTTON 1</Text>
  </View>

</View>

Answer 3

我使用了react-native-gesture-handler。安装并替换

import { TouchableOpacity } from 'react-native';

使用

import { TouchableOpacity } from 'react-native-gesture-handler';

示例

<View>

  <TouchableOpacity 
  onPressIn={()=>this.console('Button 2 pressed')}
  onPressOut={()=>this.console('Button 2 released')}>
    <View>
      <Text>BUTTON 2</Text>
    </View>
  </TouchableOpacity>

  <TouchableOpacity 
  onPressIn={()=>this.console('Button 1 pressed')}
  onPressOut={()=>this.console('Button 1 released')}>
    <View>
      <Text>BUTTON 1</Text>
    </View>
  </TouchableOpacity>

</View>

链接：https://software-mansion.github.io/react-native-gesture-handler/docs/component-touchables.html

该库还提供了可以直接使用的按钮组件，而不是使用TouchableOpacity包装文本

Answer 4

尝试：

import { TouchableOpacity } from 'react-native';

代替：

import { TouchableOpacity } from 'react-native-gesture-handler';

将帮助您使用多个按钮。

例如，如果您在TouchableOpacity中有一个TouchableWithoutFeedback，则当TouchableOpacity is touched时，它将仅调用TouchableOpacity的onPress，而不会调用TouchableWithoutFeedback的onPress。

如何在本机中同时启用触摸多个按钮？

4 个答案: