Re: Question about the exit() command

No, for that use a if statement if($fname == "Select") { $errormessage = $error; include( "other.html" ); exit; } else { //code } A switch works like this switch($var) { case 1: //if $var==1 $thisvar = $something; break; case 2: //if $var==2 $thisvar = $something2; break; } ~Mike On Thursday, December 19, 2002, at 05:34 PM, Beauford.2002 wrote:

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" <philip@cornado.com> To: "Beauford.2002" <beauford.2002@rogers.com> Cc: "PHP General" <php-general@lists.php.net> 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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