将iphone应用程序连接到mysql数据库

我有一个应用程序我想连接到iphone应用程序我已经做了代码和php代码的问题是，我总是得到不正确的密码警报视图。我正在输入正确的用户名和密码，但它再次显示错误警报视图

      NSString *post =[NSString stringWithFormat:@"UserName=%@&UserPassword=%@",userNameTextField.text, userPasswordTextFiled.text];

     NSString *hostStr = @"http://www.myurl.com/emrapp/connect.php?";
    hostStr = [hostStr stringByAppendingString:post];
     NSData *dataURL =  [NSData dataWithContentsOfURL: [ NSURL URLWithString: hostStr ]];    
     NSString *serverOutput = [[NSString alloc] initWithData:dataURL encoding: NSASCIIStringEncoding];
     if([serverOutput isEqualToString:@"Yes"]){
        UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Congrats" message:@"You are authorized "
                                                          delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
    [alertsuccess show];
    [alertsuccess release];


    } else {
        UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Error"  message:@"Username or Password Incorrect"
                                                          delegate:self     cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
         [alertsuccess show];
        [alertsuccess release];

     }

我的服务器端代码是

    <?php
     $con = mysql_connect("emriphone.db.6420177.hostedresource.com","emriphone","Light12-");
   if (!$con)
    {
   die('Could not connect: ' . mysql_error());
  }

  mysql_select_db("emriphone", $con);

   $u=$_GET['UserName'];
   $pw=$_GET['UserPassword'];


   $query = sprintf("SELECT UserName,UserPassword from appUsers WHERE UserName='%s' AND  UserPassword='%s'", mysql_real_escape_string($u),mysql_real_escape_string($pw));



  $login=mysql_query($query,$con) or die(mysql_error());


  if(mysql_num_rows($login)==1){

   $row =mysql_fetch_assoc($login);
    echo 'YES'; exit;
   }

  else{
   echo'NO';exit;
  }

  mysql_connect($con);


  ?>

---

尝试使用以下代码

           NSString *data = [[NSString stringWithFormat:@"UserName=%@&UserPassword=%@",userNameTextField.text, userPasswordTextFiled.text];

           NSData *postData = [data dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];  

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

    // preaparing URL request to send data.
NSMutableURLRequest *request = [[[NSMutableURLRequest alloc] init] autorelease];

NSString *url = [NSString stringWithFormat:@"http://www.myurl.com/emrapp/connect.php?"];

            [request setURL:[NSURL URLWithString:url]];
            [request setHTTPMethod:@"POST"];
            [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
            [request setHTTPBody:postData];
            [request setTimeoutInterval:7.0];

NSURLResponse *response;// = [[NSURLResponse alloc] init];
NSError *error;// = [[NSError alloc] init;

NSData *urlData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

    NSString *str=[[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];


     NSLog(@"Login response:is %@",str);

此链接可能会帮助您这是我的链接亲爱的，但我有相同的问题，我得到的字符串中的服务器输出是@“”；只有这样，这就是为什么我认为它给出了错误，你能帮我解决这个问题吗？你从服务器得到的响应是字符串“serverOutput”？@；这是当我登录时，但当我直接给php脚本提供用户名和密码值并与数据库匹配时，它返回了是的，我给你这个结果只是尝试回显你的查询，看看它响应了什么？这与我告诉你的打印空间相同，只是因为有@“”在strI中，我对php了解不多，我认为php代码中存在一些问题？意味着您试图从服务器回显的任何内容都是空白的？