Are there any advantages to converting this 'working' query below to
use INNER JOIN ?
If so, what would the correct syntax be ?
Many thanks
SELECT category.name, page.name, content.title, content.body
FROM category, page, content
WHERE content.page_id = page.id
AND page.category_id = category
Hi Luke..
Try this
SELECT ObjectId FROM
(SELECT f15.Form15SampleTube1RnaBarcode AS ObjectId,
f15.Form15PatientID AS PtId FROM form15 f15
WHERE f15.Form15SampleTube1RnaBarcode IN ('01D2V','01DH6')
UNION
SELECT f15.Form15SampleTube6RnaBarcode AS ObjectId,
f15.Form15PatientID AS PtId
Best group member,
My computer stopped working. And I was able to install Windows on another
partition and now I face one problem. I was running Windows with MySQL 4.1.
I have all C: working, and can access all files.
The first question: Can I recover that data from MySQL?
The second question:
Goethals
If I got u properly,
Follow these easy steps..
1. Send the output of your first query to another table
say t1
2. Send the output of finding min(Created) date to another
table t2
3. Replace the value to TYPE field with for
Hello!
I have a problem using UNIONs inside subqueries. I have simplified my
query to make it more readable/understandable.
The question is about the right syntax.
1.
This works fine /UNION/
(SELECT f15.Form15SampleTube1RnaBarcode AS ObjectId,
f15.Form15PatientID AS PtId FROM form15 f1
Jan Gomes wrote:
The query is:
SELECT * FROM table
where id_table2 in (1652272,1652293,1652316,1652362);
+-+---+---+--+
| select_type | table | type | possible_keys |
+-+---+---+--+
| SIMPLE | table | r
Hi! gerald_clark,
But when I call mysql_pconnect("localhost:3306","root","root_password"),it
return the same error. How can I get the permission to open
/var/mysql/lib/mysql.sock?
Fang
>fool.ben wrote:
>
>>Hi everybody!
>>I've install a mysql server on m
Hy Jay,
> You don't show the query you are explaining.
The query is:
SELECT * FROM table
where id_table2 in (1652272,1652293,1652316,1652362);
+-+---+---+--+
| select_type | table | type | possible_keys |
+-+---+---+---
Hi,
everyone,
This is something I
can't seem to solve. Does anyone know how to do this ?
I don't know whether I described this right. Anyhow, jumping right
in, I have the following query which gives me all the LicenseNumbers that
have a different Host-ID, but the same packageID (and w
there is also a nice tool for Mac OSX called CocoaMySQL.
http://cocoamysql.sourceforge.net/
George Law
> -Original Message-
> From: Rajesh Mehrotra [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, May 23, 2006 3:58 PM
> To: Rithish Saralaya; mysql@lists.mysql.com
> Subject: RE: Looking for f
Check out http://www.mysql.com/products/tools/ for some good stuff...
Raj Mehrotra
[EMAIL PROTECTED]
-Original Message-
From: Rithish Saralaya [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 23, 2006 1:21 PM
To: mysql@lists.mysql.com
Subject: RE: Looking for free MySQL Administrator
> I
Jan Gomes wrote:
Hy Jay,
Thanks for you attention.
Show you my structure and EXPLAIN:
CREATE TABLE `table` (
`id_table1` int(10) unsigned NOT NULL default '0',
`id_table2` int(10) unsigned NOT NULL default '0',
`field1`smallint(5) unsigned NOT NULL default '0',
`field2`mediumint(8
Hi folks. Me again.
I finally got this all up and running under crypt of 'cleartext'.
So, even though I am going to be on the same box as the server, how
do I set up an MD5 or password entry?
MYSQLCrypt password()
MYSQLCrypt
Thanks for the tip.
Simple problem, my innodb data file was created with the default my.cnf.
When I started it with the large_table version, it used different innodb
table space size. Therefore would not start :)
Cheers,
Ben
Dan Buettner wrote:
Ben, looks like you've either got it disable
Hy Jay,
Thanks for you attention.
Show you my structure and EXPLAIN:
CREATE TABLE `table` (
`id_table1` int(10) unsigned NOT NULL default '0',
`id_table2` int(10) unsigned NOT NULL default '0',
`field1`smallint(5) unsigned NOT NULL default '0',
`field2`mediumint(8) unsigned NOT N
> I'm looking for a MySQL administrator for 4.x/5.x that will allow me to
>
> Any suggestions? TIA
http://www.webyog.com/
Regards,
Rithish.
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Thanks for the excellent reference, this gives me a lot to go on.
My server is in bits at the moment, I'll let you know when it's up again!
Ben
Dan Buettner wrote:
Ben, looks like you've either got it disabled in my.cnf or with a
startup flag, or you've not set all the needed options for InnoD
Greetings,
I played around with load data from master (ldfm) and it worked fine in
test environment. Now I want to replicate our actual db to a slave. When
I issue the ldfm command, it starts the replication. I get Query OK, but
only about 5% of the db is replicated. Apparently all tables th
Ben, looks like you've either got it disabled in my.cnf or with a
startup flag, or you've not set all the needed options for InnoDB.
See http://dev.mysql.com/doc/refman/5.0/en/mysqld-max.html, near the
bottom of the page it explains what DISABLED means and refers you to the
error log for messa
Hi Dan,
This is what I have. What does this mean with regards to InnoDB?
++--++--+-++
| Engine | Support | Comment
| Transactions | XA | Savepoints |
+---
Please post your exact table schema using SHOW CREATE TABLE, and your
exact query, along with an EXPLAIN SELECT for the query.
Thanks!
-jay
Jan Gomes wrote:
Hy Guys,
I have a simple structure of tables, howewer has 50 million of registers and
2,5 GB of data.
The table is MyIsam and has 4 in
John,
Union the 2 together.
select r.room_num, count(p.peopid)
from rooms r
left join people p on r.roomid=p.roomid1
group by r.room_num
UNION
select r.room_num, count(p.peopid)
from rooms r
right join people p on r.roomid=p.roomid1
group by r.room_num
or something like that. Should give you 2
I need to write a query that essentially does both a left and right join. I
have a list of people and the rooms they occupy. Some rooms have no people.
Some people have no room assigned.
This gives me a list of rooms and how many people are in each room
including any rooms with nobody in th
Ben, what does SHOW ENGINES show you? It should list all known storage
engines and indicate whether your MySQL install supports it or not.
Here's mine (5.0.21) for comparison; I was able to create a test table
as InnoDB and the SHOW CREATE showed it as InnoDB:
-> show engines;
++
Thanks for the reply, it's much appreciated. I'll have another look at
the tables, although my criteria were such that the query should have
only returned a tiny fraction (<0.1%) of the rows (which is why I was
confused). If I get any closer to a solution I'll try and post with a
bit more info.
An
Critters wrote:
>I can get it to join on either countryA or countryB but not both :|
SELECT
d.id, d.day,
c1.country, c1.id,
c2.country, c2.id
FROM days d
INNER JOIN countries c1 ON d.countryA = c1.id
INNER JOIN countries c2 ON d.countryB = c2.id
ORDER BY d.id
(Getting ready for the World Cup
Hi Gerald,
I am sure I don't have this in my my.cfg. I am using the supplied
'large table' my.cfg. The *only* innodb option I have is the command
line parameter to mysqld:
--innodb
If anybody has any other options about how to get innodb working in
5.1.9, I'd be very interested!
Thanks
Hi Gerald,
I am sure I don't have this in my my.cfg. I am using the supplied
'large table' my.cfg. The *only* innodb option I have is the command
line parameter to mysqld:
--innodb
If anybody has any other options about how to get innodb working in
5.1.9, I'd be very interested!
Thanks
MySQL has a pluggable storage engine architecture which means that you
can use multiple storage engines within your databases. What you need
to do is investigate the different features of the engines and devide
which one best fits your system i.e. level of locking required,
InnoDB=row level My
I'm trying to find a weird performance problem in a MySQL database. I
use MySQL v5.0 but the db was migrated forward from a v4.1(?) system.
Looking at the schema in a recent backup, I was surprised to find
different engines used for different tables:
...
CREATE TABLE `comment` (
`id` int
Hi,
yes, it is happening when ever some duplicates are loading replication is
not moving further.
I mean in my experirnce, i stoped my slave from replication for a while, and
i forget the exact location where i stoped it. I resetted my binarylog to
appropriate file, and position to 0, and starte
If u need this in urgent, i think my suggestion may help you.
try to do replication in this way
Branch-A will be update in Branch-B and Branch-B will update in Branch-C
. Branch-E(which got total data of A,B,C,D) will update in Main Office.
Ben Clewett wrote:
Dear MySQL,
I've installed 5.1.9 from source on a SUSE 10 box. But I can't get
InnoDB tables respected.
I have used the correct compilation flag (--with-innodb).
SHOW VARIABLES; lists all the usual innodb variables.
The innodb table space has been created in ~/var/ibdata1
Jan, right now I would say performance sounds pretty good for the amount
of data you have. 50 million records / 2.5 GB of data is a pretty
sizable dataset, so 4 seconds to retrieve a handful of records seems decent.
Some suggestions for things you could do to possibly improve performance:
1 -
Perfect!
I tried aliasing the field names but didn't think about the table, and was
just stuck looking at that query without any idea...
Thanks a lot for your help.
melanie
From: Johan Höök <[EMAIL PROTECTED]>
To: mel list_php <[EMAIL PROTECTED]>
CC: mysql@lists.mysql.com
Subject: Re: query
this is just a test, please ignore
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MySQL General Mailing List
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Hi,
I guess you should be able to do something like:
SELECT t1.term_id, t1.name, r.type_id, t2.term_id, t2.name
FROM term t1
LEFT JOIN relationTerm r ON r.term_id1 = t1.term_id
LEFT JOIN term t2 ON r.term_id2 = t2.term_id
/Johan
mel list_php skrev:
Hi!
I'm stuck with a join query
2 tables
I think the following could be done with some sort of JOIN, but I am now
sure how:
[country]:
id, country, number
1, Germany, 27
2, Japan, 30
3, United States, 18
[days]
id, day, countryA, countryB
10, monday, 1, 3
11, tuesday, 2, 3
12, wednesday, 1, 2
[result I want]
10, monday, Germany, 27
Hi!
I'm stuck with a join query
2 tables, term and relation, the first one with definition of terms the
second one with the relations between them.
CREATE TABLE `term` (
`term_id` int(11) unsigned NOT NULL auto_increment,
`name` varchar(250) default NULL,
) ENGINE=MyISAM DEFAULT CHARSET
Very very odd.
I am compiling with the --use-innodb option. I am starting mysqld with
the --innodb option.
The table space is created for innodb. But SHOW ENGINE shows InnoDB =
DISABLED.
There are relevant no errors in the .err file.
Thanks for sending me your compilation options. I'll
Hi Ben,
I thought the InnoDB engine was included without having to set a
./configure option. On my latest build (admittedly a 5.0.18 one) InnoDB
was enabled without setting any ./configure option.
My options were
hambone /usr/dev/src/mysql-5.0.18 $ cat ../../build_scripts/mysql-5.0.18
#!/bin/bas
Logan, David (SST - Adelaide) wrote:
Hi Ben,
Try doing
SHOW ENGINES;
Here: (pertinent cols only)
++--+--+-++
| Engine | Support | Transactions | XA | Savepoints |
++--+--+-++
| CSV|
Hello MySQL Users
Is there a way to change the separator in the following example,
e.g. from "," to "'"?
SELECT FORMAT(12332.123456, 4);
-> '12,332.1235'
is this the easiest way?
SELECT REPLACE(FORMAT(12332.123456, 4), ",", "'");
-> '12'332.1235'
or can this be set somewhere?
Thank you!
Jay
Hi Ben,
Try doing
SHOW ENGINES;
and see what it says. It should say InnoDB is supported, if not then it
hasn't compiled in.
Regards
---
** _/ ** David Logan
*** _/ *** ITO Delivery Specialist -
Dear MySQL,
I've installed 5.1.9 from source on a SUSE 10 box. But I can't get
InnoDB tables respected.
I have used the correct compilation flag (--with-innodb).
SHOW VARIABLES; lists all the usual innodb variables.
The innodb table space has been created in ~/var/ibdata1.
But if I enter:
C
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