From: John Croson [mailto:[EMAIL PROTECTED]
I have a simple query:
SELECT id,year,month,day,cat_id FROM events
LEFT JOIN calendar_cat ON
events.cat=calendar_cat.cat_id
WHERE year=YEAR(CURDATE())
AND month=MONTH(CURDATE())
AND day=DAYOFMONTH(CURDATE())
AND cat_id='2' OR cat_id='5'
AND
Hi John,
I think you missed on the precedence of AND/OR
if you change to
AND (cat_id='2' OR cat_id='5' )
it should work as you want it to
/Johan
John Croson wrote:
I have a simple query:
SELECT id,year,month,day,cat_id FROM events
LEFT JOIN calendar_cat ON
events.cat=calendar_cat.cat_id
WHERE
Mike Johnson wrote:
From: John Croson [mailto:[EMAIL PROTECTED]
I have a simple query:
SELECT id,year,month,day,cat_id FROM events
LEFT JOIN calendar_cat ON
events.cat=calendar_cat.cat_id
WHERE year=YEAR(CURDATE())
AND month=MONTH(CURDATE())
AND day=DAYOFMONTH(CURDATE())
AND cat_id='2' OR
At 09:35 AM 2/4/2004, John Croson wrote:
I have a simple query:
SELECT id,year,month,day,cat_id FROM events
LEFT JOIN calendar_cat ON
events.cat=calendar_cat.cat_id
WHERE year=YEAR(CURDATE())
AND month=MONTH(CURDATE())
AND day=DAYOFMONTH(CURDATE())
AND cat_id='2' OR cat_id='5'
AND approved='1'
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On Wednesday 04 February 2004 16:03, mos wrote:
At 09:35 AM 2/4/2004, John Croson wrote:
I have a simple query:
SELECT id,year,month,day,cat_id FROM events
LEFT JOIN calendar_cat ON
events.cat=calendar_cat.cat_id
WHERE year=YEAR(CURDATE())
is a crosstab is what you want?
pls. give the result (output of the query) that you expect from the example
and what you are actually getting.
regds,
-Original Message-
From: Charles Kline [mailto:[EMAIL PROTECTED]
Sent: Saturday, March 15, 2003 20:53
To: [EMAIL PROTECTED]
Subject:
Charles-
So, you have a table unique with respect to people (a person appears at
most once), and a table with interests (some number of interests are
associated a person in the first table. This is a one-to-many
relationship. You need to join on the person in the first table.
Something like
Jeff,
Thanks. This is ALMOST doing what I need.
Suppose that the person could have from 0 - 5 records in
tbl_peoples_interests and I need to have each one of those interests as
a different name so I can use it in my PHP form?
Thanks,
Charles
On Saturday, March 15, 2003, at 11:30 AM, Jeff
Charles-
OK, let's say a person has `n' interest id entries in
tbl_people_interests. It seems like you're wanting a result set with
one row per person and (n + 2) (assuming you're selecting first and
last name with the interests). I'm no UEbergeek with respect to SQL,
but I don't think there's
On Thu, Sep 12, 2002 at 11:46:40AM -0700, Kip Krueger wrote:
I need to query a mysql db in the following fashion ...
find me all records whose column 'n' have the letters 'XY' as the first two
characters. where column 'n' is just a short string
to clarify ...
if column 'n' has the
10 matches
Mail list logo
