Form handler redirect question

I have a PHP form handler page with this line of code:

$required_errorpage = "../thank_you_error.html";

Is there any way to turn it into an if then else redirect based on where the submit page came from?  In other words, there are two forms and they both go to this page which does all the form handling.

For the error, I would like the page go to 2 different error pages based on which form was submitted.

the other error page would be "../offer_error.html"

I am using "dynaform" so I don't know if any custom code will work with it. I know very little about PHP so this solution works great other than this issue here. here is the URL for the dynaform site if that helps:
http://www.webligo.com/products_dynaform.php

-- Dan
LVL 3
danomaticAsked:
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Richard QuadlingConnect With a Mentor Senior Software DeveloperCommented:
../page for error 1.html

Ideally, you should not use relative pages.

Maybe have an /errorpages directory, so no matter where in the structure you are, the errors are ALWAYS in the same place.

$required_errorpage = "error.html";
$aOtherErrors = array('page for error 0.html','page for error 1.html');
if (isset($_POST['whicherrorpage']) && isset($aOtherErrors[$_POST['whicherrorpage']]))
     {
     $required_errorpage = '/errors/' . $aOtherErrors[$_POST['whicherrorpage']];
     }

for example.

Richard.
0
 
Richard QuadlingSenior Software DeveloperCommented:
There are several ways.

1 - Look at the referrer.
2 - Add a hidden field.

e.g

1

switch($_SERVER["HTTP_REFERER"])
      {
      case 'http://site1.com/page1.html':
            $required_errorpage = 'error1.html';
            break;
      case '...' :
            ...;
            break;
      default :
            $required_errorpage = 'defaulterror.html';
      }

or

2 For each form you want a separate error page ...

<input type="hidden" name="new_error_page" value="new_error.html">

and then in the handler php file ...

$required_errorpage = 'defaulterror.html';
if (isset($_POST['new_error_page']) && file_exists($_POST['new_error_page'] . 'html'))
      {
      $required_errorpage = $_POST['new_error_page'] . 'html';
      }


Of course, you may be using $_GET rather than $_POST.

Richard.
0
 
Richard QuadlingSenior Software DeveloperCommented:
Just checking the dynaform code.

Can you try in the dynaform code, change ...

$required_on = "yes";

and on the forms ...

<input type="hidden" name="required_errorpage" value="your_error_page.html">

Richard.
0
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Richard QuadlingSenior Software DeveloperCommented:
If this does not work, then change the $required_on back to the value already present and then alter ...

$required_errorpage = "error.html";

to

$required_errorpage = "error.html";
if (isset($_POST['new_error_page']) && file_exists($_POST['new_error_page'] . 'html'))
     {
     $required_errorpage = $_POST['new_error_page'] . 'html';
     }

This is not very secure, so try this ...

<input type="hidden" name="whicherrorpage" value="1">

in the form, changing the value for each form and then ...

$required_errorpage = "error.html";
$aOtherErrors = array('page for error 0.html','page for error 1.html');
if (isset($_POST['whicherrorpage']) && isset($aOtherErrors[$_POST['whicherrorpage']]))
     {
     $required_errorpage = $aOtherErrors[$_POST['whicherrorpage']];
     }

Richard.

I think that just about covers it!
0
 
danomaticAuthor Commented:
RQuadling:

How do I indicate to go up a directory with the error page:

$aOtherErrors = array('page for error 0.html','page for error 1.html');

would it be:

$aOtherErrors = array('page for error ../0.html','page for error ../1.html');
0
 
Richard QuadlingSenior Software DeveloperCommented:
Anything else?
0
 
danomaticAuthor Commented:
I am sorry, I have gotten held up on this project, I will resume this week.
0
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