微信小程序实现单个卡片左滑显示按钮并防止上下滑动干扰功能

<View style={`${positionStyle[index]}`}>
  <View 
    onTouchStart={this.moveTaskStart}
    onTouchMove={this.moveTask}
    onTouchEnd={this.moveTaskEnd}
  >
    这是主要的卡片部分
  </View>

  <View style="right: -260px;">这是滑动后出现的按钮部分</View>
</View>

moveTaskStart(e) {
  if (e.touches.length == 1) {
    //触摸屏上只有一个触摸点
    this.setState({
      // 开始触摸屏幕的X坐标
      startX: e.touches[0].clientX,
      startY: e.touches[0].clientY
    });
  }
}

moveTask(e) {
  if (e.touches.length == 1) {
    let moveX = e.touches[0].clientX;
    let moveY = e.touches[0].clientY;
    // 移动距离
    let disX = this.state.startX - moveX;
    let disY = this.state.startY - moveY;
    let angle = disY > 0 ? disY / disX : -disY / disX;
    let btnWidth = this.state.btnWidth;
    let txtStyle = "";
    if (disX == 0 || disX < 30) {
      //右滑动
      txtStyle = "left:0px";
    } else if(angle > 0.5) {
      return;
    } else if (disX > 30 && angle < 0.5) {
      txtStyle = `left: -${disX}px`;
      if (disX >= btnWidth) {
        //距离最大值
        txtStyle = `left: -${btnWidth}px`;
      }
    }
    let index = e.currentTarget.dataset.index;
    let list = this.state.positionStyle;
    this.setState({
      positionStyle: list
    });
  }
}

moveTaskEnd(e) {
  let txtStyle;
  if (e.changedTouches.length == 1) {
    let endX = e.changedTouches[0].clientX;
    let moveY = e.changedTouches[0].clientY;
    //移动距离
    let disX = this.state.startX - endX;
    let disY = this.state.startY - moveY;
    //移动角度
    let angle = disY > 0 ? disY / disX : -disY / disX;
    let btnWidth = this.state.btnWidth;
    //如果距离大于按钮宽度的一半，并且移动角度较小，显示按钮
    if(disX > (btnWidth / 2) && angle < 0.5) {
      txtStyle = `left:-${btnWidth}px`
    }else {
      txtStyle = "left:0px"
    }
    let index = e.currentTarget.dataset.index;
    let list = this.state.positionStyle;
    list[index] = txtStyle;
    this.setState({
      positionStyle: list
    });
  }
}