iphone - 如何使用自定义后退按钮返回上一个视图控制器？

Question

我是 iOS 开发的新手。我将使用以下语句进入下一个屏幕。

locationViewController = [[LocationViewController alloc] initWithNibName:@"LocationViewController" bundle:nil];

[navigationController pushViewController:locationViewController animated:YES];

现在我想通过点击当前屏幕顶部栏上的自定义后退按钮返回上一个屏幕。

Answer 1

用这个：

-(IBAction)back:(id)sender
{
    [self.navigationController popViewControllerAnimated:YES];
}

对于主页按钮：

NSArray *viewContrlls=[[self navigationController] viewControllers];
for( int i=0;i<[ viewContrlls count];i++)
{
    id obj=[viewContrlls objectAtIndex:i];
    if([obj isKindOfClass:[Rate_O_MeterViewController class]])
    {
        [[self navigationController] popToViewController:obj animated:YES];
        return;
    }
}

代替 Rate_O_MeterViewController 给出您的控制器名称...将其添加到主页按钮的操作中。

Answer 2

创建一个返回按钮：

    UIBarButtonItem * backButton=[[UIBarButtonItem alloc]initWithTitle:@"Back" style:UIBarButtonItemStyleDone target:self action:@selector(goBack:)];
    self.navigationItem.leftBarButtonItem=backButton;

并编写一个函数：

-(void)goBack:(id)sender
 {
      [self.navigationController popViewControllerAnimated:YES]
 }

Answer 3

试试这个，

appDelegate.m - didFinishLaunchingWithOptions 方法

self.window = [[[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]] autorelease];
// Override point for customization after application launch.
self.viewController = [[[ViewController alloc] initWithNibName:@"ViewController" bundle:nil] autorelease];
//-----------navigation view--------------
UINavigationController *navHomeController = [[UINavigationController alloc] initWithRootViewController:self.viewController];
self.window.rootViewController = navHomeController;