IOS：UIPanGesture 不适用于 ios 4.3答案

【问题标题】：IOS: UIPanGesture don't work on ios 4.3IOS：UIPanGesture 不适用于 ios 4.3
【发布时间】：2012-03-27 14:39:37
【问题描述】：

我的应用中有这个角色：

- (void) touch:(UIPanGestureRecognizer*)gesture{
if (gesture.state == UIGestureRecognizerStateBegan) 
    {   
        if (gesture.numberOfTouches == 1) 
        {
            // some code
        }
        else if (gesture.numberOfTouches == 2) //some code
        else return;
}
else if (gesture.state == UIGestureRecognizerStateChanged)
{
    if (gesture.numberOfTouches == 1){

        // some code            
     }
    else if (gesture.numberOfTouches == 2) // some code 

    else return;
}

else if (gesture.state == UIGestureRecognizerStateEnded     ||
         gesture.state == UIGestureRecognizerStateCancelled ||
         gesture.state == UIGestureRecognizerStateFailed)
{   
    // some code
}

}

它在 ios 5 中可以正常工作，但在 ios 4.3 中却不行...为什么？？？

【问题讨论】：

你为哪个控件添加了这个手势？你能在你创建这个手势的地方发布代码吗？

标签： ios xcode uipangesturerecognizer

【解决方案1】：

这是为视图添加平移手势的代码。就像您在代码中为哪个控件添加了识别器一样？

- (void)viewDidload:(BOOL)animated{
    UIRotationGestureRecognizer *panRecognizer = [[UIPanGestureRecognizer alloc]initWithTarget:self action:@selector(touch:];
    [self.view addGestureRecognizer:panRecognizer];
[super viewDidLoad:animated];
}

- (void) touch:(UIPanGestureRecognizer*)gesture{
if (gesture.state == UIGestureRecognizerStateBegan) 
    {   
        if (gesture.numberOfTouches == 1) 
        {
            // some code
        }
        else if (gesture.numberOfTouches == 2) //some code
        else return;
}
else if (gesture.state == UIGestureRecognizerStateChanged)
{
    if (gesture.numberOfTouches == 1){

        // some code            
     }
    else if (gesture.numberOfTouches == 2) // some code 

    else return;
}

else if (gesture.state == UIGestureRecognizerStateEnded     ||
         gesture.state == UIGestureRecognizerStateCancelled ||
         gesture.state == UIGestureRecognizerStateFailed)
{   
    // some code
}
}

【讨论】：