ios - 获取 sharedApplication' 在我的苹果 watch 应用程序中不可用

Question

你好，我正在开发一款适用于 Apple Watch 的应用程序，我的应用程序有一个按钮，我希望当单击该按钮以在配对的 iPhone 上打开 Safari 时，我是 iOS 开发的新手，所以这是我目前所拥有的:

接口(interface) Controller .h

//
//  InterfaceController.h
//  ToolBelt WatchKit Extension
//
//  Created by Chris on 3/12/15.
//  Copyright (c) 2015 Chris. All rights reserved.
//

#import <WatchKit/WatchKit.h>
#import <Foundation/Foundation.h>

@interface InterfaceController : WKInterfaceController
-(IBAction) internetbttn;

@end
@interface ViewController : UIViewController< UIWebViewDelegate >

@end 

InterfaceController.m

 #import "InterfaceController.h"


@interface InterfaceController()

@end


@implementation InterfaceController


-(IBAction) internetbttn: (id)sender {
    NSURL *url = [NSURL URLWithString:@"http://www.google.com"];
    [[UIApplication sharedApplication] openURL:url];

}

- (void)awakeWithContext:(id)context {
    [super awakeWithContext:context];


}

- (void)willActivate {
    // This method is called when watch view controller is about to be visible to user
    [super willActivate];
}

- (void)didDeactivate {
    // This method is called when watch view controller is no longer visible
    [super didDeactivate];
}

@end

它给我的错误是这样的:

ToolBelt WatchKit Extension/InterfaceController.m:22:21: 'sharedApplication' is unavailable: not available on iOS (App Extension) - Use view controller based solutions where appropriate instead.

对于提出这样一个菜鸟问题，任何帮助都会非常抱歉

提前致谢

Answer 1

如错误消息所示，UIApplication 类在扩展(包括 WatchKit 扩展)中不可用。无法从 WatchKit 扩展程序打开用户手机上的 URL。您应该考虑采用 Handoff，让用户可以快速从 Watch 应用过渡到手机应用。