2013-02-19 43 views
1

我有一个ViewController,它在Interface Builder中创建了几个按钮。第一个按钮将显示在IB中链接的弹出式窗口,它链接到UINavigationController并在其下有一个类别为PopupViewController的TableView。使用弹出窗口在prepareForSegue中设置属性

第二个按钮,我有一个动作设置为goToCategory和上点击的时候,我想设置在PopupViewController

ViewController.m 
//go to category 
-(IBAction)goToCategory:(id)sender{ 
    NSLog(@"GO TO CAT"); 
    //PopupViewController *popupVC = [self.storyboard instantiateViewControllerWithIdentifier:@"popoverVC"]; 
    //popupVC.currentLevel = 1; 
    [self performSegueWithIdentifier:@"popoverSegue" sender:self]; 
} 

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{ 
    NSLog(@"seg1"); 
    if ([[segue identifier] isEqualToString:@"popoverSegue"]){ 
     //PopupViewController *popupVC = (PopupViewController *)[[segue destinationViewController] visibleViewController]; 
     //PopupViewController *popupVC = [segue destinationViewController]; 
     PopupViewController *popupVC=[[[segue destinationViewController]viewControllers]objectAtIndex:0]; 
     popupVC.test = @"just a test"; 
     NSLog(@"seg2"); 
    } 
} 

PopupViewController.h 
@property (nonatomic, strong) NSString *test; 

PopupViewController.m 
@synthesize test; 
-(void)viewDidLoad{ 
    NSLog(@"test: %@", test); //returns test: (null) 
} 

属性我发现了很多SO答案,因此一些在我的prepareForSegue中注释掉的线条。但是这些都没有设定值testPopupViewController *popupVC = [segue destinationViewController];由于引用了UINavigationController而引发错误,所以我不能直接使用它,即使这似乎是在我见过的很多答案中做到的。但是不管我尝试用哪种方式做,输出总是空的?

UPDATE:

PopupViewController *popupVC = (PopupViewController *)[[segue destinationViewController] visibleViewController];PopupViewController *popupVC=[[[segue destinationViewController]viewControllers]objectAtIndex:0];从我prepareForSegue在6.1模拟器都工作上面。我的iPad的iOS是5.1.1,它不工作。我需要为iOS 5做些什么不同吗?

+0

嗯,如果你想访问你的酥料饼,这应该工作:[(UIStoryboardPopoverSegue *)Segue公司popoverController] – Leonardo 2013-02-19 10:28:30

回答

0

试试这个

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender 
{ 
    NSLog(@"seg1"); 
    if ([[segue identifier] isEqualToString:@"popoverSegue"]) 
    { 
     UINavigationController *navController = (UINavigationController*)[segue destinationViewController]; 
     PopupViewController *popupVC = [navController topViewController]; 
     popupVC.test = @"just a test"; 
     NSLog(@"seg2"); 
    } 
} 
+0

我也尝试写这样的说法,你只是做了,但我它会抛出一个错误,即'[UINavigationController setTest:]:无法识别的选择器发送到实例'因为目标视图控制器是一个UINavigationController,而不是PopupViewController – Nathan 2013-02-19 09:14:18

+0

看看更新的答案 – 2013-02-19 09:20:24

+0

刚刚尝试过你的新的更新代码,但仍然没有运气:(试图改变你的行到'PopupViewController * popupVC =(PopupViewController *)[navController topViewController];'以及xcode g ave me a warning,'不兼容的指针类型初始化PopupViewController __strong与UIViewController类型的表达式' – Nathan 2013-02-19 09:29:25