I think this idea is an excellent one, if your going for 1 huge directory.
Store the original filename in the DB and the uniquely generated filename,
so that you know what file to load and can display a meaningful filename to
the viewer. Personally I'd go with seperate directories, but that's
Hi Marc,
If I understand your question correctly, the SQL Syntax you might want to
try would be:
SELECT col_1, col_2 FROM table_name WHERE date_col = $from_date AND
date_col = $to_date
Or if you want to get each days results individually, you could do something
like this in a loop, where $loop
appreciate it!
-Original Message-
From: Szii [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 24, 2001 6:27 PM
To: Tyrone Mills; [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Can't get my head around this problem...
Perhaps...?
$x = 0;
while ($x 7)
{
if (!isset($myArray[$x]))
{
$myArray[$x
Thanks Alex!! I'll give that a go!
-Original Message-
From: Alexander Fordyce [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 24, 2001 8:07 PM
To: Tyrone Mills; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Can't get my head around this problem...
Looks to me like you'd be better off using
Hi Herman,
Try something like this...
$dbname = my_db;
$tablename = my_table;
$table_def = item_id MEDIUMINT DEFAULT '0' NOT NULL AUTO_INCREMENT,;
$table_def .= item_name VARCHAR(50) BINARY NOT NULL,;
$table_def .= lastaccessed TIMESTAMP(14),;
$table_def .= PRIMARY KEY (item_id),;
$table_def
