SQL question: group by horizontally

Question

Hi,
I'm having trouble figuring out a way to display the result of a join in the following tabular format. Can someone please assist?
SQL> select * from tab1;
GNAME SNO
ABC 5
DEF 2
ABC 3
SQL> select * from tab2;
GNAME SNO
ABC 10
XYZ 5
DEF 15
My current output using a UNION ALL:
SQL> select gname, count(*) sno1 from tab1 group by gname
2 union all
3 select gname, count(*) sno2 from tab2 group by gname;
GNAME SNO1
ABC 2
DEF 1
ABC 1
DEF 1
XYZ 1
The output I'm expecting:
GNAME SNO1          SNO2
ABC 2 1
DEF 1 1
XYZ 0 1
Thanks in advance.

Answer 1

szsl wrote:
My current output using a UNION ALL:
SQL> select gname, count(*) sno1 from tab1 group by gname
2 union all
3 select gname, count(*) sno2 from tab2 group by gname;you need this to identify all distinct gnames.
with tab1 as (
select 'ABC' as gname, 5 as sno from dual union all
select 'DEF' as gname, 2 as sno from dual union all
select 'ABC' as gname, 3 as sno from dual
), tab2 as (
select 'ABC' as gname, 10 as sno from dual union all
select 'XYZ' as gname, 5 as sno from dual union all
select 'DEF' as gname, 15 as sno from dual
select a.gname, sum( case when tab1.gname is not null then 1 else 0 end) as cnt_1,
sum( case when tab2.gname is not null then 1 else 0 end) as cnt_2
from (
  select distinct gname from (select gname from tab1 union all select gname from tab2)
) a
left join tab1 on tab1.gname = a.gname
left join tab2 on tab2.gname = a.gname
group by a.gname
order by a.gname
;output:
GNAME CNT_1 CNT_2
ABC       2     2
DEF       1     1
XYZ       0     1