android MotionEvent 获取长按压时间长
思路:
1.记录ACTION_DOWN的aX, aY坐标;
2.在ACTION_MOVE判断是否移动,移动则取消记录时间,没移动就记录;
3.记录时间,按下坐标,移动坐标分别显示在TextView aa, bb, cc;
//声明
1 public class MainActivity extends ActionBarActivity { 2 private static TextView aa, bb, cc; 3 private float atime; 4 private float aX, aY; 5 private boolean mPressBreak = false;
//指定
1 aa = (TextView) findViewById(R.id.textView1); 2 bb = (TextView) findViewById(R.id.textView2); 3 cc = (TextView) findViewById(R.id.textView3);
//获取按压时间长
1 @Override 2 public boolean onTouchEvent(MotionEvent event) { 3 // TODO Auto-generated method stub 4 super.onTouchEvent(event); 5 String str = ""; 6 switch (event.getAction()) { 7 case MotionEvent.ACTION_DOWN: 8 aX = event.getX(); 9 aY = event.getY(); 10 str = String.valueOf(aX) + " , " + String.valueOf(aY); 11 bb.setText(str); 12 mPressBreak = false; 13 break; 14 case MotionEvent.ACTION_MOVE: 15 16 atime = (event.getEventTime() - event.getDownTime()) / 1000; 17 str = String.valueOf(event.getX()) + " , " 18 + String.valueOf(event.getY()); 19 if ((Math.abs((event.getX() - aX)) > 10) 20 || (Math.abs(event.getY() - aY) > 10)) { 21 atime = 0; 22 mPressBreak = true; 23 } 24 break; 25 26 } 27 28 if (!mPressBreak) { 29 aa.setText(String.valueOf((int) (atime))); 30 cc.setText(str); 31 } 32 33 return true; 34 }
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