Question

我试图打开Apple Watch应用程序的父应用程序。

在Xcode Beta 2中，我们可以使用以下代码：

WKInterFaceController.openParentApplication

然而，在Xcode beta 3中，我无法再找到该代码。现在我不知道如何从监视应用程序打开父应用程序。请帮忙。

Answer 1

The Objective-C method是：

+ (BOOL)openParentApplication:(NSDictionary *)userInfo
                    reply:(void (^)(NSDictionary *replyInfo,
                                    NSError *error))reply

The Swift method是：

class func openParentApplication(_ userInfo: [NSObject : AnyObject]!,
                    reply reply: (([NSObject : AnyObject]!,
                                    NSError!) -> Void)!) -> Bool

因此，您需要将iPhone应用程序传递给reply（）块，以便从WatchKit扩展程序中激活它。这是可以实现的一种方式，例如：

NSString *requestString = [NSString stringWithFormat:@"executeMethodA"]; // This string is arbitrary, just must match here and at the iPhone side of the implementation.
NSDictionary *applicationData = [[NSDictionary alloc] initWithObjects:@[requestString] forKeys:@[@"theRequestString"]];

[WKInterfaceController openParentApplication:applicationData reply:^(NSDictionary *replyInfo, NSError *error) {
    NSLog(@"\nReply info: %@\nError: %@",replyInfo, error);
   }];

您的iPhone应用程序的AppDelegate需要实现以下方法：

- (void)application:(UIApplication *)application handleWatchKitExtensionRequest:(NSDictionary *)userInfo reply:(void(^)(NSDictionary *replyInfo))reply {
NSString * request = [userInfo objectForKey:@"requestString"];

if ([request isEqualToString:@"executeMethodA"]) {
    // Do whatever you want to do when sent the message. For instance... 
    [self executeMethodABC];
}

// This is just an example of what you could return. The one requirement is 
// you do have to execute the reply block, even if it is just to 'reply(nil)'.
// All of the objects in the dictionary [must be serializable to a property list file][3].
// If necessary, you can covert other objects to NSData blobs first. 
NSArray * objects = [[NSArray alloc] initWithObjects:myObjectA, myObjectB, myObjectC, nil];
NSArray * keys = [[NSArray alloc] initWithObjects:@"objectAName", @"objectBName", @"objectCName", nil];
NSDictionary * replyContent = [[NSDictionary alloc] initWithObjects:objects forKeys:keys];

reply(replyContent);
}

WKInterfaceController方法openParentApplication：reply：当iPhone（或iOS模拟器）解锁或锁定时，在后台启动包含应用程序。请注意，来自Apple的声明表明WatchKit扩展程序总是用于在后台启动您的iPhone应用程序，它只是模拟器的一个实现细节，它似乎在之前的测试版中在前台启动您的iPhone应用程序。

如果您想测试同时运行的WatchKit应用程序和iPhone应用程序，只需在Schemes菜单下从Xcode启动WatchKit应用程序，然后通过单击其跳板图标在模拟器中手动启动iPhone应用程序。

Answer 2

如果您需要在前台打开您的父应用，请使用Handoff！

示例：

某处共享两者：

static let sharedUserActivityType = "com.yourcompany.yourapp.youraction"
static let sharedIdentifierKey = "identifier"

在你的手表上：

updateUserActivity(sharedUserActivityType, userInfo: [sharedIdentifierKey : 123456], webpageURL: nil)

在App Delegate中的iPhone上

：

func application(application: UIApplication, willContinueUserActivityWithType userActivityType: String) -> Bool {
    if (userActivityType == sharedUserActivityType) {
        return true
    }
    return false
}

func application(application: UIApplication, continueUserActivity userActivity: NSUserActivity, restorationHandler: ([AnyObject]!) -> Void) -> Bool {
    if (userActivity.activityType == sharedUserActivityType) {
        if let userInfo = userActivity.userInfo as? [String : AnyObject] {
            if let identifier = userInfo[sharedIdentifierKey] as? Int {
                //Do something
                let alert = UIAlertView(title: "Handoff", message: "Handoff has been triggered for identifier \(identifier)" , delegate: nil, cancelButtonTitle: "Thanks for the info!")
                alert.show()
                return true
            }
        }
    }
    return false
}

最后（这一步很重要！）：在您的Info.plist中

enter image description here

如何从我的WatchKit应用程序在iPhone上打开父应用程序？

2 个答案: