;Arthur Fuller"
Cc: "Claudio Nanni" ; "AndrewJames"
; "mysql"
Subject: Re: database design
A) You would probably want to populate the Article.Article_Type column
with Article_Type.ID. You probably wouldn't need Article_Type table if
you're going
This is a bit of a long shot, but i really need some help and or directed to
the best reading resources.
as i begun building my database (as i went along), i now realise i have to
stop coding and sit back and design the database properly before i can go
on.
However i am still unable to wrap
Hey guys,
whenever i try to perform this function on my $variables before using them
in sql queries it deletes them and returns my variable as nothing, ''.
this is how i am using it.
my login.php form
$username = check_input($_POST['username']);
$password = check_input($_POST['password']);
m
: Per Jessen [mailto:p...@computer.org]
Gesendet: Freitag, 4. September 2009 13:05
An: mysql@lists.mysql.com
Betreff: Re: a better way, code technique?
AndrewJames wrote:
is there a better way (hopefully simpler) to code this?
i want to get the user id of the logged in user to use in my next
is there a better way (hopefully simpler) to code this?
i want to get the user id of the logged in user to use in my next statement.
$q1 = sprintf("SELECT uid FROM users WHERE users.username='$username'");
$result1 = mysql_query($q1);
$uid = mysql_fetch_array($result1);
$u = $uid['uid'];
it see
Hey,
i have a table called users which has my users in it, each have a uid field.
I also have a stories table which has stories in it each with a sid field
for each story but also a uid field so i know which user the story belongs
to.
i want to write a query that will display the story depending
