Just to make sure I have this right. I have a switch() statement like below.
If the case statement is true it should include the page indicated and exit
the PHP script without displaying the echo or executing any other code. If
the case is false the script should continue as normal.
TIA.
switch (true) {
case ($fname == "Select"):
$errormessage = $error;
include ("other.html");
exit;
}
echo "Other Code";
----- Original Message -----
From: "Philip Olson" <[EMAIL PROTECTED]>
To: "Beauford.2002" <[EMAIL PROTECTED]>
Cc: "PHP General" <[EMAIL PROTECTED]>
Sent: Thursday, December 19, 2002 4:55 PM
Subject: Re: [PHP] Question about the exit() command
>
> How about:
>
> if (!$conn = mysql_connect($host, $user, $pass)) {
> include 'static_html.inc';
> exit;
> }
>
> print "Welcome, yes the database is connected";
>
> exit ends the script, nothing after its use is executed.
>
>
> Regards,
> Philip Olson
>
>
> On Thu, 19 Dec 2002, Beauford.2002 wrote:
>
> > Hi,
> >
> > Could someone clarify this for me. I want to be able to exit out of a
PHP
> > webpage and return to the calling page if certain conditions are not
met. It
> > appears using exit() will do this, but I am unclear exactly how to use
it.
> >
> > any info is appreciated.
> >
> > TIA
> >
> >
> >
> >
> > --
> > PHP General Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
>
>
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