左侧滑动返回实现

在开发中我们经常会用到在导航控制器中实现左侧滑动返回，系统默认是支持的，但是当我们自定义NavigationController左侧按钮时，滑动返回就会失效，这个时候应该怎么办呢？

首先我们自定义一个继承自NavigationController的类，直接上代码

#import "FXQNavigationController.h"

@interface FXQNavigationController ()

@property (strong,nonatomic) id popDelegate;

@end

@implementation FXQNavigationController

- (void)viewDidLoad {

[super viewDidLoad];

self.delegate = self;

self.popDelegate = self.interactivePopGestureRecognizer.delegate;

}

- (void)navigationController:(UINavigationController *)navigationController didShowViewController:(UIViewController *)viewController animated:(BOOL)animated{

if (self.viewControllers[0] == viewController) {

self.interactivePopGestureRecognizer.delegate = self.popDelegate;

}else{

self.interactivePopGestureRecognizer.delegate = nil;

}

}

- (void)pushViewController:(UIViewController *)viewController animated:(BOOL)animated{

[super pushViewController:viewController animated:animated];

if (self.viewControllers.count > 1) {

viewController.navigationItem.leftBarButtonItem = [[UIBarButtonItem alloc] initWithTitle:@"返回" style:UIBarButtonItemStylePlain target:self action:@selector(back)];

}

}

- (void)back{

[self popViewControllerAnimated:YES];

}

自己成为自己的代理，在OC中是可以的，这样就可以监听控制器的跳转，通过判断控制器是否是根控制器来控制手势的代理是否为空。希望对大家有所帮助