iOS 滑动返回手势

一、首先创建BaseController，在viewDidLoad里：

id target = self.navigationController.interactivePopGestureRecognizer.delegate;

    // handleNavigationTransition:为系统私有API,即系统自带侧滑手势的回调方法，我们在自己的手势上直接用它的回调方法
    UIPanGestureRecognizer *panGesture = [[UIPanGestureRecognizer alloc] initWithTarget:target action:@selector(handleNavigationTransition:)];
    panGesture.delegate = self; // 设置手势代理，拦截手势触发
    [self.view addGestureRecognizer:panGesture];

    // 一定要禁止系统自带的滑动手势
    self.navigationController.interactivePopGestureRecognizer.enabled = NO;
}


// 什么时候调用，每次触发手势之前都会询问下代理方法，是否触发
// 作用：拦截手势触发
- (BOOL)gestureRecognizerShouldBegin:(UIGestureRecognizer *)gestureRecognizer
{
    // 当当前控制器是根控制器时，不可以侧滑返回，所以不能使其触发手势
    if(self.navigationController.childViewControllers.count == 1)
    {
        return NO;
    }

    return YES;
}

二、UIScrollView左右滑动手势与返回手势冲突的解决办法

解决办法是自定义ScrollView，重写- (UIView *)hitTest:(CGPoint)point withEvent:(UIEvent *)event方法，代码如下：

- (UIView *)hitTest:(CGPoint)point withEvent:(UIEvent *)event {
    UIView *hitView = [super hitTest:point withEvent:event];
    if (point.x <= 10) { //这个值是调整返回手势里屏幕的距离
        hitView = nil;
    } else {
        hitView = [super hitTest:point withEvent:event];
    }
    return hitView;
}