show your code...is the field set as a time field or what?
bastien
From: Chase [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: [PHP-DB] Issue with date() function...
Date: Fri, 10 Jun 2005 10:01:16 -0600
I have a variable:
$rec2day = date(Y-m-d);
that displays correctly when echoed to
Chase wrote:
I have a variable:
$rec2day = date(Y-m-d);
that displays correctly when echoed to the screen, but when I write it to my
DB, I get the mathmatical answer to the equasion (1989 (2005-06-10)).
What do I need to do to correct this??
Just use $rec2day= date(Y/m/d);
You
I made that modification and now it doesn't update the DB at all...
Here is the code that I am using to insert that variable into my MySQL DB...
Again, my variable is defined as $rec2day = date(Y-m-d) OR $rec2day =
date(Y/m/d)
?php
$connection = @mysql_connect($server, $user, $pass) or
Well, without seeing your query (it's much more likely some will be able
and willing to help you the more information you give) It appears that
you aren't surrounding the date in quotes? So it is being parsed as the
expression 2005 (minus) 6 (minus) 10 Instead of seeing it as the
string
need the quotes around the date
$sql = UPDATE $table1 SET last_access_date='$rec2day' WHERE
username='$dealer_number' AND password='$password';
bastien
From: Chase [EMAIL PROTECTED]
To: php-db@lists.php.net
Subject: Re: [PHP-DB] Issue with date() function...
Date: Fri, 10 Jun 2005 10:52:16