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 a list
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.
I've done all
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
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.
Anybody
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=1
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 21
Please help want is wrong.
Regards Gerlinde
BLITZ Internet Service GmbH
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 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,
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
