Good Day.
Im wondering if someone can assist me.
Ive been using replication for a while now and it tends to fail very easily.
One of my sites lost connectivity for a while and when it came back
obviously replication broke again.
How can I get it to populate all data from master again?
Load
Hi Marcel,
On Fri, Dec 5, 2008 at 2:02 PM, Marcel Grandemange
[EMAIL PROTECTED] wrote:
Good Day.
Im wondering if someone can assist me.
Ive been using replication for a while now and it tends to fail very easily.
Do you have the error messages?.
One of my sites lost connectivity for a
Hi
how can i replace, in a SELECT query, the last 3 numbers with asterisks?
from 0123456789 to 0123456***
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Hi
how can i replace, in a SELECT query, the last 3 numbers with
asterisks?
from 0123456789 to 0123456***
My psudocode for this would be something like:
Select
CONCAT(left$(`field`,(LENGTH(a)-3),'***')
From `table`
But there might be more efficient ideas out there.
Loss of connection should not break replication. It will stop it, but once
you're connected again it should pick right back up where it left off.
To restore you can either dump the entire database from the master and load
into the slave, then use CHANGE MASTER TO to set up the slave.
We
-Original Message-
From: spacemarc [mailto:[EMAIL PROTECTED]
Sent: Friday, December 05, 2008 8:50 AM
To: MySQL
Subject: SELECT with replacement chars
Hi
how can i replace, in a SELECT query, the last 3 numbers with asterisks?
from 0123456789 to 0123456***
[JS]
SELECT
I'm trying to create a trigger (5.0.45) and I've read the documentation
at mysql.com. I keep getting a syntax error, but can't figure out what
the error is. Here's the trigger I'm trying to create:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
On 12/5/08, Lola J. Lee Beno [EMAIL PROTECTED] wrote:
I'm trying to create a trigger (5.0.45) and I've read the documentation at
mysql.com. I keep getting a syntax error, but can't figure out what the
error is. Here's the trigger I'm trying to create:
delimiter //
create trigger
David Giragosian wrote:
I'm no expert, but 'old' is a table, I'm guessing, and it isn't
referenced in the 'from' clause of the query.
could it be that simple...?
David
No . . . 'old' is a virtual table that is the same as the table I'm
doing work on. See
I'm trying to create a trigger (5.0.45) and I've read the documentation
at mysql.com. I keep getting a syntax error, but can't figure out what
the error is. Here's the trigger I'm trying to create:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
can you not use referential integrity for this - assuming the tables are or
can be made to be innodb?
Does the jobposts table have a jobpost_id field, or is it just id? Maybe
it's a typo?
On Fri, Dec 5, 2008 at 11:28 AM, Lola J. Lee Beno [EMAIL PROTECTED] wrote:
David Giragosian wrote:
Jim Lyons wrote:
can you not use referential integrity for this - assuming the tables are or
can be made to be innodb?
The tables are myISAM. These could be changed to innodb but I want to
see if i can get this trigger work.
Does the jobposts table have a jobpost_id field, or is it just
Martijn Tonies wrote:
What is the exact error message?
Here's the latest query:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select count(ad.adsource_id)
from adsource ad, jobposts
Hi,
What is the exact error message?
Here's the latest query:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select count(ad.adsource_id)
from adsource ad, jobposts jp
where
Martijn Tonies wrote:
Hi,
What is the exact error message?
Here's the latest query:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select count(ad.adsource_id)
from adsource ad,
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select count(ad.adsource_id)
from adsource ad, jobposts jp
where ad.adsource_id = jp.adsource_id
and OLD.jobpost_id =
Andy Shellam wrote:
I'm guessing it's the first semi-colon in your IF statement. Does
this work...?
if @counted = 1 then SET dummy = 'Cannot delete this record' end if;
Just a guess!
Andy.
That's not it, unfortunately.
ERROR: You have an error in your SQL syntax; check the manual
ERROR: You have an error in your SQL syntax; check the manual
that corresponds to your MySQL server version for the right
syntax to use near 'if @counted = 1 then SET dummy = 'Cannot
delete this record' end if' at line 1
This works on 5.1.30 and 6.0.7:
create table jobposts(adsource_id
I'm not at all familiar with SQL Server, but you need to make sure that the
encoding/charset you use in MySQL is the same as used in SQL S.
If not the same, then you might need to transcode the data before migrating.
Collation is less important than encoding. - if you get the charset wrong,
then
Greetings,
Please see following sample code and result. Note that x is cast properly but y
isn't. (FYI 0x913386aa3cbbab5a == 10462854425033288538)
Is there a way in MySQL to convert strings to numbers such that it recognizes
decimal, hex, and optionally other bases? I'm thinking of something
Hi,
What is the exact error message?
Here's the latest query:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select count(ad.adsource_id)
from adsource ad, jobposts jp
where
thanks, it works fine (on mysql 4 and 5)
SELECT CONCAT(LEFT(myfield,LENGTH(myfield) - 3), '***') FROM table;
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All -
Thanks in advance for help with a sticky problem.
I'm attempting to create a partitioned table thus:
CREATE TABLE `my_precious_table` (
`id` bigint(20) NOT NULL AUTO_INCREMENT,
`startTimeStamp` datetime NOT NULL DEFAULT '-00-00 00:00:00',
PRIMARY KEY (`id`,`startTimeStamp`)
)
I've never created a partitioned table, but...
$ perror 13
OS error code 13: Permission denied
So I suspect some kind of file-system permissions issue...
Are you sure that the path you are giving isn't relative to the mysql data
dir?
In which case it's trying to use something more
Thanks for responding.
The CREATE TABLE docs for 5.1 say that DATA DIRECTORY and INDEX DIRECTORY
take absolute paths (not relative), and will in fact reject paths containing
the MySQL data dir. Because I'm out of other ideas, I did try creating the
directories under the MySQL data dir and it
Thanks, Martin, but that's not it. As I mentioned in my email, I'm running
as MySQL root user with all priv bits set. I tried your suggestion anyway,
but no change.
Cheers,
- Brad
On Fri, Dec 5, 2008 at 4:34 PM, Martin Gainty [EMAIL PROTECTED] wrote:
Brad-
log into mysql as admin
GRANT
On Fri, December 5, 2008 12:14, Martijn Tonies wrote:
Hi,
What is the exact error message?
Here's the latest query:
delimiter //
create trigger jobposts_control
before delete on jobposts for each row
begin
declare dummy varchar(255);
set @counted = (
select
I think you are missing the point. Where is 'OLD' or 'old' defined?
Before you try to imbed it in a trigger, try the basic query. That seems
to be what its complaining about.
OLD is a virtual table which is only present in a trigger - it's like a
table with the same layout as the table
If I have a table like this:
CREATE TABLE `Test` (
`TestId` bigint(20) default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1
Does using --safe-updates mean that it's impossible for me to ever delete
from it, since it doesn't have a key? For example:
INSERT INTO Test VALUES (1), (2);
mysql
Brad Heintz wrote:
I've googled extensively, searched the list archives, and exhausted every
other avenue I could think of before posting to the list, but am no closer
to an answer. Does anyone have any ideas? Have I missed something in the
docs?
SELinux?
--
Florin Andrei
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