xcode prepareForSegue 和 popViewController 的兼容性

prepareForSegue and popViewController compatibility

提问by Octan

The question

问题

I'm working on a iOSapp and i have 3 consecutive ViewControllers:

我正在开发一个iOS应用程序，我有 3 个连续的ViewControllers：

TableViewController--> DetailViewController--> ImageViewController

TableViewController--> DetailViewController-->ImageViewController

I perform de forward Segue with a button (Just control drag on Storyboard) and to go back I have a custom back button with

我用一个按钮执行 de forward Segue（只需控制拖动Storyboard）然后返回我有一个自定义后退按钮

[self.navigationController popViewControllerAnimated:YES];

To send data forward I use

要转发数据，我使用

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    [segue.destinationViewController someFunction];
}

To send data to the parent ViewControllerI can use prepareForSeguein the DetailViewController, but it doesn't work in the ImageViewControllerand there I have to use Notifications.

为了将数据发送到父ViewController，我可以使用prepareForSegue的DetailViewController，但它不工作ImageViewController，有我一定要使用Notifications。

It's OK if I use prepareForSegueto send data with popViewControllerAnimated?

如果我prepareForSegue用来发送数据可以popViewControllerAnimated吗？

Should I use Notifications in both cases?

在这两种情况下我都应该使用通知吗？

Some code

一些代码

What I have in the DetailViewControlleris a button that perform the Segue to the ImageViewController (just control drag on Storyboard) and a Back Button with:

我在里面DetailViewController有一个按钮，它对 ImageViewController 执行 Segue（只需控制拖动Storyboard）和一个后退按钮：

- (IBAction)backAction:(id)sender {
     [self.navigationController popViewControllerAnimated:YES];
}

and then the function:

然后是函数：

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    if([segue.identifier isEqualToString:@"forwardSegue"]) {
        [segue.destinationViewController someFuntionToImageViewController];
    } else {
        [segue.destinationViewController someFuntionToParentViewController];
    }
}

I noticed that I can't assign a segue.identifier to the popViewControlleraction.

我注意到我无法为popViewController操作分配 segue.identifier 。

回答by Tobi

Apple recommends the following way to pass data between view controllers that are navigated by segues:

Apple 建议使用以下方式在由 segue 导航的视图控制器之间传递数据：

To pass data forward:

向前传递数据：

You declare a property on the destination vc and set its value in the prepareForSeguemethod. (So kind of the way you did it)

您在目标 vc 上声明一个属性并在prepareForSegue方法中设置其值。（所以你这样做的方式）

@interface SomeViewController

@property (strong, nonatomic) VeryImportantData *data;
//...
@end

@implementation SourceVC

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    VeryImportantData *theData = ...;
    segue.destinationViewController.data = theData;
}

To pass data back

回传数据

You declare a delegate protocol. This could look like this:

您声明一个委托协议。这可能是这样的：

@protocol SomeViewControllerDelegate
- (void)someViewController:(SomeViewController *) someVC didFinishWithData:(SomeData *) data;
@end

Your destination vc provides a delegate property:

您的目标 vc 提供了一个委托属性：

@property (weak, nonatomic) id<SomeViewControllerDelegate> delegate;

The destination vc calls it's delegates method once it's ready to be closed:

一旦准备好关闭，目标 vc 就会调用它的委托方法：

@implementation SomeViewController

- (void) close
{
    [delegate someViewController:self didFinishWithData:self.data];
}

And your source controller implements the protocol and sets itself as delegate of the destination vc in the prepareForSeguemethod.

并且您的源控制器实现了协议并将自身设置为prepareForSegue方法中目标 vc 的委托。

@implementation SourceVC

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender
{
    segue.destinationViewController.delegate = self;
}


//...
- (void)someViewController:(SomeViewController *) someVC didFinishWithData:(SomeData *) data
{
    self.receivedData = data;
    [self.navigationController popViewControllerAnimated:YES];
}