1.00 |
>>> | 2 | 4 | 3 | 2.00 |
>>> | 3 | 3 | 6 | 1.00 |
>>> ++---+-----++
>>> I would like to select all rows from challenges which are NOT linked
Hi There,
If I interpreted what you are trying to do is to return all the rows from
Challenge and reference Patrocinio excluding those rows in Patrocinio where
PTRN_ID is 1?
Not sure if the below is possible in MySQL but I've used this in other places
when doing a left join and needing
1.00 |
++---+-++
I would like to select all rows from challenges which are NOT linked to a
patrocinio with the PTRN_ID -- which would be rows 2 through 6 of
challenges.
I am trying to go about this with a LEFT JOIN query but it does not seem to
be working for me.
mysql> select
6 | 1.00 |
++---+-++
I would like to select all rows from challenges which are NOT linked to a
patrocinio with the PTRN_ID -- which would be rows 2 through 6 of
challenges.
I am trying to go about this with a LEFT JOIN query but it does no
>> | 3 | 3 | 6 | 1.00 |
>> ++---+-+----+
>> I would like to select all rows from challenges which are NOT linked to a
>> patrocinio with the PTRN_ID -- which would be rows 2 through 6 of
>> challeng
_ID -- which would be rows 2 through 6 of
challenges.
I am trying to go about this with a LEFT JOIN query but it
does not seem to
be working for me.
mysql> select c.ID FROM challenge c LEFT JOIN patrocinio p ON
c.ID=p.CHLNG_ID WHERE p.PTRN_I
Hi!! Is it possible to create a left join consult with 2 tables??
I mean:
SELECT * FROM table1 LEFT JOIN (table2, table3) on table1.ID = table2.subID
and table1.ID= table3.subID
Thanks!!!
Rocío Gómez Escribano
mailto:r.sanc...@ingenia-soluciones.com r.go...@ingenia-soluciones.com
Hey there,
- Original Message -
From: Rocio Gomez Escribano r.go...@ingenia-soluciones.com
Hi!! Is it possible to create a left join consult with 2 tables??
I mean:
SELECT * FROM table1 LEFT JOIN (table2, table3) on table1.ID =
table2.subID and table1.ID= table3.subID
Pretty
Hi Johan,
I think you probably want something like this. Give the following a shot.
SELECT *
FROM table1
LEFT JOIN table2
ON table1.ID = table2.subID
LEFT JOIN table3
ON table1.ID= table3.subID
On Thu, Apr 28, 2011 at 9:41 AM, Johan De Meersman vegiv...@tuxera.be wrote:
Hey
2011/04/25 17:42 +0300, Andre Polykanine
Here is the first one.
We have two queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id`;
and the following one:
SELECT `Blogs`.* FROM `Blogs`, `Users` WHERE
`Blogs`.`UserId`=`Users
I would go with join rather than where condition.
2011/4/26 Halász Sándor h...@tbbs.net
2011/04/25 17:42 +0300, Andre Polykanine
Here is the first one.
We have two queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id`;
and the following one
://twitter.com/m_elensule
Facebook: http://facebook.com/menelion
Original message
From: Halбsz Sбndor h...@tbbs.net
To: Andre Polykanine
Date created: , 7:00:03 AM
Subject: LEFT JOIN and WHERE: identical or not and what is better?, etc.
2011/04/25 17:42 +0300, Andre
2011/04/26 17:55 +0300, Andre Polykanine
Aha. So, I should write
SELECT `Blogs`.* INNER JOIN `Users` ON `Users`.`Id`=`Blogs`.`UserId`
instead of my original WHERE clause?
Thanks!
I think so.
--
MySQL General Mailing List
For list archives: http://lists.mysql.com/mysql
To
queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id`;
and the following one:
SELECT `Blogs`.* FROM `Blogs`, `Users` WHERE
`Blogs`.`UserId`=`Users`.`Id`;
1. Are they identical?
2. Which is better (faster, more optimal
some advanced things (earlier I
accomplished those tasks with PHP), so I will be asking stupid
questions, please bear with me.
Here is the first one.
We have two queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id
stupid
questions, please bear with me.
Here is the first one.
We have two queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id`;
and the following one:
SELECT `Blogs`.* FROM `Blogs`, `Users` WHERE
`Blogs`.`UserId
The only difference once MySQL parses these two queries is the first one is
a LEFT JOIN, which will produce all records from the blogs table even if
there is no matching record in the users table. The second query produces an
INNER JOIN which means only rows with matching records in both tables
bear with me.
Here is the first one.
We have two queries:
SELECT `blogs`.* FROM `Blogs` LEFT JOIN `Users` ON
`Blogs`.`UserId`=`Users`.`Id`;
and the following one:
SELECT `Blogs`.* FROM `Blogs`, `Users` WHERE
`Blogs`.`UserId`=`Users`.`Id`;
1
Dear MySQL forum.
I have performance problems when using left join x combined with
where x.y is null, in particularily when combining three tables this
way.
Please contact me by e-mail if you are familiar with these issues and
know how to eliminate slow queries.
I would really appreciate your
2010/3/19 Olav Mørkrid olav.mork...@gmail.com
Dear MySQL forum.
I have performance problems when using left join x combined with
where x.y is null, in particularily when combining three tables this
way.
With a left join, particularly when you're using *is (not) null*, you can't
use index
A LEFT OUTER JOIN in that query in not necessary. An inner join should be used.
-Original Message-
From: Shiv [mailto:shiv...@gmail.com]
Sent: Saturday, May 30, 2009 10:18 PM
To: bharani kumar
Cc: mysql
Subject: Re: Left join query
Hi,
Along with tables, you should also provide
Hi,
Along with tables, you should also provide details on how they are
related. Assuming Code is unique in both tables and left joined on
Airport table, you can do something like this
SELECT A.Code, C.Code, A.SlNo, C.SlNo, A.Name, C.Location, A.status,
C.status, C.type
FROM Airport A
LEFT
Airport table
SlNoName Code AuthLocation status
1 ChennaiCHN Yes India 1
2. Hydarabed HYD Yes India 0
3 walkerWAK Yes uk1
common table
SlNoName Code
-Original Message-
From: shawn.gr...@sun.com [mailto:shawn.gr...@sun.com]
Sent: Saturday, April 11, 2009 3:50 PM
To: Jerry Schwartz
Cc: mysql@lists.mysql.com
Subject: Re: LEFT JOIN with third-table key
Jerry Schwartz wrote:
This is a question that I run into frequently; I might even
customer, so I want to filter on publisher. I
know that I can do this with a sub-select:
UPDATE `A` LEFT JOIN
(SELECT B.product_id FROM `C` JOIN B ON C.publisher_id = B.publisher_id
JOIN `D` ON B.product_id = D.product_id
WHERE C.publisher_code = 'Fred'
) AS `X`
ON A.pub_product_id
to filter on publisher. I
know that I can do this with a sub-select:
UPDATE `A` LEFT JOIN
(SELECT B.product_id FROM `C` JOIN B ON C.publisher_id = B.publisher_id
JOIN `D` ON B.product_id = D.product_id
WHERE C.publisher_code = 'Fred'
) AS `X`
ON A.pub_product_id = X.pub_product_id
SET
I have two simple tables. One - list of forums, second - list of
messages and i want to
get the list of forums with number of messages in each.
Here is the query:
SELECT forums.id , COUNT( forum_msg.id ) AS cnt
FROM forums
LEFT JOIN forum_msg ON forums.id = forum_msg.forum_id
ORDER
( forum_msg.id ) AS cnt
FROM forums
LEFT JOIN forum_msg ON forums.id = forum_msg.forum_id
ORDER BY forums.sorder ASC
The problem is that if a forum does not have any messages then the line
with such forums.id does not appear at all.
If i delete COUNT( forum_msg.id ) AS cnt from Select - i get
SELECT forums.id , COUNT( forum_msg.id ) AS cnt
FROM forums
LEFT JOIN forum_msg ON forums.id = forum_msg.forum_id
ORDER BY forums.sorder ASC
Missing GROUP BY.
PB
-
Artem Kuchin wrote:
I have two simple tables. One - list of forums, second - list of
messages and i want to
get the list
Following the documentation available at
http://dev.mysql.com/doc/refman/5.0/en/join.html
t1t2
a b a c
--- ---
1 x 2 z
2 y 3 w
Then a natural left join would product these results
mysql SELECT * FROM t1 NATURAL LEFT JOIN t2;
+--+--+--+
| a| b
Hi All,
Can u please let me know when should i use LEFT JOIN and when should i going
for a RIGHT JOIN. Please let me know some examples.
regards
anandkl
At 09:05 AM 7/22/2008, you wrote:
Hi All,
Can u please let me know when should i use LEFT JOIN and when should i going
for a RIGHT JOIN. Please let me know some examples.
regards
anandkl
anandkl,
Take a look at the tutorial at
http://www.keithjbrown.co.uk/vworks/mysql/mysql_p5.php
Mike
| 27 | 5 |
| Sumit | 31 | 6 |
| Anand | 29 | 55000 |
| NULL | NULL | 1 |
| NULL | NULL | 9 |
+---+--++
TABLEA LEFT JOIN TABLEB
mysql Select tableA.Name, tableA.Age, tableB.Salary from tableA left join
tableB on tableA.Name=tableB.Name
|
+---+--++
TABLEA LEFT JOIN TABLEB
mysql Select tableA.Name, tableA.Age, tableB.Salary from tableA left join
tableB on tableA.Name=tableB.Name;
+---+--++
| Name | Age | Salary |
+---+--++
| Amit | 27 | 5 |
| Sumit | 31 | 6 |
| Anand | 29 | 55000 |
| Parry
= MYISAM;
INSERT INTO prod_exists VALUES
(MCP-1018),
(MCP-1024),
...
(MCP-1031)
;
SELECT prod_exists.prod_pub_prod_id, IF(prod.prod_num IS NOT
NULL,prod.prod_num,)
as GII_prod_ID
FROM prod_exists LEFT JOIN prod ON
prod_exists.prod_pub_prod_id = prod.prod_pub_prod_id
WHERE (prod.prod_discont = 0
.keyword_trackid, K2.keyword_keyword FROM keyword K1
LEFT JOIN track T1 ON T1.track_id = K1.keyword_trackid
LEFT JOIN keyword K2 ON T1.track_id = K2.keyword_trackid
WHERE (K1.keyword_keyword LIKE '%keyword%');
I am not really worried about full text search right now, as i am more
worried about
and matching results in table
A B.
This one may be slower.
Regards,
Velen
- Original Message -
From: sbrattla [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Wednesday, May 07, 2008 6:41 PM
Subject: Any better ways that LEFT JOIN?
Hi,
My scenario is as described below, and i am
If you are getting more results, I would guess that your users table
is not a list of unique users, but a list of user logins. If that is
the case, then it's your LEFT JOIN and the count(*) that is causing
you to get more results.
If a user logs in 5 times, but only has 1 post, you will get 5
Your doing a left join which can increase the number of rows returned.
This is then GROUP BYed and run through a HAVING. Is:
posts.poster_id=users.id
a one to one relationship? If it is not, then count(*) would be a
larger number and pass the HAVING. This may not be your problem, but I
On Sun, May 4, 2008 at 5:37 PM, Patrick Aljord [EMAIL PROTECTED] wrote:
Your doing a left join which can increase the number of rows returned.
This is then GROUP BYed and run through a HAVING. Is:
posts.poster_id=users.id
a one to one relationship? If it is not, then count(*) would
hey all,
I have my query that counts posts per user:
SELECT count(*) as counted, c.user_id FROM posts c group by c.user_id
having counted1 order by counted DESC LIMIT 20
I wanted to add user login for each count so I did:
SELECT count(*) as counted, u.login FROM posts c left join users u
On Sat, May 3, 2008 at 6:00 PM, Patrick Aljord [EMAIL PROTECTED] wrote:
SELECT count(*) as counted,
u.login
FROM posts c
left join users u on posts.poster_id=u.id
group by c.user_id
having counted1
order by counted DESC
LIMIT 20
This is a bad query. You are abusing GROUP
eo_name_table LEFT JOIN prod ON eo_name_table.eo_name = prod.prod_title
AND eo_name_table.eo_pub_date = prod.prod_published
WHERE (prod.prod_discont = 0 OR prod.prod_discont IS NULL)
ORDER BY eo_name_table.eo_name;
As expected, this gives me exactly 860 rows in the result because the left
join
) AS discont,
IF(prod.prod_title IS NOT NULL, prod.prod_title, ) AS match_title
FROM eo_name_table LEFT JOIN prod ON eo_name_table.eo_name = prod.prod_title
AND eo_name_table.eo_pub_date = prod.prod_published
WHERE (prod.prod_discont = 0 OR prod.prod_discont IS NULL)
ORDER
From: Bill Newton [mailto:[EMAIL PROTECTED]
Sent: Monday, April 14, 2008 2:09 PM
To: Jerry Schwartz
Cc: 'Mysql'
Subject: Re: LEFT JOIN problem
Hi Jerry,
I think the problem is that NULL is not less than or greater than your
prod_published date. So you probably have eo_pub_date set to NULL in 56
I've found yet another oddity with this situation. If I leave the date test
off of both JOINs they give the same number of rows, but they give me the
wrong number! Neither one of them gives me 860 rows returned. I must not
understand how a LEFT JOIN works.
By the way, the EXPLAIN for both of my
.
Farmington, CT 06032
860.674.8796 / FAX: 860.674.8341
www.the-infoshop.com
www.giiexpress.com
www.etudes-marche.com
-Original Message-
From: Jerry Schwartz [mailto:[EMAIL PROTECTED]
Sent: Monday, April 14, 2008 11:26 AM
To: 'Mysql'
Subject: LEFT JOIN problem
I have a table, eo_name_table
cabeceracomprobanteventa ccv
LEFT JOIN detallecomprobanteventa dcv ON
dcv.idCabeceraComprobanteVenta=ccv.idCabeceraComprobanteVenta
LEFT JOIN articulonoauto ana ON ccv.idArticuloNoAuto=ana.idArticuloNoAuto
LEFT JOIN articulo ar ON dcv.idArticulo=ar.idArticulo
LEFT JOIN medida m ON m.idMedida
We have just moved to a new machine where we are running MySQL 5.0.45. On
the old machine we were running MySQL 4.0.15a.
The following query worked in 4.0.15a but gives an error mesage in 5.0.45.
We're pretty sure the problem is with the left join. Did anything change?
Here's the query
is with the left join. Did anything change?
Yes. Don't mix comma-join and ANSI join syntaxes. The precedences
are terribly confusing now.
To see how MySQL interprets your query, you can use EXPLAIN EXTENDED
followed by SHOW WARNINGS.
Here's the query:
select distinct depts.dept_code, depts.sci_id
is with the left join. Did
anything change?
Here's the query:
select distinct depts.dept_code, depts.sci_id,
hr.title,hr.fname,hr.lname, aptb.entry_date from depts, hr left join
crdepts aptb on depts.dept_code=aptb.dept_code where
depts.sci_id=hr.sci_id depts.dept_code=90;
which gives
Yes, that is the correct behavior of a LEFT JOIN. A left join keeps
all the record from the original/left table and will link up any
related data from the right table, but if there is no related data,
it sets it to NULL. If you want the join to act as a filter, the just
use regular JOIN
I am attempting to left join several tables. This was working at one time
but has seemed to stop just recently. I have checked and installed the
lastest version of mysql via Debian Etch apt-get. I am running version
5.0.32.
I have simplified the example down to just 3 tables but the problem
|
+--+--+
|1 | 101.94 |
|2 |47.97 |
+--+--+
2 rows in set (0.00 sec)
-- same query used in LEFT JOIN clause:
mysql SELECT po.id, po.customer_id, po.delivered, od.subtotal
- FROM purchase_order AS po
- LEFT JOIN (SELECT order_id, SUM(quantity
Andrew Carlson wrote:
Is there a purchase order (10002) with no orders in the order_details table?
No, PO 10002 is related to the order_id = 2 here:
mysql SELECT order_id, SUM(quantity * unit_price) AS subtotal
- FROM order_details GROUP BY order_id;
+--+--+
| order_id |
Hi
two tables gi_t and gi
gi_t holds 5 records. gi holds ca. 1500.
I need all from gi for each in gi_t ie LEFT JOIN!!!
but this:
SELECT gi.id AS id, overskrift,gidata FROM gi_t LEFT JOIN gi ON
gi_t.id=gi.gitref WHERE bladref=137
gives the same as
SELECT gi.id AS id, overskrift,gidata FROM
[mailto:[EMAIL PROTECTED]
Sent: Wednesday, September 05, 2007 10:39 AM
To: mysql@lists.mysql.com
Subject: Left join problem
Hi
two tables gi_t and gi
gi_t holds 5 records. gi holds ca. 1500.
I need all from gi for each in gi_t ie LEFT JOIN!!!
but this:
SELECT gi.id AS id, overskrift
Hi,
two tables gi_t and gi
gi_t holds 5 records. gi holds ca. 1500.
I need all from gi for each in gi_t ie LEFT JOIN!!!
but this:
SELECT gi.id AS id, overskrift,gidata FROM gi_t LEFT JOIN gi ON
gi_t.id=gi.gitref WHERE bladref=137
gives the same as
SELECT gi.id AS id, overskrift
with a userIDB
matching users (or leaderboard) userID
I have tried this:
SELECT users.username, gameLeaderboards.playerpoints, friends.userA
FROM gameLeaderboards
JOIN users ON gameLeaderboards.userID = users.ID
LEFT JOIN friends ON gameLeaderboards.userID = friends.userB
WHERE friends.userA = 79760
, August 22, 2007 12:23 PM
To: MySQL General
Subject: Left join is not doing what I thought it should do.
I have 3 tables
A users table (userID, userName)
A leaderboard table (userID, score)
A friends table (userIDA, userIDB)
I would like to produce the following result:
userName, score
It works if I do AND instead of WHERE
Go figure
LEFT JOIN friends ON gameLeaderboards.userID = friends.userB
AND friends.userA = 79760
--
Dave
Jerry Schwartz wrote:
I think your problem is that you can't have a missing friends record that
also has a non-null value for friends.userA
By using AND, you've moved the test for 79760 into the JOIN condition. I
wouldn't have thought of that, either. This seems to be a peculiarity of a
LEFT JOIN. Normally, if you compare a NULL value against anything, even
another NULL, the result is NULL (neither true nor false). I don't
understand
hi
i am experimenting with sql for getting lists of friends.
select friend.*
from user
left join link on user.id=link.id and user.id = $MYID
right join user friend on link.friend_id = friend.id
where user.name is null;
on my local windows machine running mysql 5 it works fine
hi
i am experimenting with sql for getting lists of friends.
select friend.*
from user
left join link on user.id=link.id and user.id = $MYID
right join user friend on link.friend_id = friend.id
where user.name is null;
on my local windows machine running mysql 5 it works
some more info on this:
putting user AS friend improves the query, but does not eliminate the problem.
also, here are the version numbers for each sql server:
mysql Ver 12.22 Distrib 4.0.22, for portbld-freebsd4.9 (i386)
mysql.exe Ver 14.12 Distrib 5.0.24, for Win32 (ia32)
the query even
I've used the syntax, UPDATE ... LEFT JOIN a few times in the past, and
today I used it again for a new employer that I work for.
Several of my associates were unaware that the UPDATE ... LEFT JOIN
syntax is valid in MySQL. After I demonstrated that the query does
indeed work fine on our MySQL 5
Incorporated
195 Farmington Ave.
Farmington, CT 06032
860.674.8796 / FAX: 860.674.8341
-Original Message-
From: Jonathan Langevin [mailto:[EMAIL PROTECTED]
Sent: Wednesday, January 03, 2007 4:13 PM
To: mysql@lists.mysql.com
Subject: Question regarding Update ... LEFT JOIN
I've used
; mysql@lists.mysql.com
Subject: RE: Question regarding Update ... LEFT JOIN
I hope it is locking both tables. Even if you aren't changing any fields
in
the right-hand column, you don't want anyone changing it under you. I
hope
that, if you aren't actually modifying the right-hand table, that MySQL
uses
the total time used for a SELECT lock)
-Original Message-
From: Jonathan Langevin [mailto:[EMAIL PROTECTED]
Sent: Wednesday, January 03, 2007 4:49 PM
To: Jerry Schwartz; mysql@lists.mysql.com
Subject: RE: Question regarding Update ... LEFT JOIN
That would be fine with me, but then conversely
, S.last_name,
P.college, avg(T.score)
FROM students S
LEFT JOIN previous_institutions P USING (studentid)
LEFT JOIN test_scores T USING (campus_id)
GROUP BY S.campus_id
This query actually works but it makes me nervous because I believe it
wouldn't even be legal in Oracle, for example. The P.college
Hi,
select * from t1,t2
left join t3 on t3.itemid=t1.itemid
left join t4 on t4.f2=t1.f2
mysql4.1 works, but mysql5.0 shows Unknown column 't1.itemid' in 'on clause'
If change into
select * from t2,t1
left join t3 on t3.itemid=t1.itemid
left join t4 on t4.f2=t1.f2
mysql4.1
This is makes more sense because it causes a SQL developer
to group the joined tables in a logical manner.
- Original Message -
From: wang shuming [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Friday, November 3, 2006 8:04:20 PM GMT-0500 US/Eastern
Subject: left join , Unknown column
Hi,
select * from t1,t2
left join t3 on t3.itemid=t1.itemid
mysql4.1 works, but mysql5.0 shows Unknown column 't1.itemid' in 'on clause'
If change into
select * from t2,t1
left join t3 on t3.itemid=t1.itemid
mysql5.0 works
Shuming Wang
Hi.
Hope this is the correct list.
I'm strugling with a right and left join query:
SELECT category.name AS cname, service.name AS tname, service.id AS sid
FROM provider
LEFT JOIN providerservice ON provider.id = providerservice.provider_id
RIGHT JOIN service ON providerservice.service_id
Worked under version 4. Does not work under version 5.
$Qstr = select distinct replace(reftraffic.refurl,'http://',''),\
count(*)\
from reftraffic,site\
left join links\
on
left(replace(reftraffic.refurl,'http://',''),locate('/',replace(reftraffic.refurl,'http
Mailing List Receiver wrote:
Worked under version 4. Does not work under version 5.
$Qstr = select distinct replace(reftraffic.refurl,'http://',''),\
count(*)\
from reftraffic,site\
left join links\
on
left(replace(reftraffic.refurl,'http://',''),locate
Mailing List Receiver wrote:
Worked under version 4. Does not work under version 5.
$Qstr = select distinct replace(reftraffic.refurl,'http://',''),\
count(*)\
from reftraffic,site\
left join links\
on
left(replace(reftraffic.refurl,'http://',''),locate
unsigned NOT NULL default '0', ...
KEY `Country1` (`Country1`), KEY `Country2` (`Country2`)
When I run then next query with LEFT join is takes approx 1 minute.
UPDATE data AS db
LEFT JOIN countries AS c1 ON db.Country1=c1.ID
LEFT JOIN countries AS c2 ON db.Country2=c2.ID
SET db.Expr = ...;
But when
On Fri, 2006-07-28 at 07:08 +0100, C.R.Vegelin wrote:
But when I run it with INNER join is takes more than 2 hours !!!
In both cases the query applies to 9.571.220 rows matched with 0 changed.
Any idea why INNER join is so much slower ?
Inner Joins joins everything, it's like a cartesian
I'm trying to upgrade from MySQL 4.1 to MySQL 5.0.
A query that works in MySQL 4.1 does not work in MySQL 5.0 and I'm at a loss
as to how to proceed.
SELECT tab_a.item1, tab_b.item2, tab_c.item3, tab_d.item4, tab_e.item5,
tab_f.item6 FROM db.tab_a, db.tab_e, db.tab_f LEFT JOIN db.tab_b
FROM db.tab_a, db.tab_e, db.tab_f LEFT JOIN db.tab_b ON tab_b.id
= tab_a.id LEFT JOIN db.tab_c ON tab_c.id = tab_a.id LEFT JOIN db.tab_d ON
tab_d.id = tab_a.id WHERE tab_a.id = 'value' AND tab_a.id2 = tab_e.id2 AND
tab_e.id3 = tab_f.id3 LIMIT 1;
The above query now generates this error: Unknown
, db.tab_f LEFT JOIN db.tab_b ON tab_b.id
= tab_a.id LEFT JOIN db.tab_c ON tab_c.id = tab_a.id LEFT JOIN db.tab_d ON
tab_d.id = tab_a.id WHERE tab_a.id = 'value' AND tab_a.id2 = tab_e.id2 AND
tab_e.id3 = tab_f.id3 LIMIT 1;
The above query now generates this error: Unknown column 'tab_a.id in 'on
clause
On 2006-07-21 1:01 PM, Gerald L. Clark [EMAIL PROTECTED]
wrote:
It is a join precedence issue. Use INNER Join instead of a comma.
Thanks Gerald.
Paul DuBois' polite suggestion to read the manual helped. Upon re-reading
the section about the change in precedence with
the comma operator and
td
INNER JOIN tbl_ARTST as art
ON?
INNER JOIN artist_tourdate artd
ON?
INNER JOIN tbl_VENUES tv
ON?
INNER JOIN tbl_VENUE_CAPACITY tvc
ON?
INNER JOIN tbl_VENUE_AGE_XREF tvax
ON?
INNER JOIN tbl_VENUE_AGES tvage
ON?
LEFT JOIN tbl_VENUE_CAPACITY
ON (tv.ID=tvc.VENUE_ID)
LEFT JOIN tbl_VENUE_AGE_XREF
,tv.TICKETAGNCY1 as tix0, tv.TICKETAGNCY2 as tix1
FROM tourdates td, tbl_ARTST as art, artist_tourdate artd , tbl_VENUES tv,
tbl_VENUE_CAPACITY tvc
,tbl_VENUE_AGE_XREF tvax, tbl_VENUE_AGES tvage
LEFT JOIN tbl_VENUE_CAPACITY ON (tv.ID=tvc.VENUE_ID)
LEFT JOIN tbl_VENUE_AGE_XREF ON (tv.ID
as capacity,
tvage.TYPE as age,tv.TICKETAGNCY1 as tix0, tv.TICKETAGNCY2 as tix1
FROM tourdates td, tbl_ARTST as art, artist_tourdate artd , tbl_VENUES tv,
tbl_VENUE_CAPACITY tvc
,tbl_VENUE_AGE_XREF tvax, tbl_VENUE_AGES tvage
These 5 tables are not joined on anything.
LEFT JOIN tbl_VENUE_CAPACITY
ON ?
JOIN tbl_VENUE_AGE_XREF tvax ON ?
JOIN tbl_VENUE_AGES tvage ON ?
LEFT JOIN tbl_VENUE_CAPACITY ON (tv.ID=tvc.VENUE_ID)
LEFT JOIN tbl_VENUE_AGE_XREF ON (tv.ID=tvax.VENUE_ID)
LEFT JOIN tbl_VENUE_AGES ON (tvax.VENUE_ID = tvage.PKEY)
WHERE
td_date NOW()
AND
(td.td_id = artd.td_id AND artd.artist_id
, and tvage are not joined
at all, producing Cartesian Products.
LEFT JOIN tbl_VENUE_CAPACITY ON (tv.ID=tvc.VENUE_ID)
This on condition does not include the table being joined.
LEFT JOIN tbl_VENUE_AGE_XREF ON (tv.ID=tvax.VENUE_ID)
This on condition does not include the table being joined
Paul,
SELECT ...
FROM
tourdates td,
tbl_ARTST as art,
artist_tourdate artd ,
tbl_VENUES tv,
tbl_VENUE_CAPACITY tvc ,
tbl_VENUE_AGE_XREF tvax,
tbl_VENUE_AGES tvage
LEFT JOIN tbl_VENUE_CAPACITY ON (tv.ID=tvc.VENUE_ID)
LEFT JOIN tbl_VENUE_AGE_XREF ON (tv.ID=tvax.VENUE_ID)
LEFT JOIN
ON (artd.artist_id = art.PKEY)
JOIN tbl_VENUES tv ON (td.venue_id = tv.ID)
LEFT JOIN tbl_VENUE_CAPACITY tvc ON (tvc.VENUE_ID = tv.ID)
LEFT JOIN tbl_VENUE_AGE_XREF tvax ON (tvax.VENUE_ID = tv.ID)
LEFT JOIN tbl_VENUE_AGES tvage ON (tvage.PKEY = tvax.VENUE_ID)
LEFT JOIN tbl_VENUE_CAPACITY ON (tv.ID
Hello, my name is Takanobu Kawabe.
I have some questions about left join SQL statement.
I want to join some tables, and using left join statement .
the system is the following. I use two machines.
I created the same databases and tables on both machines.
OS ： Debian
Takanobu Kawabe schrieb:
Hello, my name is Takanobu Kawabe.
こんいちわ孝信さん：）
I have some questions about left join SQL statement.
I want to join some tables, and using left join statement .
the system is the following. I use two machines.
I created the same databases
Hi,
2006/6/16, Takanobu Kawabe [EMAIL PROTECTED]:
[snip]
I tried this statement without error.
But Ihave some questions.
1.How many left join keywords can I use in one SQL statement if
there
are 5000 datas in one table?
as this blog point out, its 31 or 61 depending
to get
the data (since it's optional).
SELECT *
FROM UserInfo u, DslInfo d
LEFT JOIN DslExtra e ON d.DslID = e.DslID
LEFT JOIN ExtraAddr a ON a.UserID = u.UserID
WHERE u.UserID = d.UserID;
However it appears this syntax is not valid in MySQL 5.x (It works
on 4.x). I need to LEFT JOIN *two* tables
Scott
SELECT *
FROM UserInfo u, DslInfo d
LEFT JOIN DslExtra e ON d.DslID = e.DslID
LEFT JOIN ExtraAddr a ON a.UserID = u.UserID
WHERE u.UserID = d.UserID;
However it appears this syntax is not valid in MySQL 5.x
Right, as the 5.x docs say, 5.x wants ISO-compatible explicit joins, ie
SELECT
Perfect. This is exactly what I needed. Is there some place I get
some more documentation on the specifics of the ISO-compatible
queries? Might save me some hair-pulling-out in the future.
Scott
Peter Brawley wrote:
Scott
SELECT *
FROM UserInfo u, DslInfo d
LEFT JOIN DslExtra e ON d.DslID
/join.html).
PB
-
Scott
Peter Brawley wrote:
Scott
SELECT *
FROM UserInfo u, DslInfo d
LEFT JOIN DslExtra e ON d.DslID = e.DslID
LEFT JOIN ExtraAddr a ON a.UserID = u.UserID
WHERE u.UserID = d.UserID;
However
, etc...
from
master_info as r
inner join general_info as g
left join table_1 as t1 on t1.id = r.id
left join table_2 as t2 on t2.id = r.id
left join table_3 as t3 on t3.id = r.id
left join table_4 as t4 on t4.id = r.id
left join table_5 as t5 on t5.id = r.id
where
g.id
At 16:09 +0100 11/5/06, I wrote:
Not long ago, some highly knowledgeable people here kindly helped me
out with a fairly complex query...
...
That's all fine and dandy, but now I need to extend this to a
further four tables...
What I should have added is that for the moment this has to be
Following my post about this complex search I'm trying to do...
In the initial post I said I'd tried adding:
left join table_ga as tga on tga.id = r.id
left join table_a as ta on ta.ida = tga.ida
or:
left join (table_ga as tga inner join table_a as ta) on
(tga.id = r.id
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