Hi Neil,
I would actually like to get a sample of this code as well if you don't
mind
K-
-Original Message-
From: Neil Smith [MVP, Digital media]
[mailto:[EMAIL PROTECTED]
Sent: 25 June 2004 16:22
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Date help neede
0
Content-Type: text/plain;
charset="us-ascii"
Content-Transfer-Encoding: 7bit
Subject: RE: [PHP-DB] Date help needed
One thing he wanted which I didn't know how to do (Javascript I guess which
I don't know much about) was to preload a database of email address, and as
Hi there,
>A drop down with 365 days !?!? Isn't that a "little" big?
Actually it's Fridays, Sundays and Tuesdays for 2 years (365 was an example)
it's for an Admin for a client, and he asked that it be in a dropdown box
and he's the boss, so he gets what he wants :-)
One thing he wanted which
A drop down with 365 days !?!? Isn't that a "little" big?
> I have a problem, I currently have some code which populates a dropdown
> box
> - this code gives me every day for the next x amount of days (EG: a years
> worth of days), however what I really need to be able to do, is to find a
> way
Hi there,
Just got back and tried it and it works perfectly, thank you so much for
your help, you've got me out of a bind here ;-)
Chris
You could loop through the weeks and put those 3 specifically in:
$days = array();
for($i = 0; $i < 365; $i +=7) {
$days[] = strtotime('next Monday', strtoti
You could loop through the weeks and put those 3 specifically in:
$days = array();
for($i = 0; $i < 365; $i +=7) {
$days[] = strtotime('next Monday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Friday', strtotime('+ '.$i.' days'));
$days[] = strtotime('next Sunday', strtotime('+ '.
Hi there everyone,
I have a problem, I currently have some code which populates a dropdown box
- this code gives me every day for the next x amount of days (EG: a years
worth of days), however what I really need to be able to do, is to find a
way to display this data in the dropdown box but ONL
