Group by alias (Oracle)

Question

How to \'group by\' a query using an alias, for example:

select count(*), (select * from....) as alias_column 
from table 
group by alias_column

Answer 1

If you don't have to use an alias you could do it this way:

select  
EXTRACT(year from CURRENT_DATE), count(*) from something
group by EXTRACT(year from CURRENT_DATE)
order by EXTRACT(year from CURRENT_DATE)

Instead of using alias and subquery.

Answer 2

select
  count(count_col),
  alias_column
from
  (
  select 
    count_col, 
    (select value from....) as alias_column 
  from 
    table
  ) as inline
group by 
  alias_column

Grouping normally works if you repeat the respective expression in the GROUP BY clause. Just mentioning an alias is not possible, because the SELECT step is the last step to happen the execution of a query, grouping happens earlier, when alias names are not yet defined.

To GROUP BY the result of a sub-query, you will have to take a little detour and use an nested query, as indicated above.

Answer 3

select count(*), (select * from....) as alias_column 
from table 
group by (select * from....)

In Oracle you cannot use an alias in a group by clause.

Answer 4

To use an alias in Oracle you need to ensure that the alias has been defined by your query at the point at which the alias is being used.

The most straightforward way to do this is to simply treat the original query as a subquery -- in this case,

select count(*), (select * from....) as alias_column 
from table 
group by (select * from....)

becomes

select count, alias_column 
from
  (select count(*) as count, (select * from....) as alias_column 
  from table)
group by alias_column 

I can't speak to the performance implications, but it's very quick to write if you're trying to re-use an alias in your query - throw everything in parentheses and jump up a level...

Answer 5

Nest the query with the alias column:

select count(*), alias_column
from
( select empno, (select deptno from emp where emp.empno = e.empno) as alias_column
  from emp e
)
group by alias_column;