php - HTTP post 请求 objective-c

Question

我正在尝试向我的服务器发出一个 http 请求，然后将其插入到 mysql 表中。我没有收到任何错误。Objective-c 返回连接成功。

在 php 脚本中没有错误，但它没有将行添加到 mysql 服务器。我已经仔细检查了表和属性名称。怎么没有加入mysql？

  if (FBSession.activeSession.isOpen) {
        [[FBRequest requestForMe] startWithCompletionHandler:
         ^(FBRequestConnection *connection, NSDictionary<FBGraphUser> *user, NSError *error) {
             if (!error) {

                NSString *currentId = [user objectForKey:@"id"];
                NSString *currentName = user.name;


                    NSString *post = [NSString stringWithFormat:@"&name=%@&id=%@",currentName,currentId];
                    NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

                    NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];

                    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
                    [request setURL:[NSURL URLWithString:[NSString stringWithFormat:@"http://www.ratemyplays.com/api/post_user.php"]]];

                    [request setHTTPMethod:@"POST"];
                    [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
                    [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
                    [request setHTTPBody:postData];
                    NSURLConnection *conn = [[NSURLConnection alloc]initWithRequest:request delegate:self];
                    if(conn)
                    {
                            NSLog(@"Connection Successful");
                    }
                    else
                    {
                            NSLog(@"Connection could not be made");
                    }


                }
         }];

PHP 代码。

 <?php

 $link = mysqli_connect("mysql12.gigahost.dk","dirts","passwordexample","dirts_api");

 $name = (string)$_POST['name'];
 $id = (int)$_POST['id'];

 mysqli_query($link, "INSERT INTO Users (name, userID) VALUES ('$name', '$id')";


 $result = mysqli_query($link,$query);

 ?>

Answer 1

 $name = (string)$_GET['name'];
 $id = (int)$_GET['id'];

你说的是 POST？那么应该是

 $name = (string)$_POST['name'];
 $id = (int)$_POST['id'];