Question

我正在使用UrbanAirship向我的应用程序发送推送消息。我的设置适用于开发和生产。我需要发送web url作为推送消息。当用户打开消息时，我希望它重定向到我添加的网址。我将此代码添加到了我的appdelegate。

`- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo{
      NSLog(@"userInfo:%@",[userInfo description]);
      NSLog(@"alert:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"alert"]);
      NSLog(@"alert:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"url"]);
}

并试图像

一样发送推送

{
    "aps": 
    {
        "alert": "take a look at this site ",
        "url": "www.mysite.com"
    }
}

我收到了提醒信息但又打开了应用程序，而不是网址。您能否告诉我如何使用网址发送推送消息并将其打开该网址？

Answer 1

有两种方法可以做到这一点

使用safari（未测试代码）打开网址：

     - (void)application:(UIApplication *)application didReceiveRemoteNotification: (NSDictionary *)userInfo
    { 
     NSLog(@"userInfo:%@",[userInfo description]); 
     NSLog(@"alert:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"alert"]);
     NSLog(@"url:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"url"]); 
     webViewController.url = [NSURL URLWithString:[[userInfo objectForKey:@"aps"] objectForKey:@"url"]];
     [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSURL URLWithString:[[userInfo objectForKey:@"aps"] objectForKey:@"url"]];
    }

或者你必须在你的应用上处理它：

- (void)application:(UIApplication *)application didReceiveRemoteNotification: (NSDictionary *)userInfo{ 
     NSLog(@"userInfo:%@",[userInfo description]); 
     NSLog(@"alert:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"alert"]);
     NSLog(@"url:%@",[[userInfo objectForKey:@"aps"] objectForKey:@"url"]); 
     webViewController.url = [NSURL URLWithString:[[userInfo objectForKey:@"aps"] objectForKey:@"url"]];

}

例如，在你的WebViewController中需要以下方法

    - (void)viewWillAppear:(BOOL)animated { 
    [super viewWillAppear:animated];
    NSURLRequest *requestObj = [NSURLRequest requestWithURL:url];
    [webView loadRequest:requestObj];
    }

当然，在你的WebViewController.h上必须是

    IBOutlet UIWebView *webView;

全屏，或者你想要的......

UrbanAirship在IOS上使用URL推送

1 个答案: