iOS 7 通过点击推送通知查看特定的View Controller

Question

我已经为我的应用程序设置了一个推送通知，这样当我点击推送通知时，应用程序就会转到主控件 View 。但是，我想根据添加到我的应用程序的内容查看特定的 View Controller 。我该怎么做？

我的应用委托(delegate)代码。

     - (BOOL)application:(UIApplication *)application didFinishLaunchingWithOptions:(NSDictionary *)launchOptions
      {

          [[UIApplication sharedApplication] registerForRemoteNotificationTypes:(UIRemoteNotificationTypeAlert | UIRemoteNotificationTypeBadge | UIRemoteNotificationTypeNone)];
          return YES;
      }

    -(void)application:(UIApplication *)application didRegisterForRemoteNotificationsWithDeviceToken:(NSData *)deviceToken
     {

        const char* data = [deviceToken bytes];
        NSMutableString * token = [NSMutableString string];

        for (int i = 0; i < [deviceToken length]; i++) {
        [token appendFormat:@"%02.2hhX", data[i]];
      }


         NSString *urlString = [NSString stringWithFormat:@"url"?token=%@",token];

         NSURL *url = [[NSURL alloc] initWithString:urlString];
         NSLog(@"token %@",urlString);


         NSURLRequest *urlRequest = [NSURLRequest requestWithURL:url];
         NSLog(@"request %@ ",urlRequest);
         NSData *urlData;
         NSURLResponse *response;
         urlData = [NSURLConnection sendSynchronousRequest:urlRequest returningResponse:&response error:nil];
         NSLog(@"data %@",urlData);

        //  NSLog(@"token ",sendUserToken);

      }

我的 php 推送通知脚本。

 <?php



    token ="my token"

          $payload = '{
         "aps" :
     {
        "alert" :"'.$message.'",
        "badge" : 1,
        "sound" : "bingbong.aiff"
      }  
   }';
     $ctx = stream_context_create();
     stream_context_set_option($ctx,'ssl', 'local_cert','ck.pem');
     stream_context_set_option($ctx,'ssl','passphrase', 'balpad');
     $fp = stream_socket_client('ssl://gateway.sandbox.push.apple.com:2195',$err,$errstr,60,STREAM_CLIENT_CONNECT,$ctx);
     if(!fp){
       print "Failed to connect $err $errstrn";
       return;
     }else{
        print "notifications sent!";
     }
       $devArray = array();
       $devArray[] = $deviceToken;

      foreach($deviceToken as $token){
      $msg = chr(0) . pack("n",32) . pack("H*", str_replace(' ',' ',$token)).pack("n",strlen($payload)) . $payload;
      print "sending message:" .$payload . "n";
      fwrite($fp,$msg);
     }
     fclose($fp);
     }

  ?>

这是我第一次使用推送通知，我还没有找到合适的解决方案。我找到了一些建议 ( link1 link2 )，但我发现它们有点令人困惑，而且我没有任何想法。请有人指导我如何制作这个。

Answer 1

我有一个解决方案给你。请引用下面我的示例代码:

-(void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo
{       
   UIApplicationState state = [application applicationState];
   if (state == UIApplicationStateActive) {

       // Below Code Shows Message While Your Application is in Active State

       NSString *cancelTitle = @"Ok";

       NSString *message = [[userInfo valueForKey:@"aps"] valueForKey:@"alert"];
       UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"App Start"
                                                        message:message
                                                       delegate:nil
                                              cancelButtonTitle:cancelTitle
                                              otherButtonTitles: nil];
       [alertView show];
       [alertView release];

    } else {

       // Do stuff that you would do if the application was not active
       // Please add your code to go to specific view controller here.
    }
}