弹出视图时传递参数

Question

我正在构建一个ios应用程序,其中有两个视图之间的导航A和B.

导航模式是:

ViewController A >>> PushViewController >>> ViewController B

ViewController A <<< PopViewController <<< ViewController B

我希望当B弹出时A,A会相应地更新一些UI元素.例如,A视图控制器显示一些带有文本的标签,在B用户中修改文本,当视图弹出时,我想A更新并反映更改.

问题是:怎么A知道它什么时候开始B？如何A获取数据,B以便更新内容？解决这类问题的好方法是什么？

谢谢

Answer 1

您可以通过以下方式轻松完成NSNotificationCenter:

第一视图控制器:

// 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];
}

第二视图控制器:

- (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;
}