使用touchesBegan和touchesEnded的IOS多点触控序列

Question

这是用于音乐ap的

用户任务是用手指1、2和3触摸和按住屏幕，两次触摸之间有一段短暂的时间。

如果用户按(1,2,3)的顺序抬起手指，那么一切都会按预期进行。因此，看起来mulitpletouchenable、userinteractionenable等标志必须是正确的。

这是一个报告(n是返回的触摸次数，x和y是位置)(我编辑了输出文件以识别手指1、2、3

    2012-06-27 07:22:36.589 Bowing[757:907]  finger 1 Began n=0,x=190, y=860
    2012-06-27 07:22:37.207 Bowing[757:907]  finger 2 Began n=0,x=346, y=704
    2012-06-27 07:22:37.875 Bowing[757:907]  finger 3 Began n=0,x=580, y=708

    2012-06-27 07:22:38.587 Bowing[757:907]  finger 1 Ended n=0,x=191, y=854
    2012-06-27 07:22:39.252 Bowing[757:907]  finger 2 Ended n=0,x=346, y=722
    2012-06-27 07:22:40.019 Bowing[757:907]  finder 3 Ended n=0,x=585, y=712

生活是美好的

如果用户以相反的顺序(3,2,1)抬起手指，那么直到手指1被抬起，才会发送消息；然后发送所有三个手指的消息

    2012-06-27 07:22:36.589 Bowing[757:907] finger 1 Began n=0,x=190, y=860
    2012-06-27 07:22:37.207 Bowing[757:907] finger 2 Began n=0,x=346, y=704
    2012-06-27 07:22:37.875 Bowing[757:907] finger 3 Began n=0,x=580, y=708

    2012-06-27 07:22:38.587 Bowing[757:907] finger 1 Ended n=0,x=191, y=854
    2012-06-27 07:22:39.252 Bowing[757:907] finger 2 Ended n=0,x=346, y=722
    2012-06-27 07:22:40.019 Bowing[757:907] finger 3 Ended n=0,x=585, y=712

生活并不是那么美好。为了反映屏幕上发生的情况，消息应该按照finger 3、finger 2和finger 1的顺序到达

下面是响应器；不管有没有给super的消息，都是一样的

- (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event{
//    [super touchesBegan:touches withEvent:event];
    NSEnumerator *enumerator = [touches objectEnumerator];
    UITouch  *aTouch;
    int counter=0;
    while ((aTouch = [enumerator nextObject])) {
        /* code that acts on the set’s values */
        CGPoint  where=[aTouch locationInView:nil];
        NSLog(@"Began n=%i,x=%3.0f, y=%3.0f",counter,where.x,where.y);
        counter++;
    }

}
// Sent to the receiver when a system event (such as a low-memory warning) cancels a touch   event.
- (void)touchesCancelled:(NSSet *)touches withEvent:(UIEvent *)event{

//    [super touchesCancelled:touches withEvent:event];
    NSEnumerator *enumerator = [touches objectEnumerator];
    UITouch  *aTouch;
    int counter=0;
    while ((aTouch = [enumerator nextObject])) {
        /* code that acts on the set’s values */
        CGPoint  where=[aTouch locationInView:nil];
        NSLog(@"Cancelled n=%i,x=%3.0f, y=%3.0f",counter,where.x,where.y);
        counter++;
    }

}

//当从视图或窗口中举起一个或多个手指时，告知接收者。

- (void)touchesEnded:(NSSet *)touches withEvent:(UIEvent *)event{
//    [super touchesEnded:touches withEvent:event];
    NSEnumerator *enumerator = [touches objectEnumerator];
    UITouch  *aTouch;
    int counter=0;
    while ((aTouch = [enumerator nextObject])) {
        /* code that acts on the set’s values */
        CGPoint  where=[aTouch locationInView:nil];
        NSLog(@"Ended n=%i,x=%3.0f, y=%3.0f",counter,where.x,where.y);
        counter++;
    }

}   

Answer 1

问题是:结束的触摸正在等待所有的触摸完成，这就是为什么当它结束时，所有三个触摸都被发送。对此的解决方案可以是一个nstimer，它以1/32的时间间隔检测[touch count];，并检测它是增加还是减少。