Group by as part of a where clause?

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Group by as part of a where clause?

Rob-350472

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Points: 7197
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June 26, 2008 at 10:00 am

#193739

I have a table, which simplified boils down to something like this:
Year | Data
1992 string1
1992 string2
1992 string3
1993 stringN
1993 stringX
1993 stringZ
1994 string P
1994 string X
1995 stringN
1996 stringN
I should point out that the stuff in the data column is insignificant - it will never be null and is simply a varchar, the entries above are just what I felt like putting in :p
Now, I have a where clause when selecting from this table where by I specify that the other fields in the table must be > a specific value, non null etc to get the universe of data I want, this works great.
However, I want do do one more thing...
If I do the following:
SELECT
Year, count(data)
FROM mytable
Group by year
Then I get something like
1992| 4
1993| 2
Etc...
Now I basically want to modify the the where clause for the first table to say that there must be > say 2 entries of data for each year - i.e. something which would with the above data set would only return the rows for 1992 and 1993.
I've had a go with a few ideas I had but all basically didn't work what so ever...
Any ideas?!

Viewing 11 posts - 1 through 10 (of 10 total)

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Josh B SSCrazy Points: 2426 More actions · Answer 1

How about using the Having clause to pull only those with a count > 2...

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 2

I'd need a little more specifics. The table DDL, smaple data (in the form of insert statements that can be cut, pasted, and executed to populate the table), and the expected output based on the sample data.

😎

Steve Jones - SSC Editor SSC Guru Points: 736212 More actions · Answer 3

HAVING is probably what you want, but please post some examples and let us know if that works or doesn't.

Rob-350472 SSCertifiable Points: 7197 More actions · Answer 4

Lynn Pettis (6/26/2008)
I'd need a little more specifics. The table DDL, smaple data (in the form of insert statements that can be cut, pasted, and executed to populate the table), and the expected output based on the sample data.
😎

Excuse my ignorance, but...

DDL = ? just the column names?!

What's the easiest way for me to generate the insert statements which you could copy / paste?!!

Matt Miller (4) SSC Guru Points: 124210 More actions · Answer 5

DDL = data definition language. i.e. the CREATE TABLE statement(s) that would allow us to easily recreate your table structures for testing.

Easiest way to do that is to right-click on the table, and use the "script table as CREATE" to a new query window, then paste the results here.

That being said - HAVING is your answer (that's been mentioned before)

as in -

...

HAVING count(*)>2

----------------------------------------------------------------------------------
Your lack of planning does not constitute an emergency on my part...unless you're my manager...or a director and above...or a really loud-spoken end-user..All right - what was my emergency again?

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 6

Read this article, it will help you with everything you need to allow us to help you:

http://www.sqlservercentral.com/articles/Best+Practices/61537/">

http://www.sqlservercentral.com/articles/Best+Practices/61537/

😎

Rob-350472 SSCertifiable Points: 7197 More actions · Answer 7

OK, based on that, here is my attempt! (please excuse me if I am wrong - there's a first time for everything n all that :p!)

To create the table:

SET ANSI_NULLS OFF

GO

SET QUOTED_IDENTIFIER ON

GO

SET ANSI_PADDING ON

GO

CREATE TABLE [dbo].[#tmpRR_Testing](

[fund_vintage] [char](4) NULL,

[fund_id] [int] NULL,

[fund_name] [varchar](60) NULL

) ON [PRIMARY]

GO

SET ANSI_PADDING OFF

Now let's populate it with some data:

INSERT INTO #tmprr_testing

(fund_vintage, fund_id, fund_name)

SELECT '1993','333','XYZ Fund II' UNION ALL

SELECT '1989','15433','TTT Partners Fund I' UNION ALL

SELECT '1996','29133', 'ABC Fund' UNION ALL

SELECT '1998','31133','BGF I' UNION ALL

SELECT '1989','45833','TRE Partners' UNION ALL

SELECT '1999','48133','TTYE Partners IV' UNION ALL

SELECT '2000','50333','CMYK Partners' UNION ALL

SELECT '1995','55533',' PTYER Fund II' UNION ALL

SELECT '1999','55833','PSPSDS Partners' UNION ALL

SELECT '1995','59033','CVCVVVVS Partners I' UNION ALL

SELECT '1999','59133','BNBNBN Partners II' UNION ALL

SELECT '1997','63633','BNBNBNBNB Fund I'

This is a reduced set of data, both in terms of rows of data and fields present, as the others don't come in to the equation really.

I tried the above and it works, except it creates duplicates, doh! - but a select distinct gets round this... and produces 12 rows.

You will notice there are 3 entries for fund vintage 1999, the others all have less.

I basically want to set a value (2 in this example I suppose) which the count of the funds must exceed.

I.e. if you run this:

SELECT

fund_vintage, count(fund_vintage) as NumFunds

FROM #tmpRR_testing

GROUP BY fund_vintage

(OK so you get 6 due to the duplicates, but you get the idea)

So my results set from the first query would only return the 1999 vintage funds.

Does that make sense?!

Thanks!

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 8

Steve was right, all you need is a HAVING clause as shown below:

SELECT

fund_vintage,

count(fund_vintage) as NumFunds

FROM

#tmpRR_Testing

GROUP BY

fund_vintage

HAVING

count(fund_vintage) > 2

I should have been able to see that as well. I guess I was having a Missouri moment (show me).

😎

Rob-350472 SSCertifiable Points: 7197 More actions · Answer 9

I could see that, it was how exactly I obtained the rest of the data in the table, so instead I did the above and dumped it into another table - joined this table to the main table where the count > X so as to remove the fund_vintages which had less than X funds.

Perhaps a roundabout way to do it, but it seemed to work!

Ah well, at least I know how to post table creation and population code for next time I run in to a problem!! 🙂

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 10

Which is a good thing. I wish more people needing help would read that article and follow the procedures outlined. Jeff Moden did an great job with that article.

😎