UIAlertView 上的 iOS 解析通知消息(空)

Question

我正在使用 Parse.com 将推送通知发送到一个简单的 iOS Web View 应用程序，但我似乎无法让它显示 JSON 负载中的消息文本:

{
    "alert": "Push Message goes here.",
    "url": "http://www.google.com"
}

这是我的AppDelegate.m:

- (void)application:(UIApplication *)application didReceiveRemoteNotification:(NSDictionary *)userInfo {

    NSString* notifURL = [userInfo objectForKey:@"url"];
    NSLog(@"Received Push Url: %@", notifURL);

    NSString* message = [userInfo objectForKey:@"alert"];
    NSLog(@"Received Message: %@", message);

    NSLog(@"UserInfo: %@", userInfo);

    if (application.applicationState == UIApplicationStateActive) {

        UIAlertView *alertPush = [[UIAlertView alloc]initWithTitle:@"My Webview App"
                                                       message:message
                                                      delegate:self
                                             cancelButtonTitle:@"View"
                                             otherButtonTitles:@"Cancel", nil];
        [alertPush show];
        [alertPush release];

        objc_setAssociatedObject(alertPush, &aURL, notifURL, OBJC_ASSOCIATION_RETAIN_NONATOMIC);

    }

}

这是我的日志返回的内容:

Received Push Url: http://www.google.com

Received Message: (null)

UserInfo: { aps = { alert = "Push Title goes here"; }; url = "http://www.google.com"; }

我是不是漏掉了什么？

Answer 1

如果您格式化UserInfo，您会看到它有两个键aps 和url。 url 的值为 http://www/google.com 而 aps 的值又是一个字典。这本字典有一个键 alert 和值 Push Title goes here

UserInfo: { 
           aps = { 
                   alert = "Push Title goes here";
           }; 
           url = "http://www.google.com"; 
}

所以你需要通过以下方式提取它:

NSString *message = userInfo[@"aps"][@"alert"];

//First extract the dictionary with key : aps
//then extract the string with key : alert