try something like this:
select site_id, max(service_date), name from
site_service left join site on site.id =
site_service.site_id group by site_id;
--- Tom Beidler [EMAIL PROTECTED] wrote:
I¹m having problems with a left join. I¹m not even
sure if I should be using
a left join.
I have
Well I actually noticed a press release a few days ago
saying that the latest version (4.0.10??) is
supporting sub-slelect... please check the website.
this following should be working:
select nome from categorias, categoriamembro
where categoriamembro.id_categoria =
categorias.categoria and
Assuming your server supports innodb, you can use:
Alter table table_name TYPE = InnoDB;
if it failed, simply create a table with the new type,
populate it with the data in your table (with the
wrong type), drop the old table and rename the new
table.
--- [EMAIL PROTECTED] wrote:
Are you
here's another example from the manual...
SELECT * FROM pet WHERE name REGEXP ^b;
see: 3.3.4.7 Pattern Matching
--- Dobromir Velev [EMAIL PROTECTED] wrote:
Hi,
I use it in queries like
SELECT * FROM table WHERE field REGEXP patern;
It is the same as using the LIKE operator wit more
my 2 cents:
SQL in the What is SQL? phrase threw me... I thought
you were talking about MS SQL..
If I were you, I would explain relational databases
and normalization (at list to normal form III).
Usage?? well whoever goes to buy your book, must have
some ideas in head for the usage.. I don't
drop database_name;
make sure no ones is using the database_name.
--- Calvin Lam [EMAIL PROTECTED] wrote:
Hi,
I am a newbie with this and I need help with Mysql
How can I delete a database in Mysql
thanks!
-
assuming you have a table with two columns id and town
then here's one solution:
Create temporary table address (ad varchar(30));
Insert into address select concat(id, ' ', town)
from your_original_table_name;
select * from address order by ad;
--- Nicolas JOURDEN [EMAIL PROTECTED]
wrote:
refer to 1.7.4.1 Subselects in the manual for the
answer. here's a brief example:
The queries:
SELECT * FROM table1 WHERE id NOT IN (SELECT id FROM
table2);
SELECT * FROM table1 WHERE NOT EXISTS (SELECT id FROM
table2 WHERE table1.id=table2.id);
Can be rewritten as:
SELECT table1.* FROM
Folks
I have written a C# GUI for MySQL server which works
pretty nicely on my rather limited LAN at home. the
purpose of this project is to make the source code
available to other folks who are in the learning
process and might want to see some code examples.
Before uploading the code or the
Create a symlink for mysqld.sock in /tmp.
ln -s /path/to/your/mysql.sock /tmp
This should resolve the problem in over 90% of the
cases.
once this is fixed it is likely that you have other
problems. but we'll discuss them when we get there ;-)
--- Mike [EMAIL PROTECTED] wrote:
Hi
Could it be that the table types are different?
the innodb table types seem to ignore auto_increment
while MyISAM are more accommodating... so it might be
that your table on Linux is MyISAM type while the
table on XP is Innodb type.
--- Alan [EMAIL PROTECTED] wrote:
Okay, I've seen just
This actually is more tricky than it sounds.
Firstly the table already exists. So create table is
an option only if you were going to recreate a new
table with an auto_increment column, then move the
rest of the data (except for the index column) in to
the new table, drop the old table and
Auto_Increment will increment the last_INSERT_ID
(which in your case is 10)... so the num field of the
new entry will be 11.
--- Robert Mena [EMAIL PROTECTED] wrote:
Hi, I have been using autoincrement fields for some
time but was wondering how does it work in some
special situations.
Ex.
Well it all depends which version of mysql you are using... versions prior to 4 don't
support sub-select..
-- Stijn Van Rompaey [EMAIL PROTECTED] writes:
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you can get all you need from here.
http://www.mysql.com/articles/dotnet/index.html
-- Jeffrey Powell [EMAIL PROTECTED] writes:
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I thought sub-select is a 4.x feature..
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use double with specific precision like
create table b (id double (5, 3));
then:
insert into b values (1.123, 3.14);
and then
select * from b where id 1.123;
shows only one record (3.14).
select * from b where id 1.123;
empty set.
select * from b where id = 1.123;
returns a result
You probably have created your table without specifying the format of the double...
the following statement would be more appropriate to your expectation:
mysql create table testTable (id int, val double(12, 9));
this will set the display width of the double to 12 and the number of digits
try 'substring' instead of 'substr'.
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[EMAIL
Shutdown you mysqld sever and restart it with
mysqld -u root -Sg
then grant all privileges to root, make sure that your root password is okay. Make
sure that there is an entry in mysql.user table for root@localhost. Then, restart
your server and see how it goes...
Nasser.
--
1. Check if you really need an unsigned column - if not just alter the column to
signed
this is the most appropriate answer. i.e your table design should cater for the
limited set or modulus arithmatics.
Casting wouldn't work simply because you may really want the actual unsigned value of
telnet: Unable to connect to remote host: Connection refused
I don't understand this problem. I've never had this problem.
This simply means that mysqld is not running on your server. the result of the telnet
test (i.e. Connection Refused) confirms that. You need to start mysqld before
Isn't there a typo somewhere...
Shouldn't you expect 5 instead of -5 (15 - id where id = 10)??
Nevermind, let's assume that there is a typo somewhere and the correct expected value
is -5... however, in the domain of unsigned arithmetics -5 is precisely
18446744073709551611... don't you
re: Question 1:
You are right that windows does not ship the ODBC provider but MS has actually created
one and is available for free download. You also need the MyODBC driver which is
created by MySQL folks (also free). Apart from these two there are a few other pieces
of software that you
a row gets inserted into this table
whenever a visitor uses a certain tool on the site
based on this approach your table does not have any entries for the days that no one
uses the certain tool on the site. As a result your query can not pull out records
which don't exist.
You could catter
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