Android测试,如何使用乐器模拟多点触控(放大/缩小)？

Question

我可以简单地模拟单个触摸 - 在我的测试中点击,滑动,按住等,但完全坚持使用Android 2.2在HTСDesire上模拟多点触控.

请问您能告诉我,如何重现事件链以测试多点触摸？

我想我需要使用一些棘手的MotionEvent,如MASK或类似的东西,但不知道如何做到这一点.

我在这里找到了转载缩放事件的转储:http: //www.zdnet.com/blog/burnette/how-to-use-multi-touch-in-android-2-part-3-understanding-touch- ？事件/ 1775 =标签mantle_skin;内容

 1. event ACTION_DOWN[#0(pid 0)=135,179]
 2. event ACTION_MOVE[#0(pid 0)=135,184]
 3. event ACTION_MOVE[#0(pid 0)=144,205]
 4. event ACTION_MOVE[#0(pid 0)=152,227]
 5. event ACTION_POINTER_DOWN(pid 1)[#0(pid 0)=153,230;#1(pid 1)=380,538]
 6. event ACTION_MOVE[#0(pid 0)=153,231;#1(pid 1)=380,538]
 7. event ACTION_MOVE[#0(pid 0)=155,236;#1(pid 1)=364,512]
 8. event ACTION_MOVE[#0(pid 0)=157,240;#1(pid 1)=350,498]
 9. event ACTION_MOVE[#0(pid 0)=158,245;#1(pid 1)=343,494]
 10. event ACTION_POINTER_UP(pid 0)[#0(pid 0)=158,247;#1(pid 1)=336,484]
 11. event ACTION_MOVE[#0(pid 1)=334,481]
 12. event ACTION_MOVE[#0(pid 1)=328,472]
 13. event ACTION_UP[#0(pid 1)=327,471]

这是我的问题:

5. 事件ACTION_POINTER_DOWN(pid 1)[#0(pid 0)= 153,230;#1(pid 1)= 380,538]
6. 事件ACTION_MOVE [#0**(pid 0)= 153,231**;#1**(pid 1)= 380,538**]

如何生成具有4个坐标(pid 0 x0 y0和pid 1 x1 y1)的事件？

看起来我需要找到如何使用以下事件的方式:

public static MotionEvent获取(long downTime,long eventTime,int action,int pointer,int [] pointerIds,PointerCoords [] pointerCoords,int metaState,float xPrecision,float yPrecision,int deviceId,int edgeFlags,int source,int flags)

感谢Dan的回复,我尝试了这个逻辑,但仍然遇到添加坐标的问题:

MotionEvent event = MotionEvent.obtain(downTime, eventTime, MotionEvent.ACTION_DOWN, 135, 179, 0);
            inst.sendPointerSync(event);
//                eventTime+=100;
            event = MotionEvent.obtain(downTime, eventTime, MotionEvent.ACTION_MOVE, 135, 184, 0);
            inst.sendPointerSync(event);
//                eventTime+=100;

            int pointerToMove = 1; // pointer IDs are zero-based
            event = MotionEvent.obtain(downTime, eventTime, (pointerToMove << MotionEvent.ACTION_POINTER_INDEX_SHIFT) +  MotionEvent.ACTION_POINTER_DOWN, 138, 189, 0);
            inst.sendPointerSync(event);


            event = MotionEvent.obtain(downTime, eventTime, (pointerToMove << MotionEvent.ACTION_POINTER_INDEX_SHIFT) + MotionEvent.ACTION_MOVE, 158, 220, 0);
            inst.sendPointerSync(event);

 //                eventTime+=100;
            event = MotionEvent.obtain(downTime, eventTime, (2 * 256) + MotionEvent.ACTION_MOVE, 138, 180, 0);
            inst.sendPointerSync(event);
 //                eventTime+=100;
            event = MotionEvent.obtain(downTime, eventTime, MotionEvent.ACTION_MOVE,  135, 184, 0);

这些事件序列在我的测试存根中捕获并转储如下:

(14368): event ACTION_DOWN[#0(pid 0)=135,179]
(14368): event ACTION_MOVE[#0(pid 0)=135,184]
(14368): event ACTION_POINTER_DOWN(pid 1)[#0(pid 0)=138,189]
(14368): event ACTION_MOVE[#0(pid 0)=158,220]
(14368): event ACTION_MOVE[#0(pid 0)=138,180]
(14368): event ACTION_MOVE[#0(pid 0)=135,184]

在这里你可以看到,那个(2*256)+ MotionEvent.ACTION_MOVE不会改变事件的指针ID :(和pointerToMove << MotionEvent.ACTION_POINTER_INDEX_SHIFT方法不适用于ACTION_POINTER_DOWN,可能是我不允许使用这样的POINTER_DOWN的方式？

我的问题是我无法为指针0和指针1生成2对坐标:

(14368): event ACTION_POINTER_DOWN(pid 1)[#0(pid 0)=138,189]

在这里你可以看到,使用你的逻辑我已经为事件添加了pid1,但它仍然没有坐标,因为x和y与pid 0相关联.

先感谢您.

Yahor

仍然没有想法如何实现它,有人发过多点触控事件吗？

Answer 1

我相信您只需要在传递给 MotionEvent.obtain 的“action”参数中指示指针索引。具体来说，动作的高8位是指针索引，低8位是动作（例如MotionEvent.ACTION_MOVE）。因此，如果您想移动第二个指针，这应该可行：

int pointerToMove = 1; // pointer IDs are zero-based
event = MotionEvent.obtain(downTime, eventTime, (pointerToMove << MotionEvent.ACTION_POINTER_INDEX_SHIFT) + MotionEvent.ACTION_MOVE, x0, y0, 0);
inst.sendPointerSync(event);

-担