Question

我插入用户输出如下：

  $userName= SanitizeString($userName);
  $pass= SanitizeString($pass);
  $email= SanitizeString($email);

  $userName=mysql_real_escape_string($userName);
  $pass=mysql_real_escape_string($pass);
  $email=mysql_real_escape_string($email);

  $salt = 'SHIFLETT';
  $password_hash = md5($salt . md5($pass.$salt));

 mysql_query("INSERT INTO users (user_name,pass,email,reputation,role,ban,date) VALUES ('$userName', '$password_hash', '$email', '$reputation',   '$role','false','$date')" ) or exit(mysql_error());

这是SanitizeString（$ var）函数：

   function SanitizeString($var)
   {
       $var=stripslashes($var);
       $var=htmlentities($var, ENT_QUOTES, 'UTF-8');
       $var=strip_tags($var);
       return $var;
   }

但是当我尝试使用此查询查找用户密码和名称时。它失败了：

       $user_name=SanitizeString($user_name);
   $pass=SanitizeString($pass);


  $user_name=mysql_real_escape_string($user_name);
  $pass=mysql_real_escape_string($pass);


   $salt = 'SHIFLETT';
   $password_hash = md5($salt . md5($pass.$salt));

   $result=mysql_query("SELECT COUNT(*) AS Result FROM users WHERE user_name='$user_name' AND pass='$password_hash' LIMIT 1") or die(mysql_error());

如果计数大于0意味着它找到了一个结果并且用户应该登录。但是这没有发生..为什么？

更新更多内容：

   if(mysql_num_rows($result)>0)
       {
           echo "Login successful".mysql_num_rows($result);
           return $dataArray=TRUE;
       }
       else
       {
           echo "Login unsuccessful:".mysql_num_rows($result);
       }

Answer 1

如果您想使用GROUP BY，则必须使用COUNT(*)，但您可以获得mysql_num_rows()这样的行数。

$result=mysql_query("SELECT * AS Result FROM users WHERE user_name='$user_name' AND pass='$password_hash' LIMIT 1") or die(mysql_error());
$num_rows = mysql_num_rows($result);

使用php和mysql失败执行语句

1 个答案: