Question

我正在尝试将数据从一个UITableViewController传递到另一个- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { Subject *subject = (Subject *)[self.fetchedResultsController objectAtIndexPath:indexPath]; [self showList:subject animated:YES]; [self.tableView deselectRowAtIndexPath:indexPath animated:YES]; } - (void)showList:(Subject *)subject animated:(BOOL)animated { ListsViewController *lists = [[ListsViewController alloc] initWithStyle:UITableViewStyleGrouped]; lists.subject = subject; NSLog(@"%@", lists.subject); [self performSegueWithIdentifier:@"showDetail" sender:self]; }。这是我在初始视图控制器中的代码：

subject

日志输出显示它已经传递了我想要的数据。但是，当我执行segue并记录时，ListsViewController中的{{1}}显示为空。

有什么想法吗？

Answer 1

您需要覆盖prepareForSegue:sender:方法。快速解决方案是

- (void)showList:(Subject *)subject animated:(BOOL)animated
{
     [self performSegueWithIdentifier:@"showDetail" sender:subject];
}

...

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"showDetail"]) {
        ListsViewController *controller = ([segue.destinationViewController isKindOfClass:[ListsViewController class]]) ? segue.destinationViewController : nil;
        controller.subject = ([sender isKindOfClass:[Subject class]]) ? subject : nil;
    }
}

您的代码无效的原因是您在showList:animated:方法中创建了ListsViewController实例并为其分配了subject，但此视图控制器从未显示过。相反，performSegueWithIdentifier:sender会创建另一个ListsViewController类的实例，该实例对您的subject一无所知。这就是为什么你需要等待UIStoryboardSegue从故事板中实例化目标视图控制器，然后按照你想要的方式配置它，你可以在prepareForSegue:sender:方法中进行配置。

在subject方法中使用performSegueWithIdentifier:sender作为发件人可能不是最佳选择，因为它不是发件人:)。我要做的是在视图控制器类中创建一个属性主题并使用它prepareForSegue:sender:

@interface MyViewController ()

@property (strong, nonatomic) Subject *subject;

@end


@implementation MyViewController

- (void)showList:(Subject *)subject animated:(BOOL)animated
{
    self.subject = subject;
    [self performSegueWithIdentifier:@"showDetail" sender:self];
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"showDetail"]) {
        ListsViewController *controller = ([segue.destinationViewController isKindOfClass:[ListsViewController class]]) ? segue.destinationViewController : nil;
        controller.subject = self.subject;
    }
}

...
@end

Answer 2

实施prepareForSegue并在此方法中传递日期

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
if([seque.identifier isEqualToString:@"showDetail"])
{
ListsViewController *lists = seque.destinationViewController;
lists.subject = subject;
}
}

Answer 3

这很好，但现在你需要添加这个：

首先代替：

[self performSegueWithIdentifier:@"showDetail" sender:self];

您需要发送对象：

[self performSegueWithIdentifier:@"showDetail" sender:subject];

在ListsViewController.h中添加一个属性：

@property (nonatomic, strong) Subject * subjectSegue;

现在在你的第一个视图控制器中：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"showDetail"]) {
        ListsViewController * lists = (ListsViewController *)[segue destinationViewController];
        lists.subjectSegue = sender;
}

Answer 4

您需要了解performSegueWithIdentifier:sender:创建一个新的视图控制器实例。因此，您创建的ListsViewController不会显示在屏幕上。

您需要覆盖`prepareForSegue：sender：

- (void)showList:(Subject *)subject animated:(BOOL)animated 
{
  [self performSegueWithIdentifier:@"showDetail" sender:self];
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    if ([segue.identifier isEqualToString:@"showDetail"]) {
        ListsViewController *controller = (ListsViewController *)segue.destinationViewController;
    controller.subject = self.subject;
}

使用故事板在两个控制器之间传递数据

4 个答案: