Why can't I use alias in a count(*) "column" and reference it in a having clause?

SqlSql ServerAlias

Sql Problem Overview


I was wondering why can't I use alias in a count(*) and reference it in the having clause. For instance:

select Store_id as StoreId, count(*) as _count
	from StoreProduct
	group by Store_id
		having _count > 0

Wouldn't work.. But it works if I remove _count and use count(*) instead.

Sql Solutions


Solution 1 - Sql

See the document referenced by CodeByMoonlight in an answer to your recent question.

The HAVING clause is evaluated before the SELECT - so the server doesn't yet know about that alias.

> 1. First the product of all tables in the from clause is formed. > 2. The where clause is then evaluated to eliminate rows that do not satisfy > the search_condition. > 3. Next, the rows are grouped using the columns in the group by clause. > 4. Then, Groups that do not satisfy the search_condition in the having > clause are eliminated. > 5. Next, the expressions in the select clause target list are > evaluated. > 6. If the distinct keyword in present in the select clause, duplicate rows > are now eliminated. > 7. The union is taken after each sub-select is evaluated. > 8. Finally, the resulting rows are sorted according to the columns > specified in the order by clause.

Solution 2 - Sql

The select clause is the last clause to be executed logically, except for order by. The having clause happens before select, so the aliases are not available yet.

If you really want to use an alias, not that I'd recommend doing this, an in-line view can be used to make the aliases available:

select StoreId, _count
from (select Store_id as StoreId, count(*) as _count
    from StoreProduct
    group by Store_id) T
where _count > 0

Or in SQL Server 2005 and above, a CTE:

; with T as (select Store_id as StoreId, count(*) as _count
    from StoreProduct
    group by Store_id)
select StoreId, _count
from T
where _count > 0

Solution 3 - Sql

You can use the alias for count in the select clause, you just can't use it in the having statement, so this would work

select Store_id as StoreId, count(*) as _count
    from StoreProduct
    group by Store_id
        having count(*) > 0

Solution 4 - Sql

The aliases for the field names is only for naming the columns in the result, they can never be used inside the query. You can't do like this either:

select Store_id as Asdf
from StoreProduct
where Asdf = 42

However, you can safely use count(*) in both places, and the database will recognise that it's the same value, so it won't be calculated twice.

Solution 5 - Sql

Here is my contribution (based on the code posted here):

select * from (
  SELECT Store_id as StoreId, Count(*) as StoreCount 
  FROM StoreProduct
  group by Store_id
  ) data
where data.StoreCount > 0

Solution 6 - Sql

You can use the alias for the aggregates in SQL, but that is just to show the alias in the results headers. But when you want to have a condition with the aggregate function in the having you still need to use the aggregate because it evaluates the function and not the name.

Solution 7 - Sql

In Hive 0.11.0 and later, columns can be specified by position if hive.groupby.orderby.position.alias is set to true.

set hive.groupby.orderby.position.alias=true;
select Store_id as StoreId, count(*) as _count
from StoreProduct
group by 1

I'm don't understand the purpose of your query. Given the context of the query you posted, your condition is not necessary because items that do not exist, i. e. count 0, will never be a result from a query...

Solution 8 - Sql

Probably because that's the way sql defines the namespaces. take, for example:

  select a as b, b as a
    from table
   where b = '5'
order by a

what do a and b refer to? The designers just chose to make the aliases only appear on the "outside" of the query.

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionAndre PenaView Question on Stackoverflow
Solution 1 - Sqlmartin claytonView Answer on Stackoverflow
Solution 2 - SqlShannon SeveranceView Answer on Stackoverflow
Solution 3 - SqlGlenn SlavenView Answer on Stackoverflow
Solution 4 - SqlGuffaView Answer on Stackoverflow
Solution 5 - SqlNew Dawg Learning Old TricksView Answer on Stackoverflow
Solution 6 - SqlJose ChamaView Answer on Stackoverflow
Solution 7 - SqlrafaelvalleView Answer on Stackoverflow
Solution 8 - SqlJimmyView Answer on Stackoverflow