Question

我正在使用Parse.com向一个简单的iOS网页浏览应用发送推送通知，但我似乎无法通过我的JSON有效负载显示我的消息文本：

{
    "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);

    }

}

以下是我的日志返回的内容：

收到推送网址：http://www.google.com

收到消息:( null）

UserInfo：{       aps = {           alert =“推送标题到此处”;       };       url =“http://www.google.com”;   }

我在这里错过了什么吗？

Answer 1

如果您格式化UserInfo，则会看到它有两个键aps和url。 url的值为http://www/google.com，aps的值也是字典。此词典的键alert的值为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

UIAlertView

1 个答案: