RE: [PHP] Transfer variable to next web-page

Richard already answered the variable part, but I thought I'd point out that you need to use double quotes in your HTML tags... Single quotes are incorrect... <? echo("<P> <FORM method=\"POST\" action=\"join2.php\"> text for button <INPUT type=\"submit\" value=\"Next step\"> <INPUT type=\"hidden\" name=\"htmlVariableName\" value=\"$phpVariableName\"> </FORM> "); ?> Or... <? echo('<P> <FORM method="POST" action="join2.php"> text for button <INPUT type="submit" value="Next step"> <INPUT type="hidden" name="htmlVariableName" value="'. $phpVariableName .'"> </FORM> "); ?> Or... Better yet, why not just: <P> <FORM method="POST" action="join2.php"> text for button <INPUT type="submit" value="Next step"> <INPUT type="hidden" name="htmlVariableName" value="<?=$phpVariableName?>"> </FORM> See the PHP manual for escaping HTML and string literals if you're confused by any of this.. http://www.php.net/manual/en/language.basic-syntax.php#language.basic-sy ntax.phpmode http://www.php.net/manual/en/language.types.string.php -----Original Message----- From: Olav Drageset [mailto:olav@drageset.net] Sent: Saturday, November 17, 2001 1:01 PM To: php-general@lists.php.net Subject: [PHP] Transfer variable to next web-page I just started.. How do you transfer a variable from one web-page to another I try to use a form like this: <? echo("<P> <FORM method='POST' action='join2.php'> text for button <INPUT type='submit' value='Next step'> <INPUT type='hidden' name='htmlVariableName' value='$phpVariableName'> </FORM> "); ?> and catch it in join2.php by: $phpVariableName=htmlVariableName; echo"$phpVariableName"; This do not work. Can anyone tel me what should be done Neither books, php-manual nor HTML-specification does tel me. Olav -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net