De første tre grupper med de højeste karakterer skal have specifikke point

Det var udfordrende.

For at løse det brugte jeg flere metoder:

1. En CASE-erklæring at konvertere gruppens plads i top 3 til point.
2. Rækkenummerering med en variabel .
3. INDRE OG Venstre JOIN for at flette resultaterne sammen.

Følgende forespørgsel blev testet på din violin og virker:

SELECT t1.`id`,  t1.`name`,  t1.`group1`,
       t1.`section`, t1.`MARKS`, `t_group_points`.`points`

FROM   `students` t1

#--- Join  groups' points to the students
LEFT JOIN (
    (
        #---- Join all groups and give points to top 3 avg's groups ----
        SELECT `t4`.`group1`, `t_points`.`points`
        FROM   (SELECT   `t3`.`group1`, AVG(`t3`.`marks`) AS `avg`
                FROM     `students` `t3`
                WHERE    (`t3`.`section` = 'class1') AND
                         (`t3`.`group1` IS NOT NULL)
                GROUP BY `t3`.`group1`) `t4`

        INNER JOIN (
              #---------- Select top 3 avarages ----------
              (SELECT `top`.`avg`,
                      #-- Convert row number to points ---
                      CASE @curRow := @curRow + 1  
                           WHEN '1' THEN 5
                           WHEN '2' THEN 3
                           WHEN '3' THEN 1
                           ELSE NULL END 'points'

              FROM (SELECT DISTINCT `t_avg`.`avg`
                    FROM   (SELECT   `t2`.`group1`, AVG(`t2`.`marks`) AS `avg`
                            FROM     `students` `t2`
                            WHERE    (`t2`.`section` = 'class1') AND
                                     (`t2`.`group1` IS NOT NULL)
                            GROUP BY `group1`) `t_avg`
                    ORDER BY `avg` DESC
                    LIMIT 0, 3) `top`, (SELECT @curRow:=0) r
              ) AS `t_points`)
         ON (`t_points`.`avg` = `t4`.`avg`)      
    ) AS `t_group_points`)
ON (`t_group_points`.`group1` = `t1`.`group1`)