From: | Dimitri <dimitri(at)france(dot)Sun(dot)COM> |
---|---|
To: | Jan Wieck <jwieck(at)debis(dot)com> |
Cc: | pat(at)patoche(dot)org, pgsql-sql(at)postgreSQL(dot)org |
Subject: | Re: [SQL] problem with join & count |
Date: | 1999-02-04 22:10:05 |
Message-ID: | 36BA1ABD.31E18EFC@france.sun.com |
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Thread: | |
Lists: | pgsql-sql |
Hm...
Why do be complex where you can be sample?
Your query is:
select name, count(*) as how_often
from table1, table2
where table1.number = table2.number
group by name;
and that's all, folks!
P.S. Regarding your example, I can say that you are French! Isnt it? :))
"toto, titi" - very unique as names :))
Jan Wieck wrote:
>
> > 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.
>
> Postgres does not (and v6.5 will not) support outer joins or
> subselects in the targetlist, what's required to do that in a
> single SQL statement.
>
> What you could do is using a SQL function that covers the
> count() like this:
>
> create function num_refs (int4) returns int4 as '
> select count(*) from tab2 where id = $1;
> ' language 'SQL';
>
> select item_name, num_refs(id) as how_often from tab1;
>
> I changed 'number' into id and name into item_name because
> both are reserved words.
>
> Jan
>
> --
>
> #======================================================================#
> # It's easier to get forgiveness for being wrong than for being right. #
> # Let's break this rule - forgive me. #
> #======================================== jwieck(at)debis(dot)com (Jan Wieck) #
--
=====================================================
Dimitri KRAVTCHUK (dim) Sun Microsystems
Benchmark Engineer France
dimitri(at)France(dot)Sun(dot)COM
=====================================================
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