当用户摇动iPhone时,我想检测屏幕的哪个部分被触摸。
我是按照以下方式做的:
-(void) accelerometer: (UIAccelerometer*)accelerometer didAccelerate: (UIAcceleration*)acceleration
{
float shakeStrength = sqrt( acceleration.x * acceleration.x + acceleration.y * acceleration.y + acceleration.z * acceleration.z );
if (shakeStrength >= 1.5f)
{
if (isLeftHandTouches && isRightHandTouches)
{
DebugLog(@"both hands shake");
} else if (isLeftHandTouches)
{
DebugLog(@"left hand shake");
} else if (isRightHandTouches)
{
DebugLog(@"right hand shake");
}
}
}
-(void) touchesBegan: (NSSet *)touches withEvent: (UIEvent *)event
{
NSSet *allTouches = [event allTouches];
for (int i = 0; i < [allTouches count]; i++)
{
if ([ [ [allTouches allObjects] objectAtIndex: i] locationInView: [self view] ].x <= 240.0f)
{
isLeftHandTouches = YES;
} else
{
isRightHandTouches = YES;
}
}
}
-(void) touchesEnded: (NSSet *)touches withEvent: (UIEvent *)event
{
NSSet *allTouches = [event allTouches];
for (int i = 0; i < [allTouches count]; i++)
{
if ([ [ [allTouches allObjects] objectAtIndex: i] locationInView: [self view] ].x <= 240.0f)
{
isLeftHandTouches = NO;
} else
{
isRightHandTouches = NO;
}
}
}
如果用户在再次摇动前移开双手,一切正常,但如果我双手放在屏幕上并移除其中一个,则一切都会搞砸。
即。我用双手在屏幕上摇晃,之后我想用一只手摇动iPhone。在这种情况下,摇动不会计数 - 就好像屏幕上没有触摸一样。我假设当我从屏幕上移开一只手时,两个“触摸”都被删除了。
问题是什么,我该如何解决?
感谢。
答案 0 :(得分:2)
为什么要枚举-allTouches
?只需枚举传入的touches
集。两种方法都是一样的。