What do you mean by not working? What results do you get?
-Original Message-
From: Victor Subervi [mailto:victorsube...@gmail.com]
Sent: Monday, August 16, 2010 6:59 AM
To: mysql@lists.mysql.com
Subject: Join Problem
Hi;
I have this code:
select f.id from Flights f join Planes p where
Review your join type.
From: ext Gavin Towey [gto...@ffn.com]
Sent: 16 August 2010 19:36
To: Victor Subervi; mysql@lists.mysql.com
Subject: RE: Join Problem
What do you mean by not working? What results do you get?
-Original Message-
From: Victor
On Thu, 2005-02-17 at 12:08, Albert Padley wrote:
I have the following 2 tables:
CREATE TABLE `division_info` (
`id` int(11) NOT NULL auto_increment,
`division` varchar(50) NOT NULL default '',
`spots` int(11) NOT NULL default '0',
PRIMARY KEY (`id`),
KEY `division`
\ mysql@lists.mysql.com
Subject: Re: JOIN Problem
Date: Thu, 17 Feb 2005 12:20:44 -0500
On Thu, 2005-02-17 at 12:08, Albert Padley wrote:
I have the following 2 tables:
CREATE TABLE `division_info` (
`id` int(11) NOT NULL auto_increment,
`division` varchar(50) NOT NULL default '',
`spots
Albert Padley [EMAIL PROTECTED] wrote on 02/17/2005 12:08:31 PM:
I have the following 2 tables:
CREATE TABLE `division_info` (
`id` int(11) NOT NULL auto_increment,
`division` varchar(50) NOT NULL default '',
`spots` int(11) NOT NULL default '0',
PRIMARY KEY (`id`),
KEY
with odbc.
thanks!
From: Michael Dykman [EMAIL PROTECTED]
To: Albert Padley [EMAIL PROTECTED]
CC: \MySQL List\ mysql@lists.mysql.com
Subject: Re: JOIN Problem
Date: Thu, 17 Feb 2005 12:20:44 -0500
On Thu, 2005-02-17 at 12:08, Albert Padley wrote:
I have the following 2 tables:
CREATE
Michael Dykman [EMAIL PROTECTED] wrote on 02/17/2005 12:20:44 PM:
On Thu, 2005-02-17 at 12:08, Albert Padley wrote:
I have the following 2 tables:
CREATE TABLE `division_info` (
`id` int(11) NOT NULL auto_increment,
`division` varchar(50) NOT NULL default '',
`spots`
On Feb 17, 2005, at 10:34 AM, [EMAIL PROTECTED] wrote:
Albert Padley [EMAIL PROTECTED] wrote on 02/17/2005 12:08:31 PM:
I have the following 2 tables:
CREATE TABLE `division_info` (
`id` int(11) NOT NULL auto_increment,
`division` varchar(50) NOT NULL default '',
`spots`
Thanks for the suggestion, but according to explain we are in worse shape
than before. In both cases the multi-column index is ignored. I am going to
try fiddling with the index col order to see if this helps.
Here is what it comes up with as you suggested:
Query1:
explain SELECT COUNT(*) as
O. I've got a headache trying to understand joins. I'm definitely
NOT a database guru.
Why in the world doesn't this work?
SELECT dacspriv_name
FROM dacspriv
WHERE dacspriv_id not in (SELECT dacspriv_id
FROM dacs_access JOIN users
ON dacs_access.user_id=users.user_id
WHERE
Short answer is mysql does not do sub-selects (i.e., a select inside of a
select). The join part is not this issue.
Wouldn't this depend on the version... I thought the newest versions, 4.x+,
supported sub-selects.
Ryan
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: Friday, May 30, 2003 12:11 PM
To: mysql
Subject: Re: Join problem
Short answer is mysql does not do sub-selects (i.e., a select inside of a
select). The join part is not this issue.
Wouldn't this depend on the version... I thought the newest versions, 4.x+,
supported sub-selects.
Ryan
Short answer is mysql does not do sub-selects (i.e., a select inside
of a select). The join part is not this issue.
Wouldn't this depend on the version... I thought the newest versions,
4.x+, supported sub-selects.
Ryan
__
Do you Yahoo!?
Yahoo! Calendar -
]
Sent: Friday, May 30, 2003 12:11 PM
To: mysql
Subject: Re: Join problem
Short answer is mysql does not do sub-selects (i.e., a select inside
of a select). The join part is not this issue.
Wouldn't this depend on the version... I thought the newest versions,
4.x+, supported sub-selects.
Ryan
, 2003 3:42 PM
To: [EMAIL PROTECTED]
Cc: Susan Ator
Subject: RE: Join problem
Well, I'm running 3.23.54 on Red Hat 7.3. Given this, how in the world
do I accomplish the following:
I have these tables:
dacspriv - with dacspriv_id,dacspriv_name,short_name
users
Sorin,
If there are entries which are:
60 / amg / 5
120 / amg / 5
(in this example '5' is the id of the calculation-subproject)
Then the result should be:
+---+---+---+---+
| login | name | name |
Stefan Hinz wrote:
Okay, here for another wild guess:
SELECT u.login, p.name, sp.name, SUM(t.time)
FROM t_user u
LEFT JOIN t_project p ON 1
LEFT JOIN t_subproject sp ON p.id = sp.project_id
LEFT JOIN t_time t ON ?? = ??
WHERE u.login = 'amg'
GROUP BY p.name
Where I'm not sure if you will need
Defryn,
Monday, July 08, 2002, 5:09:51 AM, you wrote:
D Can anyone have a look at my join.
D It returns errors
D Select O.name, O.amount , P.Productname, P.price
D From Orders AS O
D JOIN Products AS P ON O.product=P.Productid
D Where O.cluster= 'ANP';
D It works fine when I use
D Select
Dave Butler wrote:
I am struggling with a join query using MySQL 3.23.31 under AIX 4.3.3.
Here are the tables involved:
select CAT.linenum, CAT.acct, FD.amount
from sched_acct_cat CAT LEFT JOIN fd FD
ON CAT.acct = FD.acct
where CAT.sched_acct = 'INC_STMT'
AND FD.entity='FMCI'
AND
Neil,
I did a quick test of your data and this is what I got to work.
SELECT item.item, titles.title FROM item LEFT JOIN titles ON
item.title_id = titles.title_id;
The left join selects all the items from the 'left' table even if there
are no matching entries in the 'right' table. A bit of a
You could try using an AND in the WHERE statement.
EG. SELECT DISTINCT R.ID, R.CatCode, R.Title
FROM Recipes R, Ingredients I
WHERE I.Description LIKE %$SearchKey%
AND R.ID = I.ID;
Hope this helps.
Andrew Murphy
-Original Message-
From: Urb LeJeune [mailto:[EMAIL
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