Q:

SQL Group By with an Order By

I have a table of tags and want to get the highest count tags from the list.
Sample data looks like this
id (1) tag ('night')
id (2) tag ('awesome')
id (3) tag ('night')

using
SELECT COUNT(*), `Tag` from `images-tags`
GROUP BY `Tag`

gets me back the data I'm looking for perfectly. However, I would like to organize it, so that the highest tag counts are first, and limit it to only send me the first 20 or so.
I tried this...
SELECT COUNT(id), `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY COUNT(id) DESC
LIMIT 20

and I keep getting an "Invalid use of group function - ErrNr 1111"
What am I doing wrong?
I'm using MySQL 4.1.25-Debian

A:

In all versions of MySQL, simply alias the aggregate in the SELECT list, and order by the alias:
SELECT COUNT(id) AS theCount, `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY theCount DESC
LIMIT 20

A:

MySQL prior to version 5 did not allow aggregate functions in ORDER BY clauses.
You can get around this limit with the deprecated syntax:
SELECT COUNT(id), `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY 1 DESC
LIMIT 20

1, since it's the first column you want to group on.

A:

I don't know about MySQL, but in MS SQL, you can use the column index in the order by clause.  I've done this before when doing counts with group bys as it tends to be easier to work with.
So
SELECT COUNT(id), `Tag` from `images-tags`
GROUP BY `Tag`
ORDER BY COUNT(id) DESC
LIMIT 20

Becomes
SELECT COUNT(id), `Tag` from `images-tags`
GROUP BY `Tag`
ORDER 1 DESC
LIMIT 20

