Mark A Matte
markamatte at hotmail.com
Sat Feb 2 11:38:37 CST 2008
John, If I understand what your asking...this isn't quite a normal PIVOT situation. A PIVOT uses data(the values) as row headers and column headers. Your example sounds like you want to use data as column headers and column names as row headers? If I had to do this is access I would use iif statements to determine what field the data came from. Good luck, Mark A. Matte ************************SQL********************* TRANSFORM Count(*) AS Expr1 SELECT IIf(IsNull([55_64])=True,IIf(IsNull([65_74])=True,"45_54","65_74"),"55_64") AS FieldName FROM tblJC GROUP BY IIf(IsNull([55_64])=True,IIf(IsNull([65_74])=True,"45_54","65_74"),"55_64") PIVOT IIf(IsNull([55_64])=True,IIf(IsNull([65_74])=True,[45_54],[65_74]),[55_64]); ************************SQL********************* > From: jwcolby at colbyconsulting.com > To: accessd at databaseadvisors.com; dba-sqlserver at databaseadvisors.com > Date: Sat, 2 Feb 2008 11:47:53 -0500 > Subject: [AccessD] SQL Server Pivot > > I have a bunch of columns that break out ages into bands. For example: > > Presence_of_adults_age_65_74_specific > Presence_of_adults_age_55_64_specific > Presence_of_adults_age_45_54_specific > etc > > Each column has codes 1,2,3 > > I need counts of each column, for each value > > column Cnt1 Cnt2 Cnt3 > 65_74 45 3 23 > 55_64 103 48 211 > 45_54 20 1 49 > > etc > > I think the Pivot statement is going to do that for me but I can't seem to > wrap my mind around the syntax. Can anyone point me to a web page that > clearly explains this or simply write a SQL statement that does this? > > Thanks > > John W. Colby > Colby Consulting > www.ColbyConsulting.com > > -- > AccessD mailing list > AccessD at databaseadvisors.com > http://databaseadvisors.com/mailman/listinfo/accessd > Website: http://www.databaseadvisors.com _________________________________________________________________ Climb to the top of the charts! Play the word scramble challenge with star power. http://club.live.com/star_shuffle.aspx?icid=starshuffle_wlmailtextlink_jan