500 Internal server error

[sachinz2]

I am providing the user an http link so that he can download the file from my server

this link is link "http://my.server.name/cgi-bin/Output/Waybill/waybills01_29_14.csv"

If I give the Wrong filename ,I get an error "The requested URL /cgi-bin/Output/Waybill/waybills01_29_15.csv was not found on this server."

but then for the correct filename also, I get 500 Internal server error

"The server encountered an internal error or misconfiguration and was unable to complete your request."

I am clueless, and all the other work is pending,coz the user need's to process this file for further work

Do i need to change any thing in Apache settings??

Thanks

[Anonymous Monk]

What does the error log say?

this and many more tips in the checklists :) Basic debugging checklist , brian's Guide to Solving Any Perl Problem, CGI Help Guide , Troubleshooting Perl CGI scripts

[sachinz2]

Hi The error log says

[Thu Feb 06 11:16:17 2014] [error] [client 198.38.94.79] (13)Permissio
+n denied: exec of '/var/www/cgi-bin/Output/Waybill/waybills01_29_14.c
+sv' failed
[Thu Feb 06 11:16:17 2014] [error] [client 198.38.94.79] Premature end
+ of script headers: waybills01_29_14.csv

[Anonymous Monk]

Is that file a cgi script?

[robby_dobby]

Hello sachinz2,

Please pay more attention next time :-). You're exposing "http://my.server.name/cgi-bin/Output/Waybill/waybills01_29_14.csv" but request "/cgi-bin/Output/Waybill/waybills01_29_15.csv". Remember this:

• Server:- 404, you messed up
• Server:- 500, I messed up.

[sundialsvc4]

Here’s the skinny on this one:   Apache has been configured to believe that anything which is sent to <Location /cgi-bin/> corresponds to the name of a file that it should execute, in order to return to you the output from that file being executed.   (You have not specified a Handler ...)   Which is exactly what it is trying to do.   It found a file by that name at that location, but of course it is not executable.   So, Apache throws up its hands.

What you need to do is to tell Apache, say, that <Location /cgi-bin/Output/Waybill/> should cause a particular Handler, not a file in this location, to be executed.   Upon recognizing this pattern, Apache will (maybe) launch the Handler, and pass the URL-string as input to it.   That script which will be responsible for extracting whatever parameters from it and carrying-out the request.

An elementary Apache-configuration problem ... and not Perl-specific ... but, confusing nonetheless.

[Anonymous Monk]

in c++ the solution is:
int main(){
printf("Content-type: text/html\n\n");