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  • 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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  • 原文地址:https://www.cnblogs.com/jenson138/p/4263724.html
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