ios - Segue 不传递值

Question

我正在创建一个segue，同时尝试将联系人(loggedinUser)的信息传递给destinationViewControllers的联系人(currentUser)，但由于某种原因它没有通过。下面的代码有问题吗？

原始 View :

...
self.loggedInUser.firstName = @"first name";
self.loggedInUser.lastName = @"last name";
self.loggedInUser.company = @"company"
self.loggedInUser.position = @"position";
[self performSegueWithIdentifier:@"profileSegue" sender:self];
...

-(void) prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
ProfileViewController *pvc = (ProfileViewController *) [segue destinationViewController];
pvc.currentUser = self.loggedInUser; }

然后我在标签中显示值 目标 View Controller :

- (void)viewDidLoad
{
    [super viewDidLoad];
    // Do any additional setup after loading the view.
    NSString *fullName = [NSString stringWithFormat:@"%@ %@", currentUser.firstName, currentUser.lastName];
    fullNameLabel.text = fullName;
    companyLabel.text = currentUser.company;
    positionLabel.text = currentUser.position;
}

loggedInUser 和 currentUser 都是“联系人”类型。但标签最终显示为空。我是否遗漏了一些明显的东西？

编辑

这是 self.loggedinUser 未传递到prepareForSegue 的问题。

标题代码:

#import "Contact.h"

@interface LoginViewController : UIViewController
{
}

...

@property (nonatomic, retain) Contact *loggedInUser;

Answer 1

我假设viewDidLoad在prepareFortSegue之前被调用，即当currentUser仍然是nil时。我建议将您的代码从 viewDidLoad 移至 viewWillAppear。或者您可以设置断点来检查调用顺序。