Re: modules that work with both modperl1 and 2

[speeves]

Stas Bekman wrote:

Shannon Eric Peevey wrote:

Perrin Harkins wrote:

On Mon, 2003-06-09 at 13:57, Shannon Eric Peevey wrote:

To answer your original question, Apache::Peek on CPAN now works with 
both mod_perl versions. And while it uses separate implementations for 
each version, the test suite uses the same code to test both.

Yeah, I've been messing with that, but it seems to me that I need 
something similar to a preprocessor directive, where I can load the 
appropriate "use MODULE" lines into the module bases upon which 
version of modperl they have installed.  Is it possible to use the 
BEGIN {} subroutine as this?
  


You don't need a preprocessor.  You can call require inside an if/else
conditional, unlike use() statements which run at compile time and skip
the conditional logic.
For example:

if ($mod_perl::VERSION >= 1.99) {
 require Apache2::Module;
 import Apache2::Module;
} else {
 require Apache1::Module;
 import Apache1::Module;
}
You can put that whole construct in a BEGIN block if you want to.  That
will make it run before the rest of the code in the current package.
- Perrin

Ok, I'm back... :(  Here is the code, modified to use "require":

use mod_perl;

# setting the constants to help identify which version of mod_perl
# is installed
use constant MP2 => ($mod_perl::VERSION >= 1.99);
# test for the version of mod_perl, and use the appropriate libraries
# if (!MP2) { use Apache::Constants qw(:common); }
if (MP2) {
       require Apache::Const;
       import Apache::Access;
       require Apache::Access;
       import Apache::Access;
       require Apache::Connection;
       import Apache::Connection;
       require Apache::Log;
       import Apache::Log;
       require Apache::RequestRec;
       import Apache::RequestRec;
       require Apache::RequestUtil;
       import Apache::RequestUtil;
}


You don't need to import anything, since none of these modules import 
anything. Just require will do.

Here is the code that is coughing up the error:

 102   #Look for user based on UIDAttr
   103
   104    my $attrs = ['dn'];
   105   $mesg = $ldap->search(
   106                   base => $basedn,
   107                   scope => 'sub',
   108                   filter => "($uidattr=$user)",
   109                   attrs => $attrs
   110                  );
   111
   112     if (my $error = $mesg->code())
   113    {
   114         $r->note_basic_auth_failure;
   115         MP2 ? $r->log_error("user $user: LDAP Connection 
Failed: $error",$r->uri) : $r->log_reason("us
er $user: LDAP Connection Failed: $error",$r->uri);
   116         MP2 ? return Apache::Log->HTTP_UNAUTHORIZED : return 
Apache::Constants->HTTP_UNAUTHORIZED;


this should be:

return MP2 ? Apache::HTTP_UNAUTHORIZED : 
Apache::Constants::HTTP_UNAUTHORIZED;

and before the handler (at the top of your module) you need to add:

BEGIN {
    if (MP2) {
        require Apache::Const;
        Apache::Const->import(-compile => 'HTTP_UNAUTHORIZED');
    }
    else {
        require Apache::Constants;
        Apache::Constants->import('HTTP_UNAUTHORIZED');
    }
}
See:
http://search.cpan.org/src/STAS/Apache-Peek-1.01/t/response/TestApachePeek/basic.pm 


   117    }
   118
   119    unless ($mesg->count())
   120    {
   121         $r->note_basic_auth_failure;
   122         MP2 ? $r->log_error("user $user: user entry not found 
for filter: $uidattr=$user",$r->uri) : $
r->log_reason("user $user: user entry not found for filter: 
$uidattr=$user",$r->uri);
   123         MP2 ? return Apache::Log->HTTP_UNAUTHORIZED : return 
Apache::Constants->HTTP_UNAUTHORIZED;
   124    }

And, here is the error that I am getting in the Apache2 logs, (using 
mod_perl 1.99_09)

[Mon Jun 09 13:45:30 2003] [error] user asdf: user entry not found 
for filter: uid=asdf/
[Mon Jun 09 13:45:30 2003] [error] [client 127.0.0.1] Can't locate 
object method "HTTP_UNAUTHORIZED" via package "Apache::Log" (perhaps 
you forgot to load "Apache::Log"?) at 
/usr/local/share/perl/5.6.1/Apache/AuthNetLDAP.pm line 123.

As you can see, it is running everything up to the HTTP_UNAUTHORIZED 
constant...

If you have any ideas, I would greatly appreciate it. :)

Thanks,
speeves
cws
PS.  As a matter of fact, I get the same error when using "use" 
instead...  (Possibly a screwed perl or mod_perl installation?  I'm 
running debian woody, apache 2.0.46, perl 5.6.1, mod_perl 1.99_09.)




Hi!

That did it!!!  Thank you so much for your patience and help with all 
that I am working on here.  After I test these changes on modperl 1 
tomorrow, I should be able to upload it to CPAN, and it will be ready 
for prime-time...  (fingers crossed ;) )

Most humbly yours,
speeves
cws