Hi,
How do I access hierarchical data under PostgreSQL?
Does it have SQL command similar to Oracle's CONNECT BY?
Any help is appreciated
Eric
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I have two tables, users and options. The common element between the
tables is "username". I want to select the "email" from "user" but only
if the "mailing_list" option is set to true in the "options" table.
Here are the tables:
select username, email from users;
username| e
Metnetsky <[EMAIL PROTECTED]> writes:
> It's for a class and my professor has a thing for brain teaser type
> questions.
Incidentally, TAs and Profs aren't stupid, and have been known to check on
newsgroups and mailing lists for students asking for people to do their
homework for them.
--
greg
On Monday, Mar 3, 2003, at 15:32 US/Pacific, Josh Berkus wrote:
Check out the thread: Re: [SQL] OUTER JOIN with filter
in today's list; this relates to your problem. Then try your query as:
I'll read through this, thank you.
LEFT OUTER JOIN (SELECT * FROM Customer_Month_Summary
WHRE CMS.Mon
One suggestion I'll make about your data model -- I'm not sure it would
actually help this query, but might help elsewhere:
WHERE ( C.Disabled > '2003-02-28'
OR C.Disabled IS NULL
)
Don't use NULL values like this. Most databases don't index NULLs (Oracle) or
even if they do, don't
> -> Merge Join (cost=6106.42..6335.30 rows=2679 width=265)
(actual time=859.77..948.06 rows=1 loops=1)
Actually another problem, notice the big discrepancy between the estimated row
and the actual rows. That's because you have the big OR clause so postgres
figures there's
Need some help with schema's.
I want to be able make a user have a default schema other than public.
I want to be able to have several schema's with the same group of tables
define in each one.
(Example)
schema a (users 1,2,3)
table a
table b
table c
schema b (users 4,5,6
"Hepworth, Mike" <[EMAIL PROTECTED]> writes:
> I want to be able make a user have a default schema other than public.
If the schema name is the same as the user name then this happens
automatically.
Otherwise, you want to set a search_path value for the user --- see
ALTER USER and read up on the
Dawn Hollingsworth <[EMAIL PROTECTED]> writes:
> Nested Loop (cost=0.00..466.51 rows=1 width=24) (actual
> time=226.39..1018072.99 rows=9353 loops=1)
> -> Subquery Scan t1 (cost=0.00..0.01 rows=1 width=0) (actual
> time=225.99..345.86 rows=9353 loops=1)
> -> Result (cost=0.00..0.01
