Ios5 未调用prepareforsgue

我正在使用模板主详细信息应用程序。我已经从SplitViewController添加了一个模式segue，并给它一个标识符“DisplayLoginView”

我从我的detailViewController调用以下命令：

- (void)viewDidAppear:(BOOL)animated
{
    [super viewDidAppear:YES];
    [self.splitViewController performSegueWithIdentifier:@"DisplayLoginView"    sender:self.splitViewController];
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    NSLog(@"Source Controller = %@", [segue sourceViewController]);
    NSLog(@"Destination Controller = %@", [segue destinationViewController]);
    NSLog(@"Segue Identifier = %@", [segue identifier]);
    if ([segue.identifier isEqualToString:@"DisplayLoginView"])
    {
        PrometheusLoginViewController *loginViewController = (PrometheusLoginViewController *)segue.destinationViewController;
        loginViewController.delegate = self;
    }
}

我还在detailViewController中定义了prepareForSegue方法：

- (void)viewDidAppear:(BOOL)animated
{
    [super viewDidAppear:YES];
    [self.splitViewController performSegueWithIdentifier:@"DisplayLoginView"    sender:self.splitViewController];
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    NSLog(@"Source Controller = %@", [segue sourceViewController]);
    NSLog(@"Destination Controller = %@", [segue destinationViewController]);
    NSLog(@"Segue Identifier = %@", [segue identifier]);
    if ([segue.identifier isEqualToString:@"DisplayLoginView"])
    {
        PrometheusLoginViewController *loginViewController = (PrometheusLoginViewController *)segue.destinationViewController;
        loginViewController.delegate = self;
    }
}

---

知道为什么不调用它吗？

您要求splitViewController执行该步骤，但您在detailViewController中定义了prepareForSegue。它们必须在同一个对象上才能触发PrepareForsgue。

啊，愚蠢的错误。从另一个线程中，我认为我必须从splitViewController中呈现它，否则它将无法工作。将segue从detailViewController更改为go，并将其更改为[SelfPerformsgueWithIdentifier:@“DisplayLoginView”发送方：self]；它是有效的。谢谢很好，你真的发布了有用的东西。将使用此选项使用segues在视图之间存储数据。