返回 IOS 上未访问的页面
Go back to unvisited page on IOS
我不熟悉 IOS 开发,所以可能是一个简单的问题:用户是否有可能按下顶部的 "back" 按钮,然后转到(到目前为止)非 -访问过的页面?
这是图片:
在window“1”用户按下一个按钮,下一个window是“2”。但是在这里,如果用户按下 "back" 按钮,转到页面“3”(到目前为止从未打开过)?
谢谢
您可以通过在第二个视图控制器中创建一个新按钮来简单地重新定义后退按钮,如下所示:
- (void)viewDidLoad {
[super viewDidLoad];
UIButton *backButton = [[UIButton alloc] initWithFrame: CGRectMake(0, 0, 60.0f, 30.0f)];
[backButton setTitle:@"Back" forState:UIControlStateNormal];
[backButton setTitleColor:self.view.tintColor forState:UIControlStateNormal];
[backButton addTarget:self action:@selector(pushToNextController) forControlEvents:UIControlEventTouchUpInside];
UIBarButtonItem *backButtonItem = [[UIBarButtonItem alloc] initWithCustomView:backButton];
self.navigationItem.leftBarButtonItem = backButtonItem;
}
- (void)pushToNextController {
UIViewController *thirdViewController = [self.storyboard instantiateViewControllerWithIdentifier:@"thirdViewController"];
[self.navigationController pushViewController:thirdViewController animated:YES];
}
我不熟悉 IOS 开发,所以可能是一个简单的问题:用户是否有可能按下顶部的 "back" 按钮,然后转到(到目前为止)非 -访问过的页面?
这是图片:
在window“1”用户按下一个按钮,下一个window是“2”。但是在这里,如果用户按下 "back" 按钮,转到页面“3”(到目前为止从未打开过)?
谢谢
您可以通过在第二个视图控制器中创建一个新按钮来简单地重新定义后退按钮,如下所示:
- (void)viewDidLoad {
[super viewDidLoad];
UIButton *backButton = [[UIButton alloc] initWithFrame: CGRectMake(0, 0, 60.0f, 30.0f)];
[backButton setTitle:@"Back" forState:UIControlStateNormal];
[backButton setTitleColor:self.view.tintColor forState:UIControlStateNormal];
[backButton addTarget:self action:@selector(pushToNextController) forControlEvents:UIControlEventTouchUpInside];
UIBarButtonItem *backButtonItem = [[UIBarButtonItem alloc] initWithCustomView:backButton];
self.navigationItem.leftBarButtonItem = backButtonItem;
}
- (void)pushToNextController {
UIViewController *thirdViewController = [self.storyboard instantiateViewControllerWithIdentifier:@"thirdViewController"];
[self.navigationController pushViewController:thirdViewController animated:YES];
}