Android 如何获取listview中的触摸位置?
我试图在触摸listview时获得该位置,以便在触摸特定位置时,它应该显示一个祝酒词 以下是我的listview触控监听器:-Android 如何获取listview中的触摸位置?,android,Android,我试图在触摸listview时获得该位置,以便在触摸特定位置时,它应该显示一个祝酒词 以下是我的listview触控监听器:- customListFilter.setOnTouchListener(new View.OnTouchListener() { @Override public boolean onTouch(View arg0, MotionEvent event) { int actionX = (int) event.g
customListFilter.setOnTouchListener(new View.OnTouchListener() {
@Override
public boolean onTouch(View arg0, MotionEvent event) {
int actionX = (int) event.getX();
int actionY = (int) event.getY();
int extraTapArea = 13;
int x = (int) (actionX + extraTapArea);
int y = (int) (actionY - extraTapArea);
x = getWidth() - x;
if(x <= 0){
x += extraTapArea;
}
if (y <= 0)
y = actionY;
if ((actionX,actionY).contains(x, y)) {
// Show Toast Here
}
return false;
}
});
customListFilter.setOnTouchListener(新视图.OnTouchListener(){
@凌驾
公共布尔onTouch(视图arg0,运动事件){
int actionX=(int)event.getX();
int actionY=(int)event.getY();
int-extratapea=13;
INTX=(int)(actionX+extraTapArea);
int y=(int)(actionY-外区);
x=getWidth()-x;
如果(x尝试以下方法:
Display display = getWindowManager().getDefaultDisplay();
Point size = new Point();
display.getSize(size);
int width = size.x;
customListFilter.setOnTouchListener(new View.OnTouchListener() {
@Override
public boolean onTouch(View arg0, MotionEvent event) {
int actionX = (int) event.getX();
int actionY = (int) event.getY();
int middlePoint = width/2 ;
if(actionX >middlePoint ){
// you are touching the right side
}else if (actionX <middlePoint ){
// you are touching the left side
}
return false;
}
});
Display Display=getWindowManager().getDefaultDisplay();
点大小=新点();
display.getSize(size);
int width=size.x;
customListFilter.setOnTouchListener(新视图.OnTouchListener(){
@凌驾
公共布尔onTouch(视图arg0,运动事件){
int actionX=(int)event.getX();
int actionY=(int)event.getY();
int中点=宽度/2;
如果(actionX>中点){
//你触到了右边
}else if(actionX如果你否决投票,请发表评论,并说明理由。回答很好。谢谢。告诉我,我的问题被否决有那么糟糕吗?不是我做的否决投票,所以我不知道:)欢迎你:)