Dear Group, I am posting this here even though it probably belongs on the apache list because I suspect other haskell users will be able to find it here more easily. I am playing around with hack, and am having trouble with configuring apache with fastcgi to make things work. My understanding of the "hack" concept is that it provides a stardardized interface that lets you glue together web "Applications". It also provides several front-ends, such as happs, and fastcgi etc. Now based on looking at the Middleware supplied with hack, it seems to be trying to dispatch based on the contents of the pathInfo field of the Env record. So, my question is, how do we configure Apache2 with the fastcgi handler so that something appears in the pathInfo field? I have tried several things, the most recent being: RewriteEngine on RewriteRule ^/(.*)$ /hackTest?input=$1 [T=application/x-httpd-cgi] <Location /> SetHandler fastcgi-script Options ExecCGI FollowSymLinks </Location> but the pathInfo field is always null. Env {requestMethod = GET, scriptName = "/lambda", pathInfo = "", queryString = "input=lambda", serverName = "127.0.0.1", serverPort = 80, http = [("FCGI_ROLE","RESPONDER"),("SCRIPT_URL","/lambda"), ("SCRIPT_URI","http://127.0.0.1/lambda"),("User-Agent","curl/7.18.2 (x86_64-pc-linux-gnu) libcurl/7.18.2 OpenSSL/0.9.8g zlib/1.2.3.3 libidn/1.8 libssh2/0.18"), ("Host","127.0.0.1"),("Accept","*/*"),("PATH","/usr/local/bin:/usr/bin:/bin"), ("SERVER_SIGNATURE","<address>Apache/2.2.9 (Debian) mod_fastcgi/2.4.6 proxy_html/3.0.0 mod_apreq2-20051231/2.6.0 mod_perl/2.0.4 Perl/v5.10.0 Server at 127.0.0.1 Port 80</address>\n"), ("SERVER_SOFTWARE","Apache/2.2.9 (Debian) mod_fastcgi/2.4.6 proxy_html/3.0.0 mod_apreq2-20051231/2.6.0 mod_perl/2.0.4 Perl/v5.10.0"), ("SERVER_NAME","127.0.0.1"),("SERVER_ADDR","127.0.0.1"), ("SERVER_PORT","80"),("REMOTE_ADDR","127.0.0.1"), ("DOCUMENT_ROOT","/home/henry/maztrave2/www/fcgi"), ("SERVER_ADMIN","[no address given]"), ("SCRIPT_FILENAME","/home/henry/maztrave2/www/fcgi/hackTest"), ("REMOTE_PORT","44936"), ("GATEWAY_INTERFACE","CGI/1.1"),("SERVER_PROTOCOL","HTTP/1.1"), ("REQUEST_METHOD","GET"),("QUERY_STRING","input=lambda"), ("REQUEST_URI","/lambda"),("SCRIPT_NAME","/lambda")], hackVersion = [2009,5,19], hackUrlScheme = HTTP, hackInput = Empty, hackErrors = HackErrors, hackHeaders = []}% I think what I want is to have all URLS, such as: http://127.0.0.1/lambda be dispatched though my hackTest executable, without having to go through the rewrite, but I can't convice apache to do that. In the interest of completeness, my hackTest.hs file is the following: import Hack import Hack.Handler.FastCGI import Data.ByteString.Lazy.Char8 (pack) import Hack.Contrib.Middleware.Lambda app :: Application app = \env -> return $ Response { status = 200 , headers = [ ("Content-Type", "text/plain") ] , body = pack $ show env } main = runFastCGIorCGI $ lambda app ------------------------------------------------------ One final comment to the authors of hack. Would you please consider renaming this project. hack is such a common word that has nothing to do with this project that it make searching the web with google, etc. almost useless. I realize it is a clever respelling of the ruby version "rack", but please consider naming it something more unique, while it is still relatively new on the web. Thanks in advance for your help. Best wishes, Henry Laxen