Tob*_*nin 2 uisplitviewcontroller ios5 uistoryboard segue
我正在使用模板主详细信息应用程序.我在SplitViewController中添加了一个模态segue,并为其指定了标识符"DisplayLoginView".
我从detailViewController中调用以下内容:
- (void)viewDidAppear:(BOOL)animated
{
[super viewDidAppear:YES];
[self.splitViewController performSegueWithIdentifier:@"DisplayLoginView" sender:self.splitViewController];
}
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我还在detailViewController中定义了prepareForSegue方法:
- (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;
}
}
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为什么没有调用它的任何想法?
您要求splitViewController执行segue,但您要在detailViewController中定义prepareForSegue.他们需要在同一个对象上才能触发prepareForSegue.
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