Groups | Search | Server Info | Keyboard shortcuts | Login | Register [http] [https] [nntp] [nntps]
Groups > comp.databases.ms-sqlserver > #1542
| Newsgroups | comp.databases.ms-sqlserver |
|---|---|
| Date | 2013-08-14 13:27 -0700 |
| References | <492c4449-7052-4337-a9b1-ddee6371ba20@googlegroups.com> <b189b500-6a6e-49bf-abac-3be46e0164fe@googlegroups.com> |
| Message-ID | <8b5a1edb-43f2-4975-a211-b7bdf7aa5fb7@googlegroups.com> (permalink) |
| Subject | Re: how many new IDs participated |
| From | migurus <migurus@yahoo.com> |
On Wednesday, August 14, 2013 2:18:21 AM UTC-7, Ross Presser wrote: > On Tuesday, August 13, 2013 9:43:44 PM UTC-4, migurus wrote: > > > I have a series of Customer ID / Date records showing participation of customers on the daily basis. I need to find out how many new customers participated each day. > > [snip] > > > Any ideas? I was trying to self-join but can not quite wrap my brain around it. > > > > > > Thanks in advance. > > > > WITH X1 AS ( > > SELECT ID, MIN(DT) AS FirstDT > > FROM @T AS T > > GROUP BY ID > > ), > > X2 AS ( > > SELECT DISTINCT DT FROM @T AS T > > ) > > SELECT X2.DT, COUNT(X1.ID) > > FROM X2 > > LEFT JOIN X1 ON X2.DT = X1.FirstDT > > GROUP BY X2.DT > > > > DT > > ---------- ----------- > > 2013-08-07 2 > > 2013-08-08 0 > > 2013-08-09 1 > > 2013-08-12 0 > > 2013-08-13 1 > > Warning: Null value is eliminated by an aggregate or other SET operation. > > > > (5 row(s) affected) > > > > The value for 2013-08-09 is accurate; look closely at your data, 332551 is represented twice on that day. Thank you! Works perfectly well.
Back to comp.databases.ms-sqlserver | Previous | Next — Previous in thread | Find similar
how many new IDs participated migurus <migurus@yahoo.com> - 2013-08-13 18:43 -0700
Re: how many new IDs participated Ross Presser <rpresser@gmail.com> - 2013-08-14 02:18 -0700
Re: how many new IDs participated migurus <migurus@yahoo.com> - 2013-08-14 13:27 -0700
csiph-web