Question

我已在我的应用程序中实现了推送通知。

我接受了以下的aps：

{
    aps =     {
        alert = "Bhai kaisa hai notification a rha hai!";
        customtitle = "Customtitle here";
        keyid = 8;
        sound = default;
    };
}

我想将keyid和customtitle传递给视图控制器我的代码：

  NSString * titletopass = [userInfo objectForKey:@"customtitle"];
    NSString * keytopass = [userInfo objectForKey:@"keyid"];

    //NSString *titletopass=[getpass objectForKey:@"customtitle"];
    PostsDetailsViewController *obj = [[PostsDetailsViewController alloc]init];

    obj.Viewtitle = titletopass;
    obj.key=keytopass;

   NSString *identifier;
    identifier=@"postsdetails";
    UIStoryboard *    storyboardobj=[UIStoryboard storyboardWithName:@"Main" bundle:nil];
    obj = [storyboardobj instantiateViewControllerWithIdentifier:identifier];
    [self.window setRootViewController:obj];

它不会将数据传递给viewcontroller，而keytopass + titletopass也是空的。

提前致谢..

Answer 1

将obj设置为PostDetailsViewController的实例后，您将obj设置为从故事板中实例化的postdetails视图控制器。

如果您打算使用故事板，则只应该从中创建实例。获得实例后，您可以设置属性。

例如：

NSString * titletopass = [[userInfo objectForKey:@"aps"] objectForKey:@"customtitle"];
NSString * keytopass = [[userInfo objectForKey:@"aps"] objectForKey:@"keyid"];

UIStoryboard *storyboard = [UIStoryboard storyboardWithName:@"Main" bundle:nil];
PostsDetailsViewController *obj = (PostDetailsViewController *)[storyboard instantiateViewControllerWithIdentifier:@"postdetails"];

obj.Viewtitle = titletopass;
obj.key=keytopass;

[self.window setRootViewController:obj];

远程通知Ios

1 个答案: