Hello list
I have a problem adding a user with host '%' ...
*** If I add a user with host '%' when trying to connect get this error:
ERROR 1045 (28000): Access denied for user 'myuser'@'localhost' (using
password: YES)
*** If I add the same user with localhost it connects and works fine...
My
Hi,
Miguel Cardenas wrote:
Hello list
I have a problem adding a user with host '%' ...
*** If I add a user with host '%' when trying to connect get this error:
ERROR 1045 (28000): Access denied for user 'myuser'@'localhost' (using
password: YES)
*** If I add the same user with localhost
I have a table of properties that is linked to a table f images with a one
property to many images relationship. I have manged this with nested queries
but want to try and do it on one line. My current query
$query = SELECT * FROM images, properties WHERE images.property_id =
I have a table of properties that is linked to a table f images with a one
property to many images relationship. I have manged this with nested
queries but want to try and do it on one line. My current query
$query = SELECT * FROM images, properties WHERE images.property_id =
On Wed, May 09, 2007 at 07:14:38PM +0200, Martijn Tonies wrote:
I have a table of properties that is linked to a table f images with a one
property to many images relationship. I have manged this with nested
queries but want to try and do it on one line. My current query
$query = SELECT *
No I want all the properties only one regardless of how many images are
attached to them. Think I need a distinct in there somewhere,
- Original Message -
From: Jon Ribbens [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Wednesday, May 09, 2007 6:56 PM
Subject: Re: Query problem
how do I return a single row per property even if it has 3 or 4 images
attached to it.
Please reply to the list instead of directly to me.
You could do a:
select p.from properties p where exists (select i.* from images i
where i.property_id = p.property_id)
I have a table of properties
version is 5.0.22. The script requires 4.x. If Fedora
core 6 allows without dependency issues I have no problem with 4.x.
Anyways I think it would be better if the script can be corrected
because it will be useful for the majority newbies like me.
On 5/7/07, Michael Dykman [EMAIL PROTECTED] wrote
that corresponds to your MySQL server version for the right
syntax to use near 'release char(255) bunary not null,
primary key (fileid,relea' at line 3
MySQL server version is 5.0.22. The script requires 4.x. If Fedora
core 6 allows without dependency issues I have no problem with 4.x.
Anyways I think
This portion of the script gives error.
reate table lxr_releases
(fileid int not null references lxr_files,
release char(255) binary not null,
primary key (fileid,release)
The script is present as attachment if needed. I suspect this script
is for mysql
with that version
on my test setup, so that I can be sure whatever procedure I come up
with will definitely work on the production server.)
I can repeat the problem with this procedure on the test db:
- Import a full mysqldump file from the prodution db
- flush logs
- run a full mysqldump
Mathieu Bruneau [EMAIL PROTECTED] wrote:
Ofer Inbar a écrit :
I can repeat the problem with this procedure on the test db:
- Import a full mysqldump file from the prodution db
- flush logs
- run a full mysqldump with --flush-logs --master-data=2
- do a bunch of stuff that writes
Ofer Inbar a écrit :
Mathieu Bruneau [EMAIL PROTECTED] wrote:
Ofer Inbar a écrit :
I can repeat the problem with this procedure on the test db:
- Import a full mysqldump file from the prodution db
- flush logs
- run a full mysqldump with --flush-logs --master-data=2
- do a bunch
can be sure whatever procedure I come up
with will definitely work on the production server.)
I can repeat the problem with this procedure on the test db:
- Import a full mysqldump file from the prodution db
- flush logs
- run a full mysqldump with --flush-logs --master-data=2
- do a bunch
Dear all,
I create a database called test_aaa to a user. And give him
a permission to create and drop on this database. But I found
he can create a database called test-aaa by his own. Can MySQL
identify this problem?
Also, can I limit user can only drop table in test_aaa but CAN NOT
drop
a production version.
Cheers,
Jay
-Original Message-
From: Jay Pipes [mailto:[EMAIL PROTECTED]
Sent: Wednesday, April 18, 2007 10:04 PM
To: Brent Baisley
Cc: He, Ming Xin PSE NKG; mysql@lists.mysql.com
Subject: Re: Problem on millions of records in one table?
Brent Baisley wrote
.
- Original Message -
From: He, Ming Xin PSE NKG [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Tuesday, April 17, 2007 11:03 PM
Subject: Problem on millions of records in one table?
Hi,all
The number of the records in one table increase constantly. As
evaluated, the amount would
[EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Tuesday, April 17, 2007 11:03 PM
Subject: Problem on millions of records in one table?
Hi,all
The number of the records in one table increase constantly. As
evaluated, the amount would increase to at least 30 millions within one
year. So we
Message-
From: Brent Baisley [mailto:[EMAIL PROTECTED]
Sent: Wednesday, April 18, 2007 9:37 PM
To: He, Ming Xin PSE NKG; mysql@lists.mysql.com
Subject: Re: Problem on millions of records in one table?
It all depends on how complicated your data and searches are. I've got
tables that add 2-3
: Problem on millions of records in one table?
Brent Baisley wrote:
It all depends on how complicated your data and searches are. I've got
tables that add 2-3 million per day and I don't have performance
problems. Although we only retain at most 500 millions records, not a
full years worth
this problem ,such as using Partition, dividing a big table and etc. Any
help or idea would be greatly appreciated.
Best Regards
mingxin
. Or need we some other
solutions to avoid this problem ,such as using Partition, dividing a
big table and etc. Any help or idea would be greatly appreciated.
30 million records is not a problem, assuming you have enough memory
and disk. You may still want to condider partitioning, if only
Hi,
I want to backup my databases with mysqldump, but mysqldump won't run
because I use the 'local-infile=1' option in the my.cnf file:
[client]
port= 3306
socket = /tmp/mysql.sock
local-infile= 1
This is because I want php and other clients to use local-infile. This
It looks to me that local-infile is a command-line parameter
to mysql client
mysql --local-infile -u user dbname
I've not been able to find this option elsewhere.
--
Later
Mogens Melander
+45 40 85 71 38
+66 870 133 224
On Mon, April 16, 2007 12:14, Mark van Herpen wrote:
Hi,
I want to
Mark van Herpen [EMAIL PROTECTED] writes:
Hi,
I want to backup my databases with mysqldump, but mysqldump won't run
because I use the 'local-infile=1' option in the my.cnf file:
[client]
port= 3306
socket = /tmp/mysql.sock
local-infile= 1
This is because I want
Balaraju,
When the line
set pack=''
is removed in favour of
declare pack text default '';
MySQL 5.0.37 accepts the procedure. That looks like a MySQL bug. Will
you report it, or shall I?
When the sproc runs, the first NULL passed to CONCAT() sets pack to
NULL. CONCAT_WS() ignores NULLs.
Hi all,
I am getting problem with this Procedure, i am reading the values using a
cursor, and appending them to variable. but i am getting null as output. Can
anybody please tell where is the wrong.
create procedure activity1()
begin
declare done int default 0;
declare pack text;
declare name
Hello, I've got a problem with mysql 5 on my debian server.
I've got a forum on this server and untill the database reached about 60
Mo I could dump the database with either phpmyadmin or with the command :
mysql -u user -p'password' databasename backup_date.sql
My last backup that worked
Richard, there's no inherent problem around 60 MB - I routinely dump data
ranging from a few KB to a few GB.
One thing I note is that you are using the mysql command, which is the
interactive database client. You want to use mysqldump, the client program
that dumps data from the database in SQL
Mogens Melander
+45 40 85 71 38
+66 870 133 224
On Wed, April 11, 2007 21:11, Richard wrote:
Hello, I've got a problem with mysql 5 on my debian server.
I've got a forum on this server and untill the database reached about 60
Mo I could dump the database with either phpmyadmin
Hello,
I'm having some problems with views.
Could please someone give me some help?
I created two tables and a view.
I created the view using WITH CASCADED CHECK OPTION
because I need to update some columns of the view.
When I do update 2007_marche set data_invio_teramo = current_date where
:16 p.m.
To: mysql@lists.mysql.com
Subject: Re: Problem with authentication
Mahmoud Badreddine wrote:
Hello to all
I had an old MySQL 4.0 running on a Windows Machine.
I removed that version and I installed the MySQL 5.0 .
When I went to run phpMyAdmin this is the error I receive.
#1251
Hello to all
I had an old MySQL 4.0 running on a Windows Machine.
I removed that version and I installed the MySQL 5.0 .
When I went to run phpMyAdmin this is the error I receive.
#1251 - Client does not support authentication protocol requested by server;
consider upgrading MySQL client
Mahmoud Badreddine wrote:
Hello to all
I had an old MySQL 4.0 running on a Windows Machine.
I removed that version and I installed the MySQL 5.0 .
When I went to run phpMyAdmin this is the error I receive.
#1251 - Client does not support authentication protocol requested by
server;
consider
. I'm tuning
it.
Stephen Liu
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--- Chris [EMAIL PROTECTED] wrote:
Retried as follows;
[EMAIL PROTECTED] ~]# mysqladmin -u root --password mypassword
Enter password:
Typing either mypassword or YES prompted;
mysqladmin: connect to server at 'localhost' failed
error: 'Access denied for user
satimis wrote:
Typing either mypassword or YES prompted;
mysqladmin: connect to server at 'localhost' failed
error: 'Access denied for user 'root'@'localhost' (using password: YES)'
* end *
That is asking for your *old* password.
[/quote]
I did not create password before. This is a new
--- Chris [EMAIL PROTECTED] wrote:
- snip -
Try this:
mysqladmin -u root password xyz
That will change your password to 'xyz'.
Before received this email I tried;
[EMAIL PROTECTED] ~]# mysqladmin -u root -p
mysqladmin Ver 8.41 Distrib 4.1.20, for redhat-linux-gnu on x86_64
Copyright
[EMAIL PROTECTED] ~]# mysqladmin -u root password xyz
mysqladmin: connect to server at 'localhost' failed
error: 'Access denied for user 'root'@'localhost' (using password: NO)'
* end *
So you *have* set the root password before.
From your previous emails:
# mysqladmin -u root password
In news:[EMAIL PROTECTED],
satimis [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] ~]# /etc/init.d/mysqld restart
Stopping MySQL: [FAILED]
Starting MySQL:[ OK ]
Ok, lets try it a different way, using only the
--- Chris [EMAIL PROTECTED] wrote:
- snip -
So you *have* set the root password before.
From your previous emails:
# mysqladmin -u root password yourrootsqlpassword
That set the password to yourrootsqlpassword.
If you need to reset it, see documentation:
On Thursday 29 March 2007 11:38:21 Stephen Liu wrote:
http://dev.mysql.com/doc/refman/4.1/en/resetting-permissions.html
I tried learning resetting the password according to above URL. I
can't find set command on this box. I don't know which package
provides it so finally I surrendered.
`cat /mysql-data-directory/host_name.pid`
mysqld_safe –skip-grant-tables
mysqladmin -u root flush-privileges password “newpassword”
OR is there any other way to fix this problem. TIA
B.R.
satimis
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In news:[EMAIL PROTECTED],
satimis [EMAIL PROTECTED] wrote:
Pls advise how to locate mysql=E2=80=99s hostname.pid file?
Try looking in standard directories such as:
/var/run/mysqld or /var/run/mysql or /var/run
The filename can be in form of 'hostname.pid', but it will rather be named
this message in context:
http://www.nabble.com/Problem-on-creating-root-password-tf3479106.html#a9712041
Sent from the MySQL - General mailing list archive at Nabble.com.
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To unsubscribe:http://lists.mysql.com/[EMAIL
In news:[EMAIL PROTECTED],
satimis [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] /]# mysqld_safe =E2=80=93skip-grant-tables
[1] 5119
[EMAIL PROTECTED] /]# Starting mysqld daemon with databases from
/var/lib/mysql STOPPING server from pid file
/var/run/mysqld/mysqld.pid 070328 19:39:12 mysqld
.log.old
* end *
Read /var/log/mysql.log
Can't find anything related.
satimis
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MySQL General Mailing List
Depending on the version you use MySQL will see a definition of
varchar(25) as 25 bytes or 25 characters. I believe this changed from
4.1 to 5.0 respectively but I am not sure.
THis could be the root of the problem
Boyd
CONFIDENTIALITY NOTICE: This email attached documents may
In news:[EMAIL PROTECTED],
satimis [EMAIL PROTECTED] wrote:
[EMAIL PROTECTED] /]# mysqld_safe =E2=80=93skip-grant-tables
[1] 5119
[EMAIL PROTECTED] /]# Starting mysqld daemon with databases from
/var/lib/mysql STOPPING server from pid file
/var/run/mysqld/mysqld.pid 070328 19:39:12 mysqld
34
070328 23:56:09 [Note] /usr/libexec/mysqld: Shutdown complete
070328 23:56:09 mysqld ended
* end *
satimis
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Sent from the MySQL - General mailing list archive at Nabble.com
;
mysqladmin: connect to server at 'localhost' failed
error: 'Access denied for user 'root'@'localhost' (using password: YES)'
* end *
Still failed
satimis
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Sent from the MySQL - General
Retried as follows;
[EMAIL PROTECTED] ~]# mysqladmin -u root --password mypassword
Enter password:
Typing either mypassword or YES prompted;
mysqladmin: connect to server at 'localhost' failed
error: 'Access denied for user 'root'@'localhost' (using password: YES)'
* end *
That is asking
://www.softtools.com
-Original Message-
From: Micah Stevens [mailto:[EMAIL PROTECTED]
Sent: Sunday, March 25, 2007 9:23 PM
To: [EMAIL PROTECTED]; mysql@lists.mysql.com
Subject: Re: Sorting Problem
This doesn't work?
SELECT businesses.name from businesses
left join links using (businessID)
left join
Micah,
I think I understand, here is my query, however I get a syntax error report
and it is not clear what the problem is. Any suggestions:
select * from business_names
left join business_entries using business_entries.bus_id
left join business_categories using
the column must be in both tables.
A.
On 3/26/07, Sid Price [EMAIL PROTECTED] wrote:
Micah,
I think I understand, here is my query, however I get a syntax error
report
and it is not clear what the problem is. Any suggestions:
select * from business_names
left join business_entries using
PROTECTED]
Cc: Micah Stevens; mysql@lists.mysql.com
Subject: Re: Sorting Problem
Try:
select * from business_names
left join business_entries using (bus_id)
left join business_categories using (bcat_id)
where business_categories.bcat_id=17
order by business_names.organisation
You must have
On 3/26/07, Anil D [EMAIL PROTECTED] wrote:
Varchar = 0 bytes
I don't think this is right, see below.
Charset used: UTF8
UTF8 means that some characters may be two bytes, see below.
Note: When consider even the size Varchar(m) = m+1 bytes, the size of row
has reached 35,000 bytes.
; mysql@lists.mysql.com
*Subject:* Re: Sorting Problem
Try:
select * from business_names
left join business_entries using (bus_id)
left join business_categories using (bcat_id)
where business_categories.bcat_id=17
order by business_names.organisation
You must have brackets around column name
anything BUT root, even though the usernames I
assigned didn't even exist. So I don't think I was really connected.
I've seen some sources that talk about symlinking MySQL (or phpMyAdmin?) if
Apache is on a different server. Could that be the problem? If not, does anyone
know how I can troubleshoot
be in multiple
categories.
The problem I am having is after having queried the entries table for all
the entries for a given category I query the names table for each entry to
display the business name and address, I can not figure a way to sort the
displayed data by company name, The entries table
[mailto:[EMAIL PROTECTED]
Sent: Sunday, March 25, 2007 9:23 PM
To: [EMAIL PROTECTED]; mysql@lists.mysql.com
Subject: Re: Sorting Problem
This doesn't work?
SELECT businesses.name from businesses
left join links using (businessID)
left join categories using (categoryID)
where category.name
Hello,everyone!
I want to use Mysql C API to connect to Mysql server. When I use the
following command :
gcc test.c -o test.bin -I/usr/local/mysql/include -L/usr/local/mysql/lib
-lmysqlclient -lz,
I got the erro information:
-5.0.37/cmd-line-utils'
make[1]: *** [all-recursive] Error 1
make[1]: Leaving directory `/tmp/ben/mysql-5.0.37'
I tried looking in the config.log file, i *think* these are the problem
parts...
configure:4135: gcc -c -O3conftest.c 5
conftest.c:2: error: syntax error before me
configure:4141
BY from_user_id;
In your original query I think you meant to select from, not to, since to will
be 1;
- Original Message -
From: John Kopanas [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Sunday, March 11, 2007 12:59 PM
Subject: Re: Finding a Solution To A Distinct Problem of Mine
I
to will
be 1;
- Original Message -
From: John Kopanas [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Sunday, March 11, 2007 12:59 PM
Subject: Re: Finding a Solution To A Distinct Problem of Mine
I think I got it:
SELECT * FROM (SELECT * FROM messages ORDER BY created_at DESC) as
messages
:
SELECT DISTINCT to_user_id, created_at FROM messages WHERE to_user_id
= 1 ORDER BY created_at;
But the problem is that I only get distincts when I only have
to_user_id in the SELECT clause. Any suggestions.
I need to return everything on the latest row that has a distinct
from_user_id :-).
--
John
, created_at FROM messages WHERE to_user_id
= 1 ORDER BY created_at;
But the problem is that I only get distincts when I only have
to_user_id in the SELECT clause. Any suggestions.
I need to return everything on the latest row that has a distinct
from_user_id :-).
--
John Kopanas
[EMAIL PROTECTED]
http
want to return the rows with the
newest created_at.
I thought this would work:
SELECT DISTINCT to_user_id, created_at FROM messages WHERE to_user_id
= 1 ORDER BY created_at;
But the problem is that I only get distincts when I only have
to_user_id in the SELECT clause. Any suggestions.
I
SQLServer2000:Item ColorQuantity
--
TableBlue 124
TableRed 223
20Hi;
I had a working installation of MySQL, and then I zapped it. I found out the
hard way that when you delete symlinks, you also delete the files to which they
are linked :( I wiped my entire /usr/local/include dir. I'm using ports on
FreeBSD 6.2. As it happens, I had a complete copy of the
The log files are named hostname.err and must be located inside your
data dir, which can be inside your /usr/local directory or in the /var/
Carlos
Scott Johnson wrote:
20Hi;
I had a working installation of MySQL, and then I zapped it. I found out the
hard way that when you delete
Hi;
[After waiting 4 hours, I don't think my post posted, so I'm re-posting]
I
had a working installation of MySQL, and then I zapped it. I found out
the hard way that when you delete symlinks, you also delete the files
to which they are linked :( I wiped my entire /usr/local/include dir.
I'm
I have a database online and have a strange mysql problem.
When I connect remotely from my desktop (same browse_database.php page) I
get over 1000 results with the query below but when I run the same page on
the remoted server this value is almost halved to 520.
any ideas what is hapening
Scott Hamm wrote:
Line 48:
48, 14.729606, 10.1.1.22, 10.182.167.209, TCP, pop3
[SYN,
ACK] Seq=0 Ack=1 Win=16384 Len=0 MSS=1460
Is the line 48 is different than other lines?
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I've been trying to import fields that contains the comma character ','
inside double quotes '', with the results following:
code
mysqlimport --fields-optionally-enclosed-by= --fields-terminated-by=,
--lines-terminated-by=\r\n --ignore-lines=1 --user=root --password shark
c:\documents and
Hey Scott -
I dont think you can use , with mysqlimport as a field separator if it is
part of the data. use something else - I used the pipe | character...
This is what worked for me:
C:\mysqlimport --fields-enclosed-by= --fields-terminated-by=|
--lines-terminated-by=\r\n --ignore-lines=1
Hi,
I want to search from a table , I have a table with a column that each
record filled with different character set ,is it possible
to know what are their character set and then is it possible to change all
of them
to a character set ,how can I do that?
these records show like below that I
Hi,
on server B you need to set log-slave-updates to pass statements from A
over B to D.
BTW: I can recommend you to use replicate-wild-do-table=db_name.*
instead of replicate-do-db, otherwise statements with db prefix before
tables won't replicate over B to D. RTM please
Filip
I'm
Hi all,
I have a serious problem with MySQL 4.1.22 on Solaris 9, SPARC. I have a
database server with around 10 or so databases, each with roughly 20-50
tables. At fairly consistent intervals, MySQL gets stuck with 100s of
queries in the Opening tables state. I have checked the value of my
You should have:
log-slave-updates
on the slave/master host(s)
Best regards,
Irek Slonina
Irek,
Thanks for the follow up. Someone else has mentioned it but I just noticed
that it was offlist.
Hi,
on server B you need to set log-slave-updates to pass statements from
A
over B to D.
BTW: I can recommend you to use replicate-wild-do-table=db_name.*
instead of replicate-do-db, otherwise statements with db prefix before
tables won't replicate over B to D. RTM please
Filip
I'm working with two distinct databases on four different boxes.
Databases A on Server A needs to be present everywhere (Server B, C and
D). Database B needs to be present on Server C.
So I setup replication from Server A to Server B and Server D and then I
setup replication from Server B to
CASCADE, SET NULL and NO ACTION, as foreign key constraints of
the table that I want to update.
Any help or idea of how can I solve this problem?
Thank you very much in advance,
Alicia
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can I solve this problem?
Thank you very much in advance,
Alicia
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Fortunately I have solved this problem. I think it was something about
being a two field composed foreign key and updating these two fields at
the same time gave some kind of error.
I solve it by adding a temporary row in the referenced table with one of
the fields already changed
remain hanging. Mysqld is totally using one of the
processors. The strange thing is that the load moves from one processor
to the other. But this might be some Windows weirdness. I know about
the undecideability of the stopping problem, but I find the increase of
the execution time from
MySQL 4.1.22 seems to treat the following characters as equal
(comparing them as varchar values):
U+03B7 (206 183) greek small letter eta
U+1F75 (225 189 181) greek small letter eta with accent oxia
U+1FC4 (225 191 135) greek small letter eta with accent persispomeni
and accent
Sven Fuchs wrote:
These characters are stored/retrieved correctly. But they are wrongly
regarded the same character by statements like SELECT * FROM tablename
WHERE fieldname LIKE '[greek small eta]'
The database's character-set is set to UTF-8 Unicode (utf8) and the
table's and varchar
Am 05.02.2007 um 18:11 schrieb Chris White:
SELECT * FROM tablename WHERE fieldname LIKE BINARY '[greek small
eta]'
that *should* ( see disclaimer ;) ) give you what you need
Yes, it does.
I should have also asked for SELECT DISTINCT fieldname ... in the
first place, but looking at your
PM
To: mysql@lists.mysql.com
Subject: WHERE (NOT) EXISTS problem
Having a very bad time with the subject sorts of queries.
Here is a simple reproduction of the problem for me.
Perhaps I'm blind/stupid while looking at the docs,
or there's a bug...
mysql version 5.0.24-standard
simple schema
@lists.mysql.com
Subject: Re: WHERE (NOT) EXISTS problem
Michael Fischer wrote:
Having a very bad time with the subject sorts of queries.
Here is a simple reproduction of the problem for me.
Perhaps I'm blind/stupid while looking at the docs,
or there's a bug...
mysql version 5.0.24-standard
simple
Having a very bad time with the subject sorts of queries.
Here is a simple reproduction of the problem for me.
Perhaps I'm blind/stupid while looking at the docs,
or there's a bug...
mysql version 5.0.24-standard
simple schema:
mysql desc people
Michael Fischer wrote:
Having a very bad time with the subject sorts of queries.
Here is a simple reproduction of the problem for me.
Perhaps I'm blind/stupid while looking at the docs,
or there's a bug...
mysql version 5.0.24-standard
simple schema:
mysql desc people
Hello all,
I have struck with a big problem with MySQL 5.0.22 server on RHEL 3.
After an upgradation from MySQL 4.1.11 to MySQL 5.0.22 almost all the queries
are struggling to execute and the DB server is clogged. Below is an example of
what is happening.
This query usd to execute very fast
-functions.html
Thanks
ViSolve DB Team.
- Original Message - From: Gabriel Linder
[EMAIL PROTECTED]
To: MySQL List mysql@lists.mysql.com
Sent: Friday, January 19, 2007 9:43 PM
Subject: Request problem (with \\)
Hello list,
I am currently trying to fix a bug in a search function
On 1/19/07, Daniel Culver [EMAIL PROTECTED] wrote:
Are you working on a Mac? If so, logging in as root is not good
enough. You must have set up and log into the root user account of
your Mac computer or server.
The OP is talking about managing MySQL accounts with MySQL
Administrator. MySQL
Actually on my Mac; 10.4.6 ppc the MySQL Administrator will only open
to a connection through localhost. The localhost connection will
accept any authorised user to connect, but, MySQL Administrator will
only assume the privileges of the account I am working from (I have
several). It has
Hello list,
I am currently trying to fix a bug in a search function with a request
like this one :
select * from forum where topic like '%[...]%' ;
where [...] is a string escaped by mysql_real_escape_string (C API) and
topic is a varchar field (not null).
It works, but there is a bug if
Are you working on a Mac? If so, logging in as root is not good
enough. You must have set up and log into the root user account of
your Mac computer or server. This, if anything is a Apple problem and
advantage. The Administrator is opening to the account you are in,
the login in window
in as root is not good
enough. You must have set up and log into the root user account of
your Mac computer or server. This, if anything is a Apple problem and
advantage. The Administrator is opening to the account you are in,
the login in window will accept any correct password combination
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