Hi, What is the correct way to send http status code of 206 from from Modperl Registry script [In modperl2/apache2]? I searched the modperl site but did not find any examples.
I looked at the following redirect example [by setting $r->status(Apache2::Const::HTTP_PARTIAL_CONTENT) and returning Apache2::Const::HTTP_PARTIAL_CONTENT.] This gives a http status code of 200 and internal error message in the server. http://perl.apache.org/docs/2.0/user/coding/cooking.html#Sending_Cookies_in_REDIRECT_Response__ModPerl__Registry_ OK, after a while, I also tried setting $r->statusline( ) and this seems to work, but I am not sure if this the right way. I am pasting a sample code that works correctly Thanx, HB. ----Sample code that works fine ---- use Apache2::Request; use Apache2::RequestUtil; use Data::Dumper; use Apache2::Const -compile => qw(HTTP_PARTIAL_CONTENT OK HTTP_OK); use APR::Table (); use strict; use constant MOD_PERL => $ENV{MOD_PERL}; my $r = shift; $r->headers_out->set('Accept-Ranges'=> 'bytes'); $r->content_type('audio/mpeg'); $r->headers_out->set('Content-Length' => 2); $r->headers_out->set('Content-Range' => 'bytes 0-1/40000'); $r->status_line(Apache2::Const::HTTP_PARTIAL_CONTENT. ' Partial Content' ); $r->print('ZZ'); return Apache2::Const::OK;