I want to do something like this:
select id,
count(*) as total,
FOR temp IN SELECT DISTINCT somerow FROM mytable ORDER BY somerow LOOP
sum(case when somerow = temp then 1 else 0 end) temp,
END LOOP;
from mytable
group by id
order by id
I created working select:
select id,
count(*) as total,
sum(case when somerow = 'a' then 1 else 0 end) somerow_a,
sum(case when somerow = 'b' then 1 else 0 end) somerow_b,
sum(case when somerow = 'c' then 1 else 0 end) somerow_c,
sum(case when somerow = 'd' then 1 else 0 end) somerow_d,
sum(case when somerow = 'e' then 1 else 0 end) somerow_e,
sum(case when somerow = 'f' then 1 else 0 end) somerow_f,
sum(case when somerow = 'g' then 1 else 0 end) somerow_g,
sum(case when somerow = 'h' then 1 else 0 end) somerow_h,
sum(case when somerow = 'i' then 1 else 0 end) somerow_i,
sum(case when somerow = 'j' then 1 else 0 end) somerow_j,
sum(case when somerow = 'k' then 1 else 0 end) somerow_k
from mytable
group by id
order by id
this works, but it is 'static' - if some new value will be added to 'somerow' I will have to change sql manually to get all the values from somerow column, and that is why I'm wondering if it is possible to do something with for loop.
So what I want to get is this:
id somerow_a somerow_b ....
0 3 2 ....
1 2 10 ....
2 19 3 ....
. ... ...
. ... ...
. ... ...
So what I'd like to do is to count all the rows which has some specific letter in it and group it by id (this id isn't primary key, but it is repeating - for id there are about 80 different values possible).