Here's how I'd do it:
SELECT * FROM customers WHERE Site='egebjergnet' AND
TO_DAYS(NOW()) - TO_DAYS(date_column) <= '45'
where date_column contains record dates...
Martin
""Dan Eskildsen"" <[EMAIL PROTECTED]> wrote in message
9crq7l$mfb$[EMAIL PROTECTED]">news:9crq7l$mfb$[EMAIL PROTECTED]..
Hello everyone, I hope this hasn't been discussed before (I've looked but
haven't found any answers).
This seems very, very strange, and maybe it's just me but here goes:
PHP 4.0.4
mySQL 3.23.30(gamma)
assuming:
$var="1234";(type does not seem to matter)
this query returns no error, but no
IL PROTECTED]> wrote in message
9cs5ef$6js$[EMAIL PROTECTED]">news:9cs5ef$6js$[EMAIL PROTECTED]...
> In article <9crr9l$4c4$[EMAIL PROTECTED]>,
> [EMAIL PROTECTED] ("martin helie") wrote:
>
> > assuming:
> >
> > $var="1234";(type d
Cool! I'll give it a try.
Thanks,
Martin
""Gyozo Papp"" <[EMAIL PROTECTED]> wrote in message
02e601c0d3ff$d89a83a0$5145c5d5@jaguar">news:02e601c0d3ff$d89a83a0$5145c5d5@jaguar...
> I wish there were a php debugger.
maybe, this would satisfy you:
http://dd.cron.ru/dbg/home.php
best regards,
Pa