Ed Since schrieb:
> Hello, I'm wondering if this is the most effective way of doing an outer
> join with 'extra criteria' (I don't feel like it's the best way):
>
> SELECT e.EventID, ue.Contact, ut.Discount
> FROM Event e
> LEFT OUTER JOIN
> (SELECT EventID, Contact FROM UserEvent WHERE UserId =
Thanks much!
ViSolve DB Team-2 wrote:
>
> Hello,
>
> Try this...
>
> select a.rhrqid,a.rhrqsid,a.rhrssid,b.sid,b.rlsid
> from rqhistory a left join relay b
> on (a.rhrqid = b.rqid and (a.rhrqsid = b.rlsid or a.rhrqsid = b.sid or
> a.rhrssid = b.rlsid or a.rhrssid = b.sid))
> where a.rhrqsi
Hello,
Try this...
select a.rhrqid,a.rhrqsid,a.rhrssid,b.sid,b.rlsid
from rqhistory a left join relay b
on (a.rhrqid = b.rqid and (a.rhrqsid = b.rlsid or a.rhrqsid = b.sid or
a.rhrssid = b.rlsid or a.rhrssid = b.sid))
where a.rhrqsid = 101 or a.rhrssid = 101
Thanks,
ViSolve DB Team
- Ori
Thanks! Between the ERD and your descriptions I think I've got it. Most of
the others on this list who have designed a system like yours (gradebooks
or attendance taking) designed it so that it supported multiple
instructors for multiple courses each of which have their own class
schedules (at
On Tue, 21 Sep 2004 10:49:31 -0400
[EMAIL PROTECTED] wrote:
> Hmmm. you want to see a student, all of the classes they are
> enrolled in and how many times they attended? I understand the
> relationships between the student, class, and class_attended tables
> (that's all related to attendance and
Josh Trutwin <[EMAIL PROTECTED]> wrote on 09/21/2004 09:40:03 AM:
> On Tue, 21 Sep 2004 08:57:21 -0400
> [EMAIL PROTECTED] wrote:
>
>
>
>
>
> Perhaps another example would help. I've been trying to re-write
> another join query that's designed to produce an attendance record for
> each stude
On Tue, 21 Sep 2004 08:57:21 -0400
[EMAIL PROTECTED] wrote:
> There are up to three layers of record filtering that happen during
> any query. First is the JOIN filtering. That is where the ON
> conditions are used with the table declarations to build a virtual
> table that consists of all colu
Josh Trutwin <[EMAIL PROTECTED]> wrote on 09/20/2004 10:41:46 PM:
> On Mon, 20 Sep 2004 10:25:16 -0400
> [EMAIL PROTECTED] wrote:
>
> > I think you missed my point. I think the 5.0.1 behavior was correct
> > and the others are wrong. There is a known bug (or two) about mixing
> > outer joins and
On Mon, 20 Sep 2004 10:25:16 -0400
[EMAIL PROTECTED] wrote:
> I think you missed my point. I think the 5.0.1 behavior was correct
> and the others are wrong. There is a known bug (or two) about mixing
> outer joins and inner joins and it looks like it may be fixed. IF
> you want to see all of the
I think you missed my point. I think the 5.0.1 behavior was correct and
the others are wrong. There is a known bug (or two) about mixing outer
joins and inner joins and it looks like it may be fixed. IF you want to
see all of the students THAT TABLE (students) needs to be on the LEFT side
of a
On Mon, 20 Sep 2004 09:33:56 -0400
[EMAIL PROTECTED] wrote:
> Sounds like your 4.0.20 may be the buggy installation... let me see
> if I can explain.
Except this is a 5.0.1 installation. The query worked as is in 4.0.20
(and it also worked in 5.0.0), only after playing with 5.0.1 did the
results
Sounds like your 4.0.20 may be the buggy installation... let me see if I
can explain.
Let's analyze your FROM clause and imagine there is no WHERE clause, for
the moment:
FROM student s
INNER JOIN enrollment e ON e.tech_id = s.tech_id
INNER JOIN submitted_assignment sa ON sa.tech_id = s.tech_id
The (+) indicates an OUTER JOIN.
This should work:
SELECT A1.store_name, SUM(A2.Sales) SALES
FROM Georgraphy A1 LEFT JOIN Store_Information A2
ON A1.store_name = A2.store_name
GROUP BY A1.store_name;
Scott Purcell wrote:
Hello,
I am working through a sql tutorial, and would like to perform this (wr
First, you database design. You don't need to separate actresses from
actors... Why do that? They are the same entity, a person, with only one
different attribute: the genre. So, you should join them in one single
table:
...
Actually, it is possible to be female and to be an Ac
Sorry, i meant "gender", not "genre".
-Forwarded Message-
First, you database design. You don't need to separate actresses from
actors... Why do that? They are the same entity, a person, with only one
different attribute: the genre. So, you should join them in one single
table:
Actors
==
First, you database design. You don't need to separate actresses from
actors... Why do that? They are the same entity, a person, with only one
different attribute: the genre. So, you should join them in one single
table:
Actors
==
act_id
name
genre ENUM('m','f')
Then, the table DVD. If we ma
> [EMAIL PROTECTED] wrote:
>>> DVD_Actor:
>>> dvd_ID REFERENCES DVD
>>> actor_ID REFERNCES Actor
>>
>> Is this how you setup a join table ?
>
> Yes.
>
>
>> what exactly is the references keyword ?
>
> It indicates a foreign key. Full syntax is something like:
> dvd_ID CONSTRAINT dvc_fk FOREIGN KEY
[EMAIL PROTECTED] wrote:
DVD_Actor:
dvd_ID REFERENCES DVD
actor_ID REFERNCES Actor
Is this how you setup a join table ?
Yes.
what exactly is the references keyword ?
It indicates a foreign key. Full syntax is something like:
dvd_ID CONSTRAINT dvc_fk FOREIGN KEY REFERENCES DVD (dvd_ID)
Read the ma
>
> DVD_Actor:
> dvd_ID REFERENCES DVD
> actor_ID REFERNCES Actor
>
Is this how you setup a join table ? what exactly is the references
keyword ?
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Bjorn Barton-Pye wrote:
I am using a test database to teach myself MYSQL and am using my DVD
collection as the subject. I have 3 tables in this example:
Actresses
===
actr_id
name
Actors
==
acto_id
name
DVD
==
Title
acto_id
actr_id
The acto_id and actr_id in the DVD table indicates the
> so soemthing like
>
> select * from dvd left join actresses actr on actr.actr_id=dvd.actr_id
> left join actors acto on acto.acto_id=dvd.acto_id or by dvd.title
>
>
totally forgot, to get a really good query especially when you use Innodb
it doesnt like null values on foreign keys, i'd setup a r
> So, can somebody please correct the following query (and explain the
> syntax) so that it will work please? (I haven't tried putting an outer
> join in it because I don't understand the syntax.)
>
> Select
> actr.name,
> acto.name,
> dvd.title
> from
> actresses actr,
>
Try something like
select distinct S.US_FOLIO US_FOLIO
, ifnull(SD.US_FOLIO,'false') FOLIO2
from SEGUIMIENTO S LEFT OUTER JOIN SEGUIMIENTO_DETALLE
SD
on (S.US_FOLIO=SD.US_FOLIO)
-Original Message-
From: Gustavo Mejia [mai
Hi,
Yes I have made some mistakes. There was problem with the outer
join. It should have been ad outer joined to review, not the other way.
I didn't notice the first tabel person in the query.
Try the following with two tables ad and review and later add person
table. We don't know the column
At 01:43 AM 2/12/2002, you wrote:
>Hi,
>Does this work for you?
>
>select ad.id,adtype,name,sum(review.id is not null) from person
>review left join ad on ad.id=review.id
>group by ad.id;
this one returns 1 for the sum column no mater what.
>select ad.id,adtype,name,sum(if (ifnull(review.id,0)=
Hi,
Does this work for you?
select ad.id,adtype,name,sum(review.id is not null) from person
review left join ad on ad.id=review.id
group by ad.id;
OR
select ad.id,adtype,name,sum(if (ifnull(review.id,0)=0,0,1)) from person
review left join ad on ad.id=review.id
group by ad.id;
Anvar.
At 06:13
"Harlan Feinstein" <[EMAIL PROTECTED]> wrote:
> I've joined them an SQL statement that looks like this:
>
> select val, status
> from A
> left outer join B on val=id;
>
> What I'd LIKE is a 90-row result set, with the "status" field from table B
> when there was data available. What I'm getting i
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