您好,我正在为 Apple Watch 开发一个应用程序,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,所以这就是我目前所拥有的:
InterfaceController.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
接口(interface) Controller .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)awakeWithContextid)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:
ToolBelt WatchKit Extension/InterfaceController.m:22:21: 'sharedApplication' is unavailable: not available on iOS (App Extension) - Use view controller based solutions where appropriate instead.
如果提出这样一个菜鸟问题,任何帮助都会非常抱歉
提前致谢
Best Answer-推荐答案 strong>
如错误消息所示,UIApplication 类在扩展(包括 WatchKit 扩展)中不可用。无法通过 WatchKit 扩展在用户手机上打开 URL。您应该考虑采用 Handoff 来让用户快速从 Watch 应用过渡到手机应用。
关于ios - 在我的 Apple Watch 应用程序中获取 sharedApplication' 不可用,我们在Stack Overflow上找到一个类似的问题:
https://stackoverflow.com/questions/29024531/
|