Eric Covener wrote:
> On 1/12/07, bill lam <
cbill.lam@...> wrote:
>> Hello all,
>> I'm have manually started a fastcgi server listening to 127.0.0.1:8999, but I
>> cannot figure out how to use the FastCgiExternalServer for apache to make it
>> working,
>>
>> manual said:
>> FastCgiExternalServer filename -host hostname:port [option ...]
>>
>> but what is "filename" for? My fastcgi server is always running.
>>
>> PS. I've tested my fastcgi against lighty as an external server and it's ok.
>
> Apache needs the filename to figure out what requests should be sent
> to the application. IIRC in the case of remote external servers you
> have to have duplicate filesystem layouts on the webserver and fastcgi
> system
>
Hi Eric,
I retried and found that "filename" can be a directoy name in filesystem that
need not exist at all. Any URI correspond to files under this directory will be
handled by external servers.
This may be an undocumented feature, or I cannot read the documentation
correctly, or documentation itself is unclear.
Anyway it works now. Thank you for taking time to help.
--
regards,
bill
___________________________________
fastcgi-developers mailing list
http://fastcgi.com/fastcgi-developers/
