我创建了一个 onTouchListener 。不幸的是onTouch（）方法抛出我一个警告：

  COM /计算器/ activitys /计算器$ 1号onTouch应该调用查看＃performClick点击时检测
 

这是什么意思？我还没有发现有关此警告的任何信息。这里是全code：

 的LinearLayout llCalculatorContent =（的LinearLayout）fragmentView.findViewById（R.id.calculator_content）;

llCalculatorContent.setOnTouchListener（新View.OnTouchListener（）{

    @覆盖
    公共布尔onTouch（视图V，MotionEvent事件）{
        Tools.hideKeyboard（getActivity（），getView（））;
        。getView（）clearFocus（）;
        返回false;
    }
}）;
 

解决方案

在这里你去：

 公共布尔onTouch（视图V，MotionEvent事件）{
    开关（event.getAction（））{
    案例MotionEvent.ACTION_DOWN：
        //一些code ....
        打破;
    案例MotionEvent.ACTION_UP：
        v.performClick（）;
        打破;
    默认：
        打破;
    }
    返回true;
}
 

I have created a onTouchListener. Unfortunately onTouch() method throws me a warning:

com/calculator/activitys/Calculator$1#onTouch should call View#performClick when a click is detected

What does it means? I have not found any information about this warn. Here is the full code:

LinearLayout llCalculatorContent = (LinearLayout) fragmentView.findViewById(R.id.calculator_content);

llCalculatorContent.setOnTouchListener(new View.OnTouchListener() {

    @Override
    public boolean onTouch(View v, MotionEvent event) {
        Tools.hideKeyboard(getActivity(), getView());
        getView().clearFocus();
        return false;
    }   
});

Here you go:

public boolean onTouch(View v, MotionEvent event) {
    switch (event.getAction()) {
    case MotionEvent.ACTION_DOWN:
        //some code....
        break;
    case MotionEvent.ACTION_UP:
        v.performClick();
        break;
    default:
        break;
    }
    return true;
}