each into an array.
--
Jamie Saunders
Media Architect
[EMAIL PROTECTED]
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.
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Jamie Saunders
Media Architect
[EMAIL PROTECTED]
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, anyone with half a brain and a popup-stopper won't get that
page :)
--rylin
Jamie Saunders wrote:
Hi,
I'm trying to figure out a way of detecting when a user closes the
browser, or if they leave the site. If they do this I'd like to change
a value in the database, i.e:
if (browser
Hi,
I'm trying to find a way of simply checking a string for numbers.
e.g.
if ($myVar contains a number)
{
//numbers present
}
else
{
//no numbers present
}
Thanks,
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Jamie Saunders
Media Architect
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a database error every time informing me the query could not
be performed. To test the query I've pasted the printed query string into
phpMyAdmin and it works first time every time. So if there's nothing wrong
with the query string or the database, what else could be causing this
error?
--
Jamie
is one of the most versatile and user friendly programming
languages available.
Best Regards,
J
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Jamie Saunders
Media Architect
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This was indeed the problem, I was calling the mysql_fetch_array earlier in
the code.
Once removed it worked fine.
Thanks.
Jamie
Paul Burney [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
on 8/7/01 1:41 PM, Jamie ([EMAIL PROTECTED]) wrote:
while ($previews
