Question

我正在使用Objective-C中的ios 10 Rich Media推送通知。我想在按下 UNNotificationAction 按钮时打开一个URL。我在followig方法中得到了点击响应：

     -(void)userNotificationCenter:(UNUserNotificationCenter *)center didReceiveNotificationResponse:(UNNotificationResponse *)response withCompletionHandler:(void (^)())completionHandler{
        NSDictionary * userInfo = response.notification.request.content.userInfo;
        NSArray *btnArray = userInfo[@"aps"][@"btnArr"];
        NSString *urlString = [btnArray valueForKey:@"url"];    //Supose url = www.google.com
         NSURL *url = nil;
             if ([urlString hasPrefix:@"http://"] || [urlString hasPrefix:@"https://"]) {
                    url = [NSURL URLWithString:urlString];
                } else {
            url = [NSURL URLWithString:[NSString stringWithFormat:@"https://%@",urlString]];     //now url = https://www.google.com
                    }

// Here the url that will be hit = https://www.google.com
          if (![[UIApplication sharedApplication] openURL:url]) {
            if ([[url absoluteString] isEqual:[NSNull null]] || [[url absoluteString] isEqualToString:@""] || [[url absoluteString] length] == 0) {
         NSLog(@"action url is empty");
            } else {
                NSLog(@"%@%@",@"Failed to open url:",[url description]);
                    }
               }
        }

但它仅在应用程序仅处于前台状态时才起作用。我想在应用程序的所有状态（即非活动，活动和背景状态）中打开URL。任何人都可以帮助我实现我的目标。如果可能，请提供任何参考颂歌，

当应用程序处于非活动状态时，如何在浏览器中打开来自UNNotificationAction响应的URL

0 个答案: