Well, trying your example gives me the expected result.
select cust.ssn, disposition.disposition, source.source
from cust LEFT JOIN source ON (cust.source=source.id)
LEFT JOIN disposition ON (cust.disposition=disposition.id);
'123456789', 'dispo2', 'source1'
'123456780', 'dispo1', 'source2'
'1234
Hi Mike
Thanks for your input. I read the page before and it caused more confusion.
May be an example would clarify this:
create table `cust` (`ssn` varchar(10), `source` int(3) , `disposition`
int(3));
insert into cust (ssn, source, disposition) values ('123456789', 1, 2);
insert i
Can you tell us what exactly is your problem ? ie returned error, logic,
or what ?
My first impression es that your insert is wrong, because inserts cant
have "where" conditions (it makes no sense) probably you want to do a
completely new insert including the column1 or maybe you want an
u
a left join and a right join are 2 very distinct things... It is not
clear from your text what it is you exactly are going for here but I
doubt that applying either LEFT or RIGHT to ALL of your (many) joins
is going to give it to you. You need to stop and examine the
relationships between the t
I tried the following 2 SQL's and the results are less than satisfactory. The
RIGHT join does not show where disposition is NULL. The LEFT join shows
dispositions as NULL where they shouldn't be. Also the LEFT join generates more
dupes. Any way to fix this?
select cust.first as FIRST, cust
What's is the best way to save CD/DVD tree in a MySQL record ? Knowing that
it will be used frequently for displaying as a tree and for searching by
file name, file size..
--
View this message in context:
http://www.nabble.com/Describing-CD-contents-in-a-MySQL-record-tf3518847.html#a9819829
Sent
Without knowing where the values of column2, column3, and column4 are coming
from it's a little hard to say what the best technique would be. Usually one
would take the POSTed value from the select control and use it to retrieve
the related data from a table in your data base.
Regards,
Jerry Sch
Hi,
I am getting the below error in my MySQL err.log file. Can I know the
significance of this error and also how can I identify the respective query
which caused this error.
070403 6:28:28 [ERROR] <> : Sort aborted
Thanks,
Phani
Do you really mean 1:1 relationships? A song might have more than one
artist, album, or genre. Could a song have no album at all, such as
something that was released in MP3 format? Also, you might have more than
one version of a song. You might need a separate table for recordings, with
each song h
"Nikita Tovstoles" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
Do both of your SELECT statements use the same index (`token`) when setting
locks on the rows? My guess is that the situation is as follows.
TX2: SELECT ... WHERE id = ... FOR UPDATE
TX2 sets an exclusive lock for a ro
Hmmm, using "distinct email" in a query where you want
to count email won't work.
select mail, count(mail) as mailcnt from guest
where voted='yes'
group by mail
order by mailcnt;
should do the trick.
On Tue, April 3, 2007 12:51, Me2resh Lists wrote:
> hi
> i need help regarding a sql query in my
hi
i need help regarding a sql query in my php app.
the query is :
$SQL = "SELECT DISTINCT(EMail) FROM mena_guests WHERE Voted = 'yes'
LIMIT $startingID,$items_numbers_list";
i want to sort this query by the number of the repeated EMail counts.
can anyone help me with that please ?
Can you help me please?
1-I have created a while loop to populate the list box with the information
of column1.
2-I need to update (insert data) in to column2, column3, column4 base
on the selection of the list box.
echo'';
$query = "SELECT column1 FROM table";
$res
Hi Scott,
Thanks for the information. I'll look into linux-ha seriously.
Regards,
Ben
Scott Tanner wrote:
Ben,
Check out Linux-ha ( http://linux-ha.org). We are using version 1 at my company, which is fairly easy to set up and there are a lot of good articles on it. With this you can cr
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