Question

我的故事板中有很多视图控制器，我将如何将登录页面数据电子邮件ID传递给我的第三个视图控制器，但我得到了NULL值

这是第一个视图的代码

//in first view i already import my second view below code is in 1st view
//.h
@property (weak, nonatomic) IBOutlet UITextField *emailid;

//.m
-(void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"sendemailid"])
    {
        UINavigationController *navController = (UINavigationController *)segue.destinationViewController;
        UploadImageData *controller = (UploadImageData *)navController.topViewController;
        controller.newemailid = self.emailid.text;
        NSLog(@"emailid is %@",emailid.text);
    }


}
// i already created segue by identifier name "sendemailid"

现在在我的第二个视图中

// in .h 

@property (strong, nonatomic) IBOutlet UILabel *emailid;
@property (nonatomic) NSString *newemailid;

// .m in view didload method

self.emailid.text = self.newemailid;
    NSLog( @"we get the emailid %@", emailid.text );

我正在获取null值，还有一件事我正在使用NSUSER DEFAULT来保存第一个视图中的emailid。我不知道为什么价值不通过是我遗失的东西请建议...

Answer 1

我已经完成了这个

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"sendemailid"])
    {
        UploadImageData *controller=segue.destinationViewController;
        controller.newemailid=self.emailid.text;
    }
}

使用storyboard xcode6在视图控制器之间传递数据

1 个答案: