我创建了一个
onTouchListener
。不幸的是 onTouch() 方法throws
我警告:
com/calculator/activitys/Calculator$1#onTouch should call View#performClick when a click is detected
这是什么意思?我还没有找到有关此警告的任何信息。这是完整的代码:
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;
}
});
给你:
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;
}
检测到点击时,onTouch 应调用 View#performClick
您可以抑制此 Lint 警告
@SuppressLint("ClickableViewAccessibility")
您应该在
performClick()
内致电onTouchEvent()
。
@Override
public boolean onTouchEvent(MotionEvent event) {
//Logic
performClick();
return super.onTouchEvent(event);
}
或
findViewById(R.id.view1).setOnTouchListener(new View.OnTouchListener() {
@Override
public boolean onTouch(View v, MotionEvent event) {
v.performClick();
return v.onTouchEvent(event);
}
});
阅读更多这里
如果您没有使用明确覆盖
Custom View
的 onPerformClick
,则仅按照 Secko 的答案不会删除警告。
除了他的答案之外,为了在
android.widget.Button
或 Button
等类上执行相同的操作,您需要创建一个扩展目标视图的简单自定义视图。
示例:
自定义视图类:
public class UselessButton extends AppCompatButton {
public UselessButton(Context context) {
super(context);
}
public UselessButton(Context context, AttributeSet attrs) {
super(context, attrs);
}
public UselessButton(Context context, AttributeSet attrs, int defStyle) {
super(context, attrs, defStyle);
}
@Override
public boolean performClick() {
return super.performClick();
}
}
XML:
<stackoverflow.onEarth.UselessButton
android:id="@+id/left"
android:layout_width="60dp"
android:layout_height="60dp"
android:layout_marginStart="16dp"
android:layout_marginTop="16dp"
android:layout_marginEnd="8dp"
android:layout_marginBottom="8dp"
android:background="@drawable/left"
app:layout_constraintBottom_toBottomOf="parent"
app:layout_constraintEnd_toEndOf="parent"
app:layout_constraintHorizontal_bias="0.16"
app:layout_constraintStart_toStartOf="parent"
app:layout_constraintTop_toTopOf="parent"
app:layout_constraintBaseline_toBaselineOf="@+id/right"
app:layout_constraintVertical_bias="0.5" />
Java:
left.setOnTouchListener((v, event) -> {
if (event.getAction() == MotionEvent.ACTION_DOWN) {
enLeft = 1;
enRight = 0;
return true;
} else if (event.getAction() == MotionEvent.ACTION_UP) {
enLeft = 0;
v.performClick();
return false;
} else {
return false;
}
});
当前问题:IDE 解决了警告,但在真实的 Android 设备上看不到实际执行的单击操作。
编辑:修复了获取点击事件:使用
View.setPressed(boolean)
down.setOnTouchListener((v, event) -> {
if (event.getAction() == MotionEvent.ACTION_DOWN) {
enFront = 0;
enBack = 1;
left.setPressed(true);
return true;
} else if (event.getAction() == MotionEvent.ACTION_UP) {
enBack = 0;
v.performClick();
v.setPressed(false);
return false;
} else {
return false;
}
只需调用performClick方法,如下所示:
@Override
public boolean onTouch(View v, MotionEvent event) {
v.performClick();
Tools.hideKeyboard(getActivity(), getView());
getView().clearFocus();
return false;
}
我通过使用 Kotlin 扩展解决了这个警告
首先创建扩展(例如ViewExtensions.kt)
fun Button.onTouch(touch: (view: View, motionEvent: MotionEvent) -> Unit) {
setOnTouchListener { v, event ->
touch(v,event)
v.performClick()
true
}
}
其次,在您的 Fragment 或 Activity 中创建一个函数
private fun onTouchButton(v: View, event: MotionEvent) {
/* My Amazing implementation */
}
最后,使用扩展程序
myButton.onTouch { v, event ->
onTouchButton(v, event)
}
我对
MultiTouchListener
也有类似的问题,并通过实现 GestureDetector
并监听 SingleTap
来解决它(这不会删除警告,但在我看来会开始触发 onClick
事件)
class TouchListener(context: Context) : MultiTouchListener() {
private var tochedView: View? = null
private var mGestureDetector = CustomGestureDetector()
private var gestureDetector: GestureDetector = GestureDetector(context, mGestureDetector)
@SuppressLint("ClickableViewAccessibility")
override fun onTouch(view: View?, event: MotionEvent?): Boolean {
val aux = super.onTouch(view, event)
tochedView = view
gestureDetector.onTouchEvent(event)
return aux
}
private inner class CustomGestureDetector: GestureDetector.SimpleOnGestureListener() {
override fun onSingleTapUp(e: MotionEvent?): Boolean {
// this will be called even when a double tap is
tochedView?.performClick()
return super.onSingleTapUp(e)
}
override fun onSingleTapConfirmed(e: MotionEvent?): Boolean {
// this will only be called after the detector is confident that the
// user's first tap is not followed by a second tap leading to a double-tap gesture.
tochedView?.performClick()
return super.onSingleTapConfirmed(e)
}
}
}
这是一个超级晚的答案,但我希望它能对某人有所帮助。我不久前遇到了这个警告,我的方法是,创建一个扩展
View.class
的类,然后在该类中创建一个名为 closeKeyboard(View view){}
的公共方法,这就是它的样子:
public class MyTouchEvent extends View {
public MyTouchEvent(Context context) {
super(context);
}
public void closeKeyboard(View view) {
view.setOnTouchListener((view1, motionEvent) -> {
view1.performClick();
// this is a library handels the closing of keyboard
UIUtil.hideKeyboard((Activity) getContext());
return false;
});
}
@Override
public boolean performClick() {
super.performClick();
return true;
}
}
在
MainActivity
中,我调用了 closeKeyboard(view)
并使用父布局作为参数,如下所示:
new MyTouchEvent(this).collapseKeyboard(parentLayout);
用于关闭键盘的库
implementation 'net.yslibrary.keyboardvisibilityevent:keyboardvisibilityevent:3.0.0-RC2'
我不得不忽略它。我有一个 TouchListener 来引发动画按钮按下(在我的情况下改变颜色)。但我还有一个点击监听器来完成按钮的工作。所以这导致我的点击侦听器再次触发。我在触摸事件(Action_Down、Up等)切换中尝试将其作为默认情况