ios - 弹出 View 时传递参数

Question

我正在构建一个 ios 应用程序，其中有两个 View A 和 B 之间的导航。

导航模式是:

ViewController A >>> PushViewController >>> ViewController B

ViewController A <<< PopViewController <<< ViewController B

我希望当 B 返回到 A 时，A 相应地更新一些 UI 元素。例如，A View Controller 显示一些带有文本的标签，在 B 用户修改文本，当 View 弹出时，我希望 A 到更新并反射(reflect)更改。

问题是:A 如何知道它何时从 B 弹出？ A 如何获取 B 传递的数据以便更新内容？解决此类问题的良好做法是什么？

谢谢

Answer 1

您可以使用 NSNotificationCenter 轻松完成此操作:

第一个 View Controller :

// Assuming your label is set up in IB, otherwise initialize in viewDidLoad
@property (nonatomic, strong) IBOutlet UILabel *label;

- (void)viewDidLoad
{
    [super viewDidLoad];

    // Add an observer so we can receive notifications from our other view controller
    [[NSNotificationCenter defaultCenter]addObserver:self selector:@selector(updateLabel:) name:@"UpdateLabel" object:nil];
}

- (void)updateLabel:(NSNotification*)notification
{
    // Update the UILabel's text to that of the notification object posted from the other view controller
    self.label.text = notification.object;
}

- (void)dealloc
{ 
    // Clean up; make sure to add this
    [[NSNotificationCenter defaultCenter]removeObserver:self];
}

第二 View Controller :

- (void)viewDidDisappear:(BOOL)animated
{
    [super viewDidDisappear:animated];

    NSString *updateLabelString = @"Your Text Here";        
    // Posting the notification back to our sending view controller with the updateLabelString being the posted object
    [[NSNotificationCenter defaultCenter]postNotificationName:@"UpdateLabel" object:updateLabelString;
}