I 've given a value se the code : $_POST["hiddenField"] = $image_avant;
where $image_avant is a variable which contains a value !! Cdt ----- Original Message ----- From: "Michael Egan" <[EMAIL PROTECTED]> To: "Dominique ANOKRE" <[EMAIL PROTECTED]>; "Php List" <[EMAIL PROTECTED]> Sent: Friday, February 20, 2004 11:22 AM Subject: RE: [PHP] display a hiddenfield No value has been given to the hidden field. e.g. print (<input type=\"hidden\" name=\"hiddenField\" value=\"yourvalue\">); HTH. Cheers, Michael Egan > -----Original Message----- > From: Dominique ANOKRE [mailto:[EMAIL PROTECTED] > Sent: 20 February 2004 11:17 > To: Php List > Subject: [PHP] display a hiddenfield > > > I use a simple form like this : > > print("<form method=\"post\" action=\"image.php\">\n"); > print("<input type=\"hidden\" name=\"hiddenField\">\n"); > $_POST["hiddenField"] = $image_avant; > print("<input type=\"submit\" name=\"avant\" value=\"Avant\">\n"); > print("</form>\n"); > > and when i click on the submit button, i want to display the > value of my hiddenField (doing by the file image.php) : > > <?php > echo $_POST["hiddenField"]; > ?> > > but nothing is displaying !!!!!! > > Please help ! > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
