Question

当我运行项目时，我收到以下错误，在行secView.delegate=self;中指出我该如何解决上述问题？谢谢！

在'UIViewController *'
类型上找不到属性'委托'对象

myViewController.h

@interface myViewController : UIViewController <UITextFieldDelegate,UIAlertViewDelegate,secondViewControllerDelegate>
- (IBAction)popBookmarkTable:(id)sender;

myViewController.m

- (IBAction)popBookmarkTable:(id)sender {
    UIStoryboard *st = [UIStoryboard storyboardWithName:[[NSBundle mainBundle].infoDictionary objectForKey:@"UIMainStoryboardFile"] bundle:[NSBundle mainBundle]];
    UIViewController *secView = [st instantiateViewControllerWithIdentifier:@"bookmarkViewController"];
[self presentViewController:secView animated:YES completion:nil];

    secView.delegate=self;
}

- (void)passData:(NSString *)data
{

          _addressBar.text=data;

    }

bookmarkViewController.h

@protocol secondViewControllerDelegate <NSObject>

@required

- (void)passData:(NSString *)data;

@end
@property (nonatomic, weak) id<secondViewControllerDelegate> delegate;
@property (nonatomic,strong) NSString *a;
@end

bookmarkViewController.m

-(void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {
   UITableViewCell * cell = [tableView cellForRowAtIndexPath:indexPath];
              if ([indexPath row]==0) {
                   _a=(NSString *)cell.textLabel.text;
                  [_delegate passData:_a];
                  [self dismissViewControllerAnimated:YES completion:nil];


              }

}

Answer 1

当您声明secView时，您将其定义为UIViewController *secView - 即普通UIViewController。因此，就编译器而言，它没有任何自定义属性。

您需要将其声明为它在运行时的特定类类型：

BookmarkViewController *secView = ...;

Answer 2

您需要进一步确定secView对象的范围。 UIViewController未声明委托属性，但您的协议确实如此。如果您确定视图控制器将实现该协议，请尝试以下操作：

UIViewController<secondViewControllerDelegate> *secView = [st instantiateViewControllerWithIdentifier:@"bookmarkViewController"];

财产代表＆＃39;在类型＆＃39; UIViewController上找不到对象*＆＃39;

2 个答案: