iOS禁止右滑返回页面

话不多说，直接看代码

需要禁止右滑的controller里面引入代理
@interface SAL5DVC ()<UIGestureRecognizerDelegate>
@end
@implementation SAL5DVC
#pragma mark -- 禁止右滑返回
- (void)viewDidAppear:(BOOL)animated
{
    [super viewDidAppear:animated];
    if ([self.navigationController respondsToSelector:@selector(interactivePopGestureRecognizer)]) {
        self.navigationController.interactivePopGestureRecognizer.delegate = self;
    }
}
- (void)viewWillDisappear:(BOOL)animated
{
    [super viewWillDisappear:animated];
    if ([self.navigationController respondsToSelector:@selector(interactivePopGestureRecognizer)]) {
        self.navigationController.interactivePopGestureRecognizer.delegate = nil;
    }
}
- (BOOL)gestureRecognizerShouldBegin:(UIGestureRecognizer *)gestureRecognizer
{
    return NO;
}
@end

完美运行、亲测成功，之前 我是这么写的

- (void)viewDidAppear:(BOOL)animated 
{
   [super viewDidAppear:animated];
  //禁用屏幕左滑返回手势
  if ([self.navigationController respondsToSelector:@selector(interactivePopGestureRecognizer)])
    self.navigationController.interactivePopGestureRecognizer.enabled = NO;
}
- (void)viewDidDisappear:(BOOL)animated 
{
  [super viewDidDisappear:animated];
  //开启
 self.navigationController.interactivePopGestureRecognizer.enabled = YES;
}

这么写的问题在于，当进入页面之后，第一次向右滑动是可以返回的，松开第二次，才会有效果，不完美😂😂😂