Question

我想传递从以下网址收到的json数据并传递＆＃34;用户名＆＃34;数据到SecondViewController。我只想传递数据，何时连接成功。

我遇到以下问题：

无论URL是对还是错，我都会被定向到SecondViewController。我只想在连接成功并获取数据时转到SecondViewController。
当我在URL连接代码中放入以下行时，我无法在SecondViewController上接收数据：

SecondViewController * destination=[segue destinationViewController];

destination.displayName=[info objectForKey:@"username"];

如果我之前放置了以下代码，NSURL可以工作，我在SecondViewController中看到了这条消息。但我通过destination.displayName=@"Testing Message"。

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {

    NSURL *url = [NSURL URLWithString:@"http://localhost/?user_id=1"];
    NSURLRequest *request = [NSURLRequest requestWithURL:url];
    [NSURLConnection sendAsynchronousRequest:request
                                       queue:[NSOperationQueue mainQueue]
                           completionHandler:^(NSURLResponse *response,
                                               NSData *data, NSError *connectionError)
     {
         if (data.length > 0 && connectionError == nil)
         {
             NSDictionary *info = [NSJSONSerialization JSONObjectWithData:data
                                                                      options:0
                                                                     error:NULL];
             //this is my second view controller
             SecondViewController * destination=[segue destinationViewController];

             destination.displayName=[info objectForKey:@"username"];

         }//if connected

     }];//end of connection

}//end of prepareforsegue

实现这一目标的最佳方法是什么？将URL连接代码放在prepareForSegue中是否正确？

Answer 1

我建议这个解决方案：

- (void)fetchRequest {
    NSURL *url = [NSURL URLWithString:@"http://localhost/?user_id=1"];
    NSURLRequest *request = [NSURLRequest requestWithURL:url];
    [NSURLConnection sendAsynchronousRequest:request queue:[NSOperationQueue mainQueue] completionHandler:^(NSURLResponse *response, NSData *data, NSError *connectionError) {
        if (data.length > 0 && !connectionError) {
            NSDictionary *info = [NSJSONSerialization JSONObjectWithData:data options:0 error:NULL];
            [self performSegueWithIdentifier:@"SecondViewController" sender:info];
        }//if connected
    }];//end of connection
}

- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender {
    if ([segue.identifier isEqualToString:@"SecondViewController"]) {
        if ([sender isKindOfClass:[NSDictionary class]]) {
            //this is my second view controller
            SecondViewController * destination = [segue destinationViewController];
            destination.displayName=[sender objectForKey:@"username"];
        } // end of check on Sender Type
    } // end identifier check ..
}//end of prepareforsegue

祝你的应用好运。

URL连接后无法将Web服务数据传递给另一个viewcontroller

1 个答案: