Hello.
Start points for you problem:
http://dev.mysql.com/doc/refman/5.0/en/can-not-connect-to-server.html
http://dev.mysql.com/doc/refman/5.0/en/starting-server.html
> socket=/home.dbdata/mysql/mysql.sock
What is your client thinks about the location of mysqld socket?
Put the same
Hi all
First usage of MySQL. Newbie in Linux as well as MySQL. Using CentOS 4.2
with MySQL 4.1.12, rpm install.
Changed the data directory.
My /etc/my.conf
---
[mysqld]
datadir=/home.dbdata/mysql
socket=/home.dbdata/mysql/mysql.sock
# Default to using old password format for compa
without any problems and
waiting for connections.
>>ps -aux shows that mysql is running
And for the same reason I mentioned above, you are seeing this.
Hope this helps.
sujay
-Original Message-
From: Sanjay Arora [mailto:[EMAIL PROTECTED]
Sent: Monday, November 14, 2005 4:46 PM
To: MySql
Hi all
First usage of MySQL. Newbie in Linux as well as MySQL. Using CentOS 4.2
with MySQL 4.1.12, rpm install.
Changed the data directory.
My /etc/my.conf
[mysqld]
datadir=/home.dbdata/mysql
socket=/home.dbdata/mysql/mysql.sock
# Default to using old password format for compatibility with mys
Bob Cohen wrote:
Here's my question. I wrote this query:
Select * FROM name WHERE last LIKE "d" AND choice1="2" OR choice2="2" OR
choice3="2";
What I'm looking for are records that satisfy the LIKE "d" condition
But then, Only one of the three other con
hi,
you should set parentheses on it
Select * FROM name WHERE last LIKE "d" AND (choice1="2" OR choice2="2" OR
choice3="2");
without it interprets it as
like "d" and choice1="2"
or
choice2="2"
or
choice3="2"
Regards
Ingo Thierack
--On Freitag, 7. Mai 2004 13:51 -0400
Bob Cohen wrote:
Here's my question. I wrote this query:
Select * FROM name WHERE last LIKE "d" AND choice1="2" OR choice2="2" OR
choice3="2";
To predict the outcome of this query requires knowledge of the precedence of
AND vs. OR in mysql. I can't find it documented anywhere, but I think AND
Thanks to Bernard, Sunmaia, and Ingo! I knew the query wasn't quite
right.
Bob Cohen
b.p.e.Creative
http://www.bpecreative.com
Design and production services for the web
Put creative minds to work for you
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscr
You need parenthesis around cond1 OR cond2 OR cond3 otherwise effectively you
are going to get all records matching the following:
last "d" AND choice1="2" + choice2="2" + choice3="2"
Therefore, you select statement should be written:
SELECT * FROM name WHERE last LIKE "d" AND (choice1="2" OR ch
Here's my question. I wrote this query:
Select * FROM name WHERE last LIKE "d" AND choice1="2" OR choice2="2" OR
choice3="2";
What I'm looking for are records that satisfy the LIKE "d" condition
But then, Only one of the three other conditions need be t
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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(C
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'
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 cat
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 ye
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
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'
ORDER BY year,month,day ASC
Results:
+-
: Newbie Query Question
Hi all,
My first post to this list.
I have a query which I can't quite get to work. I was hoping that
someone here could educate me a bit. I have two tables.
tbl_people
| id| fname|lname|
---
| 1 |
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
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 Kil
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
SE
Hi all,
My first post to this list.
I have a query which I can't quite get to work. I was hoping that
someone here could educate me a bit. I have two tables.
tbl_people
| id| fname|lname|
---
| 1 |frank |smith |
t
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'
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 value "YR12345" don't get that record.
if column 'n' has the value "XY
At 10.08 21/11/01 -0700, Steve Meyers wrote:
>Wait for 4.1, it will have multi-table updates in it. Until then,
>you'll have to use two separate queries.
>
>Steve
i'm also new to mysql but i think you can use MERGE tables.
Couldn't you ?
==
dott
Wait for 4.1, it will have multi-table updates in it. Until then,
you'll have to use two separate queries.
Steve
On Wed, 2001-11-21 at 07:28, Charles Allen wrote:
> Hi,
>
> A v. basic question from a mySql newbie:
>
> I want to update a table based on the contents of another table. The SQL I
is correct?
Thanks,
Charles.
-Original Message-
From: Charles Allen [mailto:[EMAIL PROTECTED]]
Sent: 21 November 2001 14:29
To: Mysql (E-mail)
Subject: Newbie query
Hi,
A v. basic question from a mySql newbie:
I want to update a table based on the contents of another table. The SQL I
am
Hi,
A v. basic question from a mySql newbie:
I want to update a table based on the contents of another table. The SQL I
am trying to execute is:
"update servers set servers.responsecount=servers.responsecount+1 where
servers.gameid = lastsnapshot.gameid;"
ie: Increment a responsecount for rows
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