Re: [SQL] problem with join & count - Mailing list pgsql-sql

From Dimitri
Subject Re: [SQL] problem with join & count
Date
Msg-id 36BA1BBB.323A76EB@france.sun.com
Whole thread Raw
In response to RE: [SQL] problem with join & count  ("Jackson, DeJuan" <djackson@cpsgroup.com>)
List pgsql-sql
Jackson, DeJuan wrote:
>
> try:
> SELECT t1.name, count(t2.tag)
>   FROM t1, t2
>  WHERE t1.number = t2.number
>  GROUP BY t1.number, t1.name
______________^
your query will never work :)), you have to miss t1.number...

Best regards!
(dim)

>
>         -DEJ
>
> > -----Original Message-----
> > From: pat@patoche.org [mailto:pat@patoche.org]
> > Sent: Thursday, February 04, 1999 2:39 PM
> > To: pgsql-sql@postgreSQL.org
> > Subject: [SQL] problem with join & count
> >
> >
> > Hi,
> >
> > I don't succeed to build an SQL to do the task i describe below.
> > If someone can helps, thanks in advance.
> >
> > I have two tables
> >
> > number  | name
> > --------------
> >  1      | toto
> >  2      | titi
> >  3      | tutu
> >
> > (here the number is a primary key)
> >
> > and
> >
> > number  |  tag
> > ---------------
> >  1      | alpha
> >  1      | beta
> >  2      | gamma
> >  3      | zeta
> >  3      | epsilon
> >
> > the number in the left column is the same of the number in
> > the first table,
> > but can appear in that second table many times.
> >
> > i need an sql query that will return:
> >
> > name   |  how_often
> > -------------------
> >  toto  | 2
> >  titi  | 1
> >  tutu  | 2
> >
> > that is, the result table should contain all names present in
> > the first table
> > and then the number of times the number associated with the
> > name appears in
> > the second table.
> >
> > Thanks a lot for your help.
> >
> > Patrick
> >
> > /\//\/\/\\/\/\//\/\\/\/\\/\\/\//\/\\/\//\/\\/\//\/\\/\//\/\\
> > Patrick M.   pat@patoche.org    http://www.patoche.org/
> >
> >

--
=====================================================
 Dimitri KRAVTCHUK  (dim)           Sun Microsystems
 Benchmark Engineer                 France
 dimitri@France.Sun.COM
=====================================================

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