好。如果你有两个viewControllers,你从第一个到第二个进行模态Segue,那么你用[self dismissModalViewControllerAnimated:YES]解雇它;它似乎没有回忆起viewDidLoad。我有一个主页面(viewController),然后是各种选项页面,我希望在更改选项时更新主页面。当我刚刚做了两个模态分段(一个前进,一个后退)时,这很有用,但这似乎是非结构化的,并且可能导致大型项目中的代码混乱。
我听说过push segues。他们还好吗?
感谢。我感谢任何帮助:)。
答案 0 :(得分:10)
那是因为UIViewController已经加载到内存中了。但是,您可以使用viewDidAppear:。
或者,您可以使推送视图控制器成为推送视图控制器的委托,并在推出的控制器退出屏幕时通知它更新。
后一种方法的好处是不需要重新运行viewDidAppear:
的整个主体。例如,如果您只更新表格行,为什么要重新渲染整个表格?
编辑:仅供您使用,以下是使用代理的快速示例:
#import <Foundation/Foundation.h>
// this would be in your ModalView Controller's .h
@class ModalView;
@protocol ModalViewDelegate
- (void)modalViewSaveButtonWasTapped:(ModalView *)modalView;
@end
@interface ModalView : NSObject
@property (nonatomic, retain) id delegate;
@end
// this is in your ModalView Controller's .m
@implementation ModalView
@synthesize delegate;
- (void)didTapSaveButton
{
NSLog(@"Saving data, alerting delegate, maybe");
if( self.delegate && [self.delegate respondsToSelector:@selector(modalViewSaveButtonWasTapped:)])
{
NSLog(@"Indeed alerting delegate");
[self.delegate modalViewSaveButtonWasTapped:self];
}
}
@end
// this would be your pushing View Controller's .h
@interface ViewController : NSObject <ModalViewDelegate>
- (void)prepareForSegue;
@end;
// this would be your pushing View Controller's .m
@implementation ViewController
- (void)prepareForSegue
{
ModalView *v = [[ModalView alloc] init];
// note we tell the pushed view that the pushing view is the delegate
v.delegate = self;
// push it
// this would be called by the UI
[v didTapSaveButton];
}
- (void)modalViewSaveButtonWasTapped:(ModalView *)modalView
{
NSLog(@"In the delegate method");
}
@end
int main(int argc, char *argv[]) {
@autoreleasepool {
ViewController *v = [[ViewController alloc] init];
[v prepareForSegue];
}
}
输出:
2012-08-30 10:55:42.061 Untitled[2239:707] Saving data, alerting delegate, maybe
2012-08-30 10:55:42.064 Untitled[2239:707] Indeed alerting delegate
2012-08-30 10:55:42.064 Untitled[2239:707] In the delegate method
示例是在CodeRunner中为OS X运行的,我与之没有任何关联。