How to create columns for different fields without applying the pivoting function

Question

So i have a question for the tough hearted! I have had an issue with this concept for awhile and I need it to be cleared... The following code shows students who have more than

Answer 1

A SQL query returns a fixed number of columns. If you want just three additional columns, then you can use dynamic aggregation:

with t as (
      SELECT s.studentnumber as studentnr, p.firstname AS name,
             sl.gradenumber as gradenumber, l.text as language,
             dense_rank() over (partition by s.studentnumber, p.firstname, sl.gradenumber order by l.text) as seqnum
      FROM student s JOIN 
           pupil p
           ON p.id = s.pupilid JOIN
           pupillanguage pl 
           ON pl.pupilid = p.id JOIN
           language l 
           ON l.id = pl.languageid JOIN
           schoollevel sl
           ON sl.id = p.schoollevelid
     )
select studentnr, name, gradenumber,
       max(case when seqnum = 1 then language end) as language_1,
       max(case when seqnum = 2 then language end) as language_2,
       max(case when seqnum = 3 then language end) as language_3
from t
group by studentnr, name, gradenumber;

However, if you want a variable number, then you need crosstab or dynamic SQL. I might also suggest that you consider arrays instead.

Answer 2

If you have a fixed number of columns, you can use an array:

select studentnr, name, gradenumber, 
       languages[1] as language_1,
       languages[2] as language_2,
       languages[3] as language_3,
       languages[4] as language_4,
       languages[5] as language_5
FROM (       
  SELECT s.studentnumber as studentnr, 
         p.firstname AS name,
         sl.gradenumber as gradenumber,
         array_agg(DISTINCT l.text) as languages
  FROM student s
      JOIN pupil p ON p.id = s.pupilid    
      JOIN pupillanguage pl on pl.pupilid = p.id
      JOIN language l on l.id = pl.languageid
      JOIN schoollevel sl ON sl.id = p.schoollevelid
  GROUP BY s.studentnumber, p.firstname
) t

Note that distinct is usually not needed when you use group by