At 16:02 17.07.01 +0100, Glenn Wearen wrote:
> > select distinct PersonID
> > from PersonProject
> > where ProjectID != '2';
> >
> > Is this what you are looking for?
>
>
>No, what I'm looking for is...
>SELECT PersonID
>FROM PersonProject
>AND PersonID NOT IN (SELECT PersonID
>
Hello, can you send the datastruktur, so it is difficult to answer your
question.
You don't need, if the other answers were ok.
Regards Gerlinde Fischer
At 08:40 16.07.01 -0400, Technodium.net wrote:
>Hello there, I would need some help with a select statement for my databases.
>
>
At 09:02 16.07.01 -0700, John Hawkins wrote:
>I'm having a bit of trouble doing a LEFT JOIN. Any input would be greatly
>appreciated.
>
>table 1 (sites, has approx 30 records) has a list of sites with the
>following fields:
>ID, sitename, url
>
>table 2 has (ratings, has approx 2020 records) has
At 09:31 16.07.01 +0200, Beatriz Lapaz wrote:
>Hi!!
>
>I'm working with PHP and I use MySQL. I've just known that in MySQL
>doesn't exist NESTED SELECTS and it's a problem for me.
>I'd need a query to get all the ids from a table which are not in another
>table. Something like that:
>
>SELECT TA
Hi,
have a look at show status,
and here for the variable
wait_uptime in seconds. How long it is.
Regards Gerlinde
At 13:54 10.07.01 +0100, Andy C wrote:
>Please excuse if this is a FAQ or well known effect. I'm running mySql
>version 3.23.36 on a Winnt 4.0 Sp6a machine. It's connected to
>a
Have you filled out the tables
user, db
in mysql.
And after filling this tables
you must start mysqladmin reload.
Regards Gerlinde
At 01:11 09.07.01 -0700, Prabu Subroto wrote:
>I tried to connect to my mysql server but there was
>always an error message saying "this host is not
>allowed".
>Anyb
Hi, have you created the directory. InnoDB can't create directories.
Regards Gerlinde
At 16:56 05.07.01 -0400, Brandon Berry wrote:
>These are the options I used.
>
>[mysqld]
>innodb_data_file_path = ibdata;
>innodb_data_home_dir = /usr/local/mysql/data
>set-variable = innodb_mirrored_log_groups=
Hi
With this message:
show table status/usr/bin/safe_mysqld: line 156: 30874 Segmentation
fault nohup $ledir/mysqld --basedir=$MY_BASEDIR_VERSION
--datadir=$DATADIR --skip-locking >>$err_log 2>&1
Please help want is wrong.
Regards Gerlinde
BLITZ Internet Service GmbH
Hi,
I need a transaction Database with Mysql.
I have compiled the sourccode 3.23.39 the following on Linux.
./configure --with-innodb
after that I have stopped and restartet mysql.
but if I make "create table .. type=InnoDB";
I create a ISAM-Table. How can I make a InnoDB-Table.
Please help.
Hi,
are you sure, that mysql make the same sorting order??
I seems in the second query first all with 1, you cann't be sure whether
the to in not listet later on.
Try it again with "order by id" every three statements and compare then the
listings.
Gerlinde
At 12:50 02.07.01 +0200, you wrote:
Hi,
I have the following configuration
I have Linux
I have compile mysql with the following configure parameters:
./configure --enable-shared --with-berkeley-db
--with-mysqld-ldflags='-all-static'
I have created a BerkleyDB-table.
After that I have
set autocommit=0 and I have inserted one row
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