Echoing student ranks

Question

I echo a rank of a student whose regd is equal to $regd. In fact, this is working code. However, I was advised by a friend that the MySQL statements Distinct and Group By should not be used together. But as a newbie, I could not figure out how I could implement it without using Distinct because it does not return rows without Distinct.

Can anyone suggest how I can improve this code?

<?php 
mysql_select_db($database_dbconnect, $dbconnect);
$query_myrank = "SELECT Distinct regd, Name_of_exam, 
                    Name_of_Student, TOTALSCORE, Rank 
             FROM (SELECT *, IF(@marks = (@marks := TOTALSCORE), 
                          @auto, @auto := @auto + 1) AS Rank 
                   FROM (SELECT Name_of_Student, regd, 
                               Name_of_exam, SUM(Mark_score) AS TOTALSCORE 
                         FROM cixexam, (SELECT @auto := 0, 
                                               @marks := 0) AS init 
                                        GROUP BY regd 
                         ORDER BY TOTALSCORE DESC) t) AS result 
             HAVING (Name_of_exam='First Terminal Exam' OR 
                     Name_of_exam='First Term Test')";

$myrank = mysql_query($query_myrank, $dbconnect) or die(mysql_error());

$i = 0;
$j = 0;
$data = array();
while($row_myrank = mysql_fetch_assoc($myrank))
{
   $data[$i] = $row_myrank;
   if(isset($data[$i - 1]) 
        && $data[$i - 1]['TOTALSCORE'] == $data[$i]['TOTALSCORE'])
{
   $data[$i]['Rank'] = $j;
   }else{
     $data[$i]['Rank'] = ++$j;
}
   $i++;
}
foreach($data as $key => $value)
{
 if($value['regd'] == $regd)
    {
    echo $value['Rank'];
    }
  }
 ?>    

Answer 1

Since you use it in a subquery, I don't see the problem. If you used both in the same query, that would be redundant. But the subquery needs to use it to obtain the required data.

Did the person give a reason as to why he/she thinks this, or was it simply a general statement?

The only thing I can think of is that DISTINCT is rather slow compared to GROUP BY: stackoverflow.com/questions/1887953/