How to ORDER BY Different column than in the group by clause.

Question

Hello, I have this sql statement
SELECT COUNT([Tags]) AS CNT, [Tags]
FROM [Tags], [Images]
WHERE
([Images].[Tags] LIKE '%' + [Tags].[LongTag] + '%') AND
([Tags].AllowTagPost = 'True' )
GROUP BY [Images].[Tags]
HAVING COUNT([Tags]) > 30
ORDER BY [Tags].[LastTagPost] DESC
It returns the tags, the number of occurances sorted by count.  However, I want to sort it by a column in the Tags table named LastTagPost (DATETIME)  The idea is to get the tag that hasn't been used in the longest time and also has more occurances
than 30 in the image table.  When I try to order by this column i get an error that i'm sure most of you are all to familer with.
Error:
Column "Tags.LastTagPost" is invalid in the ORDER BY clause because it is not contained in either an aggregate function or the GROUP BY clause.
Any help would be greatly appreciated.

Answer 1

Couldn't you just include the LastTagPost as a MIN?
DECLARE @tags TABLE (LastTagPost DATETIME, LongTag VARCHAR(100), allowTagPost CHAR(5))
DECLARE @images TABLE (tags VARCHAR(10))
SELECT COUNT(i.Tags) AS CNT, i.Tags, MIN(t.LastTagPost) AS lastTagPost
FROM @Tags t
INNER JOIN @Images i
ON i.Tags LIKE '%' + t.LongTag + '%'
WHERE t.AllowTagPost = 'True'
GROUP BY i.Tags
HAVING COUNT(i.Tags) > 30
ORDER BY LastTagPost DESC
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