Start here: http://www.postgresql.org/community/lists/subscribe/ Change the drop down from SUBSCRIBE to UNSUBSCRIBE and put in the rest of the required information.
On Fri, May 22, 2015 at 8:06 AM, Tim Rowe <digi...@gmail.com> wrote: > Sorry to post this on the list, but I can't find any way of unsubscribing > -- I've looked in messages, on the community home pages and on a web > search, but all I find is a lot of other subscribers with the same problem. > > How do I unsubscribe from this list, please? > > On 22 May 2015 at 11:46, Nicklas Avén <nicklas.a...@jordogskog.no> wrote: > >> >> >> 2015-05-22 skrev Albe Laurenz : >> >> Nicklas Avén wrote: >> >> I was a little surprised by this behavior. >> >> Is this what is supposed to happen? >> >> >> >> This query returns what I want: >> >> >> >> with >> >> a as (select generate_series(1,3) a_val) >> >> ,b as (select generate_series(1,2) b_val) >> >> ,c as (select generate_series(1,1) c_val) >> >> select * from a >> >> inner join c on a.a_val=c.c_val >> >> full join b on a.a_val=b.b_val >> >> ; >> >> >> >> I get all values from b since it only has a full join and nothing else. >> >> >> >> But if I change the order in the joining like this: >> >> >> >> with >> >> a as (select generate_series(1,3) a_val) >> >> ,b as (select generate_series(1,2) b_val) >> >> , c as (select generate_series(1,1) c_val) >> >> select * from a >> >> full join b on a.a_val=b.b_val >> >> inner join c on a.a_val=c.c_val >> >> ; >> >> >> >> also b is limited to only return value 1. >> >> >> >> I thought that the join was defined by "on a.a_val=c.c_val" >> >> and that the relation between b and the rest wasn't affected by that >> last inner join. >> >> >> >> I use PostgreSQL 9.3.6 >> >> >> >> Is this the expected behavior? >> > >> >Yes. >> > >> >In >> > >> http://www.postgresql.org/docs/current/static/queries-table-expressions.html#QUERIES-JOIN >> >you can read: >> > >> > "In the absence of parentheses, JOIN clauses nest left-to-right." >> > >> >So the first query will first produce >> > >> > a_val | c_val >> >-------+------- >> > 1 | 1 >> > >> >and the FULL JOIN will add a row for b_val=2 with NULL a_val. >> > >> >The second query will first produce >> > >> > a_val | b_val >> >-------+------- >> > 1 | 1 >> > 2 | 2 >> > 3 | >> > >> >an since none but the first row matches a_val=1, you'll get only that >> row in the result. >> > >> >Yours, >> >Laurenz Albe >> >> >> Thank you! >> >> Sorry for not finding it myself, but now I understand why it behaves >> like this :-) >> >> Thanks >> >> Nicklas >> > > > > -- > Tim Rowe > -- My sister opened a computer store in Hawaii. She sells C shells down by the seashore. If someone tell you that nothing is impossible: Ask him to dribble a football. He's about as useful as a wax frying pan. 10 to the 12th power microphones = 1 Megaphone Maranatha! <>< John McKown
