Solved

Undefined Variable when loading page

Posted on 2003-11-27
3
945 Views
Last Modified: 2006-11-17
Hi everyone,

Created a php page.  First couple of lines have code that open and connect to a mysql database. The next lines of code are functions. After I finished with the function I have a switch statement as follows:

<?php
switch($action) {
   case "register":
      create_account();
   break;
   default:
      html_header();
      register_form();
      html_footer();
   break;
}
?>

This is a user registration page. the variable $action is a hidden text assigned within a form and is assigned the value of "register". If the user first opens the page the $action variable has no value thus it will create a registration form (register_form()), but if the user submits the form the variable should have the value of "register". when I load the page I get undefined Variable.

What am I doing wrong?

Thanks
Chris
0
Comment
Question by:clekkas
3 Comments
 
LVL 24

Expert Comment

by:shivsa
ID: 9833848
check the register_form()

u have to define variable like this, eg :ssn is variable.

$ssn = $_POST['ssn'];
0
 
LVL 5

Accepted Solution

by:
arjanh earned 50 total points
ID: 9833854
use the isset() function to test if $action is set:

if (isset( $action )) {
      if ( $action = "register" ) {
           create_account();
      } else {
           # probably not possible...
      }
} else {
      html_header();
      register_form();
      html_footer();
}
0
 

Author Comment

by:clekkas
ID: 9833877
Thank you very much to everyone
That isset() worked perfectly.

Chris
0

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