The manual page at http://dev.mysql.com/doc/mysql/en/crashing.html contains
information that should help you find out what is causing the crash.
130620 00:47:21 mysqld_safe Number of processes running now: 0
130620 00:47:21 mysqld_safe mysqld restarted
InnoDB: Error: tablespace size stored in header is 456832 p
Am 20.06.2013 23:47, schrieb Peter:
>> Hello,
>>>
>>> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>>> database directory) from one crashed machine to another.
>>> I find that I cannot start database to get the database data any more.
>>
>>
>>> How did you copy the d
>Hello,
>>
>>I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>>database directory) from one crashed machine to another.
>>I find that I cannot start database to get the database data any more.
>
>
>>How did you copy the database?
>>Manuel
>
>I copy the files ib_logfile
2013/6/20 Peter
>
>2013/6/20 Peter
>
>Hello,
>>
>>I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>>database directory) from one crashed machine to another.
>>I find that I cannot start database to get the database data any more.
>
>
>>How did you copy the databa
Am 20.06.2013 15:18, schrieb Peter:
>>
>>> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>>> database directory) from one crashed machine to another.
>>> I find that I cannot start database to get the database data any more.
>
>>> How did you copy the database?
>>> M
Am 20.06.2013 15:18, schrieb Peter:
>>
>>> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>>> database directory) from one crashed machine to another.
>>> I find that I cannot start database to get the database data any more.
>
>>> How did you copy the database?
>>>
2013/6/20 Peter
>
> 2013/6/20 Peter
>
> Hello,
>
> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
> database directory) from one crashed machine to another.
> I find that I cannot start database to get the database data any more.
>
>
> >How did you copy the database?
>
2013/6/20 Peter
Hello,
>
>I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>database directory) from one crashed machine to another.
>I find that I cannot start database to get the database data any more.
>How did you copy the database?
>Manuel
I copy the files ib_lo
Am 20.06.2013 15:18, schrieb Peter:
> 2013/6/20 Peter
>
> Hello,
>>
>> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
>> database directory) from one crashed machine to another.
>> I find that I cannot start database to get the database data any more.
>
>> How did y
2013/6/20 Peter
> Hello,
>
> I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
> database directory) from one crashed machine to another.
> I find that I cannot start database to get the database data any more.
How did you copy the database?
Manuel
nnoDB. The tail of the system tablespace is
> InnoDB: missing
>
> is there a way to start the database again?
> Thanks for your help in advance
restore your backups
that is one reason why replication exists to have a slave which
a) does not die with the master at a complete crash
Hello,
I copied innodb database (ib_logfile0 ib_logfile1 ibdata1 and the whole
database directory) from one crashed machine to another.
I find that I cannot start database to get the database data any more.
I tried innodb_force_recovery=1
or innodb_force_recovery=4, it doesn't help.
He
Hello,
>Many thanks for your response. Can yo u offer any advice with regards
usage
>of country_codes eg gb and regions, cities etc ? I've been reading up on
>http://en.wikipedia.org/wiki/ISO_3166 etc. Should I be looking to use a
>Surrogate key for countries ? Or the country code like fr for
Many thanks for your response. Can yo u offer any advice with regards
usage of country_codes eg gb and regions, cities etc ? I've been reading
up on http://en.wikipedia.org/wiki/ISO_3166 etc. Should I be looking to
use a Surrogate key for countries ? Or the country code like fr for France
?
Sa
Neil,
Am 21.04.2013 08:47, schrieb Neil Tompkins:
Using joins I can obtain which country each city belongs too. However,
should I consider putting a foreign key in the CITIES table referencing the
countries_id ? Or is it sufficient to access using a join ?
It depends. Adding a reference to
Hi
I'm creating the following basic tables
COUNTRIES
countries_id
name
REGIONS
region_id
countries_id
name
CITIES
cities_id
region_id
Using joins I can obtain which country each city belongs too. However,
should I consider putting a foreign key in the CITIES table referencing the
countries_i
Hello All,
Happy Friday! I know how to do the following query:
>select count(*) from sales where WEEK(sale_date)=15 AND
YEAR(sale_date)=2013;
But can someone tell me I can do a query that will give me:
the count(*) for each week of 2013 so that I end up with:
WEEK | COUNT
1 | 22
2
Perfect! Thank you Larry et all.
Have a great weekend.
2013/4/19 Larry Martell
> On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina
> wrote:
> > Hello All,
> >
> > Happy Friday! I know how to do the following query:
> >
> >>select count(*) from sales where WEEK(sale_date)=15 AND
> > YEAR(sale_dat
On Fri, Apr 19, 2013 at 8:24 AM, Richard Reina wrote:
> Hello All,
>
> Happy Friday! I know how to do the following query:
>
>>select count(*) from sales where WEEK(sale_date)=15 AND
> YEAR(sale_date)=2013;
>
> But can someone tell me I can do a query that will give me:
>
> the count(*) for each w
Hello Richard,
>the count(*) for each week of 2013 so that I end up with:
http://dev.mysql.com/doc/refman/5.1/en/group-by-functions.html
Ilya.
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
On 19.04.2013 06:49, Kapil Karekar wrote:
Though I would recommend not using such names. Some poor guy working
on your application six months down the line is going to wonder why
his queries are failing, spend a day trying to figure out and will
post the same question again to this list :-)
.
Hello,
why these db names created fail but the last one gets success?
mysql> create database 3208e1c6aa32;
ERROR 1064 (42000): 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 '3208e1c6aa32' at line 1
mysql> cre
On 2013-04-19, Doug wrote:
>
> why these db names created fail but the last one gets success?
[snips]
> mysql> create database 3208e1c6aa32;
> mysql> create database 208e1c6aa32;
> mysql> create database 08e1c6aa32;
> mysql> create database 8e1c6aa32;
> mysql> create database e1c6aa32;
https://
On 19-Apr-2013, at 9:14 AM, Doug wrote:
> why these db names created fail but the last one gets success?
>
> mysql> create database 3208e1c6aa32;
> mysql> create database 208e1c6aa32;
> mysql> create database 08e1c6aa32;
> mysql> create database 8e1c6aa32;
These are not working because MySQL is
Nope. That's just granting replication privileges so it can read updates
on all tables on all databases. It cannot select anything.
Why are you trying to connect with a replication slave user?
On 4/2/13 1:47 PM, "Richard Reina" wrote:
>I did a "GRANT REPLICATION SLAVE ON *.* TO 'user'@'19
On Tue, Apr 2, 2013 at 11:30 AM, Richard Reina wrote:
> use DBI;
> my $dbh = DBI->connect( "DBI:mysql:rushload;192.168.0.1", $usrr, $passw, {
> RaiseError => 3 } );
> my $dbs = $dbh->selectcol_arrayref("show databases");
>
> #my $dsn = "dbi:mysql:information_schema:192.168.0.1:3306";
> #my $dbh =
I did a "GRANT REPLICATION SLAVE ON *.* TO 'user'@'192.168.0.23' IDENTIFIED
BY 'psswd';
on the master. Doesn't *.* mean everything? Why would it just show me to
databases?
2013/4/2 Larry Martell
> On Tue, Apr 2, 2013 at 11:30 AM, Richard Reina
> wrote:
> > use DBI;
> > my $dbh = DBI->conn
use DBI;
my $dbh = DBI->connect( "DBI:mysql:rushload;192.168.0.1", $usrr, $passw, {
RaiseError => 3 } );
my $dbs = $dbh->selectcol_arrayref("show databases");
#my $dsn = "dbi:mysql:information_schema:192.168.0.1:3306";
#my $dbh = DBI->connect($dsn, $usrr, $passw);
my $dbs = $dbh->selectcol_arrayr
of course, "Group By"
bill
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
On 3/31/2013 7:32 AM, william drescher wrote:
I have a table, schedule:
CREATE TABLE `schedule` (
`schedule_id` mediumint(9) NOT NULL AUTO_INCREMENT,
`provider` varchar(15) NOT NULL,
`apptTime` datetime NOT NULL,
`location` varchar(10) NOT NULL,
`duration` smallint(5) unsigned NOT
I have a table, schedule:
CREATE TABLE `schedule` (
`schedule_id` mediumint(9) NOT NULL AUTO_INCREMENT,
`provider` varchar(15) NOT NULL,
`apptTime` datetime NOT NULL,
`location` varchar(10) NOT NULL,
`duration` smallint(5) unsigned NOT NULL,
`standing_script` mediumint(9) DEFAULT NULL,
2013/02/02 12:58 -0600, Peter Brawley
On 2013-02-01 10:18 PM, h...@tbbs.net wrote:
>2013/01/31 22:24 -0600, Peter Brawley
>Is this what you mean?
>
>Select,
>pricelist
>If( !IsNull(specialprice) And specialprice < unitprice And CurDate() Between
>startingDate And endingDate,
>sp
On 2013-02-01 10:18 PM, h...@tbbs.net wrote:
2013/01/31 22:24 -0600, Peter Brawley
Is this what you mean?
Select,
pricelist
If( !IsNull(specialprice) And specialprice < unitprice And CurDate() Between
startingDate And endingDate,
specialprice,
unitprice
) as used_price
>From catalog
Where
2013/01/31 22:24 -0600, Peter Brawley
Is this what you mean?
Select,
pricelist
If( !IsNull(specialprice) And specialprice < unitprice And CurDate() Between
startingDate And endingDate,
specialprice,
unitprice
) as used_price
>From catalog
Where itemid='WB314';
PB
Maybe this i
ber_format(mysql_result($result,0,"priceList"),2));
printf('Your Price: $%s',
number_format(mysql_result($result,0,"used_price"),2)); ?>
This seems rather convoluted to me and I've been struggling with it all
day. Any help would be greatly
x27;,
number_format(mysql_result($result,0,"priceList"),2));
printf('Your Price: $%s',
number_format(mysql_result($result,0,"used_price"),2)); ?>
This seems rather convoluted to me and I've been struggling with it all
day. Any help would b
Okay, panic over. I recursively stripped the ACLs and things are working.
Next time I drop a table from phpMyAdmin, I'll carefully read the little thing
that pops up saying I'm about to drop an entire database... :-( One gets so
"yea, whatever" to warning notifiers...)
Thanks to all who sent he
Hi,
It is not very surprising that the database cannot recover from a Time Machine
backup. This generally applies to any software that is running at the moment
the backup is taken. The InnoDB is especially sensitive to taking what is
called a 'dirty' backup because it has a cache. You ma
Am 09.01.2013 16:33, schrieb Jan Steinman:
> I accidentally dropped a crucial database. My only backup is via Apple's Time
> Machine.
>
> First, I stopped mysqld and copied (via tar) the database in question from
> the backup. Restarted, but drat -- most of the tables were apparently using
>
I accidentally dropped a crucial database. My only backup is via Apple's Time
Machine.
First, I stopped mysqld and copied (via tar) the database in question from the
backup. Restarted, but drat -- most of the tables were apparently using
innodb's ibdata1 file, as only the MyISAM tables showed u
presented in the subset. If I looked for 2 terms and I ended up
with hits=2, then I know that those ID values matched on both terms.
You can expand on this pattern to also do partial (M of N search terms)
or best-fit determinations.
I hope this was the kind of help you were looking for.
Reg
ta_cstimage.name column. I think I need a left
> outer join, but I've been messing with this for hours, and I can't get
> the syntax right. I've googled it, but all the examples are simple
> with just 2 tables. Can someone help me with this?
> <<<<<<<&
d is NULL I don't get that data_cst row. But
I changed the RIGHT JOIN to a LEFT JOIN and then it was doing exactly
what I wanted. Thanks for all the help!
> On 2012-12-11 5:11 PM, Larry Martell wrote:
>
> SELECT data_target.name, ep, wafer_id, lot_id,
>>
>>date_time,
On Wed, Dec 12, 2012 at 8:25 AM, Shawn Green wrote:
> On 12/11/2012 7:22 PM, h...@tbbs.net wrote:
>>
>> ... (Are all the distinct "id"s really needed? When one joins on a
>>
>> field with the same name in both tables, one may use 'USING', and
>> only the common field, with neither NULL, shows up i
On 12/11/2012 7:22 PM, h...@tbbs.net wrote:
... (Are all the distinct "id"s really needed? When one joins on a
field with the same name in both tables, one may use 'USING', and
only the common field, with neither NULL, shows up in the output.)
This is a perfectly acceptable naming convention t
>ERROR 1054 (42S22): Unknown column 'data_tool.category_id' in 'on clause'
>But category_id is a column in data_tool.
Then a bit of reordering is required ...
SELECT data_target.name, ep, wafer_id, lot_id,
date_time, data_file_id, data_cstimage.name,
bottom, wf_file_path_id, data_m
've googled it, but all the examples are simple
with just 2 tables. Can someone help me with this?
<<<<<<<<
Modern forms do not give a left join if one uses WHERE-clause to reduce a full
cross-join to an inner join. It is better to start with something like this,
FROM dat
on clause'
But category_id is a column in data_tool.
> I think I need a left
> outer join, but I've been messing with this for hours, and I can't get
> the syntax right. I've googled it, but all the examples are simple
> with just 2 tables. Can someone help me with this?
>
> TIA!
> -larry
>
>
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
tax right. I've googled it, but all the examples are simple
with just 2 tables. Can someone help me with this?
TIA!
-larry
Hi,
I'm running this query:
mysql> SELECT email FROM promoters where id NOT IN (SELECT promoter_id
FROM credits WHERE success = 1 ) and active = 1;
Empty set (31.89 sec)
its returning an empty set and take over 30 seconds to return.
mysql> describe promoters;
+---+--
Hi,
A subquery with IN clause is not a good idea. If you want to tune this
query, try adding indexes on the tables accessed in the inner query
"credits". A composite index on (success,promoter_id) would be sufficient,
then the optimizer will use this index for the where clause and as a
covering in
:00 PM, Mogens Melander
> > wrote:
> >>
> >> On Thu, November 22, 2012 15:45, Neil Tompkins wrote:
> >>> Basically I only what to return the IDs that have both types.
> >>>
> >>
> >> And that's exactly what below statement will return.
&
Claudio
This is the solution i decided to go for as provided in a previous response.
Thanks
Neil
On 23 Nov 2012, at 00:41, Claudio Nanni wrote:
> On 11/22/2012 04:10 PM, Ben Mildren wrote:
>> SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id;
> Ben you were almost there ;)
>
> SELECT id
t;> which version of MySQL you are running and
>> what class you are attending.
>>
>> All necessary information to provide a sufficient help.
>>
>>>
>>> On Thu, Nov 22, 2012 at 2:39 PM, marek gutowski
>>> wrote:
>>>
>>>> SELECT D
On 11/22/2012 04:10 PM, Ben Mildren wrote:
SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id;
Ben you were almost there ;)
SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id HAVING COUNT(id)=
The only bad is the hardcoded parameter in the HAVING, may be it might be
improved.
Anywa
2012/11/22 14:30 +, Neil Tompkins
I'm struggling with what I think is a basic select but can't think how to
do it : My data is
id,type
1000,5
1001,5
1002,2
1001,2
1003,2
1005,2
1006,1
>From this I what to get a distinct list of id where the type equals 2 and 5
Any ideas ?
2012/11/22 14:30 +, Neil Tompkins
I'm struggling with what I think is a basic select but can't think how to
do it : My data is
id,type
1000,5
1001,5
1002,2
1001,2
1003,2
1005,2
1006,1
>From this I what to get a distinct list of id where the type equals 2 and 5
Any ideas ?
u are attending.
>
> All necessary information to provide a sufficient help.
>
>>
>> On Thu, Nov 22, 2012 at 2:39 PM, marek gutowski
>> wrote:
>>
>>> SELECT DISTINCT id FROM table WHERE type IN ('2','5')
>>>
>>> should wor
e attending.
All necessary information to provide a sufficient help.
>
> On Thu, Nov 22, 2012 at 2:39 PM, marek gutowski
> wrote:
>
>> SELECT DISTINCT id FROM table WHERE type IN ('2','5')
>>
>> should work
>>
>>
>> On 22 November 20
Hmmm.
OR, IN and HAVING pops up.
On Thu, November 22, 2012 15:30, Neil Tompkins wrote:
> Hi,
>
> I'm struggling with what I think is a basic select but can't think how to
> do it : My data is
>
> id,type
>
> 1000,5
> 1001,5
> 1002,2
> 1001,2
> 1003,2
> 1005,2
> 1006,1
>
> From this I what to get
On Thu, Nov 22, 2012 at 11:58 AM, Neil Tompkins
wrote:
>
> By unique you mean that no id and type would be duplicated like
>
> 1,1
> 1,1
>
> Yes it isn't possible for duplicate id and type in more than 1 row
Yes, that's exactly what I meant.
- mdyk...@gmail.com
May the Source be with you.
--
Doing a EXPLAIN on the SELECT statement it is using "Using where; Using
temporary; Using filesort" with 14000 rows of data. How best to improve
this; when I already have indexed on id and type
On Thu, Nov 22, 2012 at 4:46 PM, Michael Dykman wrote:
> Assuming that (id,type) is unique in the so
Ignore that it does work fine. Sorry
On Thu, Nov 22, 2012 at 4:46 PM, Michael Dykman wrote:
> Assuming that (id,type) is unique in the source data, that is a pretty
> elegant method:
>
> > select id from
> > (select distinct id, count(*)
> > from my_table
> > where type in (2,5)
> > group by id
When trying this query I get
FUNCTION id does not exist
On Thu, Nov 22, 2012 at 4:46 PM, Michael Dykman wrote:
> select id from
> > (select distinct id, count(*)
> > from my_table
> > where type in (2,5)
> > group by id
> > having count(*) = 2)a;
>
By unique you mean that no id and type would be duplicated like
1,1
1,1
Yes it isn't possible for duplicate id and type in more than 1 row
On Thu, Nov 22, 2012 at 4:46 PM, Michael Dykman wrote:
> Assuming that (id,type) is unique in the source data, that is a pretty
> elegant method:
>
> > se
Assuming that (id,type) is unique in the source data, that is a pretty
elegant method:
> select id from
> (select distinct id, count(*)
> from my_table
> where type in (2,5)
> group by id
> having count(*) = 2)a;
>
--
- michael dykman
- mdyk...@gmail.com
May the Source be with you.
--
MySQ
m: Neil Tompkins [mailto:neil.tompk...@googlemail.com]
> Sent: Thursday, November 22, 2012 8:30 AM
> To: [MySQL]
> Subject: Basic SELECT help
>
> Hi,
>
> I'm struggling with what I think is a basic select but can't think how
> to do it : My data is
>
>
*HAVING typelist = 'x,y,z';
On 22 November 2012 15:25, Ben Mildren wrote:
> Ah read it quickly and misread your requirement. Joins are likely FTW
> here. The alternative would be to do something like this, but I'd opt
> for the joins if you have a reasonably sized data set:
>
> SELECT id, GROUP
Of course there is a cost for the join, each link being a distinct
lookup query but that is the same cost the INTERSECT would impose.
It is not a bad as multiple joins generally might be as all the
lookups are against the same key in the same table which should keep
that index in ram. (type is ind
Ah read it quickly and misread your requirement. Joins are likely FTW
here. The alternative would be to do something like this, but I'd opt
for the joins if you have a reasonably sized data set:
SELECT id, GROUP_CONCAT(type ORDER BY type) AS typelist FROM mytable
WHERE id IN(x,y,z) GROUP BY id H
Do you know if I had multiple joins there would be a performance issue ?
On Thu, Nov 22, 2012 at 3:06 PM, Michael Dykman wrote:
> Keep joining I think. In the absence of intersect (which incurs the cost
> of a query per type anyhow ), this join pattern is the only option I can
> think of.
>
> O
SELECT id FROM mytable WHERE type IN(x,y,z) GROUP BY id;
On 22 November 2012 15:01, Neil Tompkins wrote:
> Michael,
>
> Thanks this kind of works if I'm checking two types. But what about if I
> have 5 types ?
>
> On Thu, Nov 22, 2012 at 2:53 PM, Michael Dykman wrote:
>
>> response did not go t
Keep joining I think. In the absence of intersect (which incurs the cost of
a query per type anyhow ), this join pattern is the only option I can think
of.
On 2012-11-22 10:01 AM, "Neil Tompkins"
wrote:
Michael,
Thanks this kind of works if I'm checking two types. But what about if I
have 5 ty
Michael,
Thanks this kind of works if I'm checking two types. But what about if I
have 5 types ?
On Thu, Nov 22, 2012 at 2:53 PM, Michael Dykman wrote:
> response did not go to the list..
>
>
> I assume that you mean the id must be associated with both type=5 AND
> type=2 as opposed to type=5
How about if I have the following
SELECT DISTINCT id
FROM my_table
WHERE (type = 3 OR type = 28 OR type = 1)
In this instance, for the id 280149 it only has types 3 and 28 but *not *1.
But using the OR statement returns id 280149
On Thu, Nov 22, 2012 at 2:53 PM, Benaya Paul wrote:
> U ca
U can remove the type field it will work
On Nov 22, 2012 8:21 PM, "Neil Tompkins"
wrote:
> Basically I only what to return the IDs that have both types.
>
>
> On Thu, Nov 22, 2012 at 2:39 PM, marek gutowski >wrote:
>
> > SELECT DISTINCT id FROM table WHERE type IN ('2','5')
> >
> > should work
>
response did not go to the list..
I assume that you mean the id must be associated with both type=5 AND
type=2 as opposed to type=5 OR type=2;
in some dialect of SQL (not mysql) you can do this:
select distinct id from 'table' where type=5
intersect
select distinct id from 'table' where type=2
Basically I only what to return the IDs that have both types.
On Thu, Nov 22, 2012 at 2:39 PM, marek gutowski wrote:
> SELECT DISTINCT id FROM table WHERE type IN ('2','5')
>
> should work
>
>
> On 22 November 2012 14:30, Neil Tompkins wrote:
>
>> Hi,
>>
>> I'm struggling with what I think is a
Hi Neil
Would something like this work.
SELECT DISTINCT id,type from your_table WHERE type=2 OR type=5;
Mike
- Original Message -
From: "Neil Tompkins"
To: "[MySQL]"
Sent: Thursday, November 22, 2012 9:30 AM
Subject: Basic SELECT help
Hi,
I'm struggli
SELECT DISTINCT id FROM table WHERE type IN ('2','5')
should work
On 22 November 2012 14:30, Neil Tompkins wrote:
> Hi,
>
> I'm struggling with what I think is a basic select but can't think how to
> do it : My data is
>
> id,type
>
> 1000,5
> 1001,5
> 1002,2
> 1001,2
> 1003,2
> 1005,2
> 1006,1
Hi,
I'm struggling with what I think is a basic select but can't think how to
do it : My data is
id,type
1000,5
1001,5
1002,2
1001,2
1003,2
1005,2
1006,1
>From this I what to get a distinct list of id where the type equals 2 and 5
Any ideas ?
Neil
, 2012 1:54 PM
> To: Daevid Vincent; mysql@lists.mysql.com
> Subject: RE: Help with purging old logs for each customer ID
>
> If the 90 days is back from MAX(created_on) for a given customer...
>INDEX(customer_id, created_on)
> will probably be needed. And that should repla
.@daevid.com]
> Sent: Thursday, October 25, 2012 1:33 PM
> To: Rick James; mysql@lists.mysql.com
> Subject: RE: Help with purging old logs for each customer ID
>
> Well, the customer_id is relevant in that I want the last 90 days
> relative to each customer.
>
> customer_id
there.
Does that make more sense?
I guess I was trying to avoid looping over every customer ID and computing
if I could help it. I thought by using a GROUP BY or something it could
group all the logs for a given customer and then trim them that way.
But maybe brute force is the way to go?
> -
y, October 25, 2012 11:46 AM
> To: mysql@lists.mysql.com
> Subject: Help with purging old logs for each customer ID
>
> I have a customer log table that is starting to rapidly fill up (we
> have hundreds of thousands of users, but many are transient, and use
> the service for a f
log' in that I want to DELETE any
entries older than 90 days for EACH `customer_id`.
I'm not sure how to do that in a query? I'd rather not iterate over each
customer_id if I can help it.
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
ne 15
> > Lost connection to MySQL server during query
> >
> > Able to log in to the phpmyadmin database account and the script that the
> > query is pointing to has the right password and appears to the untrained
> > eye to be working.
> >
> > I host with myh
e working.
>
> I host with myhosting.com and the site has been down for about 8 hours.
>
> Any and all help is appreciated.
>
> Yours Sincerely,
>
> Adrian Burridge
> CanadianInvestors.com Inc.
>
>
in to the phpmyadmin database account and the script that the
> query is pointing to has the right password and appears to the untrained
> eye to be working.
>
> I host with myhosting.com and the site has been down for about 8 hours.
>
> Any and all help is appreciated.
>
>
during query
Able to log in to the phpmyadmin database account and the script that the
query is pointing to has the right password and appears to the untrained
eye to be working.
I host with myhosting.com and the site has been down for about 8 hours.
Any and all help is appreciated.
Yours
2012/09/30 11:07 -0700, Mark Phillips
The data for this table comes from a web page (charet utf8). I copy/paste word
files into gedit (on linux) and then copy/paste from gedit to a text boxes on
the web page input form. I had thought I was stripping out all the funky
characters by usin
the
single straight one. But that would be work for you to explicitly do.
If you are using PHP, see functions htmlentities() and html_entity_decode().
> -Original Message-
> From: Mark Phillips [mailto:m...@phillipsmarketing.biz]
> Sent: Sunday, September 30, 2012 11:08
> -Original Message-
> From: Rik Wasmus [mailto:r...@grib.nl]
> Sent: Monday, October 01, 2012 10:44 AM
> To: mysql@lists.mysql.com
> Subject: Re: NOT_REGEXP Query Help
>
> > I'm having an issue using regular expressions, and I was hoping
> > someon
> I'm having an issue using regular expressions, and I was hoping someone
> can see the syntax error that I'm missing (Monday morning, not enough
> coffee, etc). Running the following query returns the error:
>
> SELECT `a.custid`,`a.login` FROM ol.customers a WHERE `a.login` NOT_REGEXP
> '^ano
er version 4.1.22 (yeah, I know it's old :) )
Any help would be greatly appreciated.
--
John C. Nichel IV
System Administrator
KegWorks
http://www.kegworks.com
716.362.9212 x160
j...@kegworks.com
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To unsubscribe:http://lists.mysql.com/mysql
COLUMN BINARY (or BLOB); -- to forget any charset
> knowledge
> ALTER TABLE ... MODIFY COLUMN CHARACTER SET ...; -- coming from BINARY,
> this does not check the encoding.
> (sorry, don't have the link handy)
>
> > -Original Message-----
> > From: h...@tb
the encoding.
(sorry, don't have the link handy)
> -Original Message-
> From: h...@tbbs.net [mailto:h...@tbbs.net]
> Sent: Thursday, September 27, 2012 2:24 PM
> To: Mark Phillips
> Cc: Mysql List
> Subject: Re: Need Help Converting Character Sets
>
> >>&
To go along with what Rick is saying, this link might help you:
http://dba.stackexchange.com/questions/10467/how-to-convert-control-characters-in-mysql-from-latin1-to-utf-8
I remember doing a bunch of converting HEX() control characters (such as an
apostrophe copied from a Word document) before
2012/09/24 16:28 -0700, Mark Phillips
I have a table, Articles, of news articles (in English) with three text
columns for the intro, body, and caption. The data came from a web page,
and the content was cut and pasted from other sources. I am finding that
there are some non utf-8 characte
eclared for the column they go in. (Presumably, all the text columns will be
declared utf8 or utf8mb4.)
> -Original Message-
> From: Mark Phillips [mailto:m...@phillipsmarketing.biz]
> Sent: Monday, September 24, 2012 4:28 PM
> To: Mysql List
> Subject: Need Help Convertin
101 - 200 of 7359 matches
Mail list logo
