迅速:

navigationController!.interactivePopGestureRecognizer!.enabled = false

我找到了一个解决方案:

objective - c:

if ([self.navigationController respondsToSelector:@selector(interactivePopGestureRecognizer)]) {
    self.navigationController.interactivePopGestureRecognizer.enabled = NO;
}

斯威夫特3 +: self.navigationController?.interactivePopGestureRecognizer?.isEnabled = false

对于Swift 4，这是有效的:

class MyViewController: UIViewController, UIGestureRecognizerDelegate {

    override func viewDidLoad() {
        super.viewDidLoad()

        self.navigationController?.interactivePopGestureRecognizer?.gesture.delegate = self
    }

    override func viewWillAppear(_ animated: Bool) {
        super.viewWillAppear(true)

        self.navigationController?.interactivePopGestureRecognizer?.gesture.isEnabled = false
    }

}

EDIT

如果您想管理特定导航控制器的滑动回滚功能，请考虑使用SwipeBack。

有了这个，你可以设置navigationController。swipeBackEnabled = NO。

例如:

#import <SwipeBack/SwipeBack.h>

- (void)viewWillAppear:(BOOL)animated
{
    navigationController.swipeBackEnabled = NO;
}

它可以通过CocoaPods安装。

pod 'SwipeBack', '~> 1.0'

我很抱歉没有解释清楚。

迅速:

navigationController!.interactivePopGestureRecognizer!.enabled = false

没有一个给出的答案帮助我解决这个问题。在这里张贴我的答案;可能对某人有帮助

声明私有var popGesture: UIGestureRecognizer?作为视图控制器中的全局变量。然后在viewDidAppear和viewWillDisappear方法中实现代码

override func viewDidAppear(animated: Bool) {

    super.viewDidAppear(animated)

    if self.navigationController!.respondsToSelector(Selector("interactivePopGestureRecognizer")) {

        self.popGesture = navigationController!.interactivePopGestureRecognizer
        self.navigationController!.view.removeGestureRecognizer(navigationController!.interactivePopGestureRecognizer!)
    }
}


override func viewWillDisappear(animated: Bool) {

    super.viewWillDisappear(animated)

    if self.popGesture != nil {
        navigationController!.view.addGestureRecognizer(self.popGesture!)
    }
}

这将禁用滑动回iosv8。x起