Hi ppl,
I have setup Apache2/mod_perl and started writing perl scripts to
generate html code.
I have a /wood/ directory with two subdirs : /perl/ and /images/
Actually I have setup httpd.conf this way :
<Directory /prog/www/wood/perl>
SetHandler perl-script
PerlResponseHandler ModPerl::Registry
PerlOptions +ParseHeaders
Options +ExecCGI
</Directory>
In my perl file I have this line :
<link type="text/css" rel="stylesheet" href="../menu.css"></link>
And this line:
<img src="../images/banner.jpg"></img>
The problem is it don't go at the right place to find my image and
stylesheet. [Tue Feb 03 15:08:58 2004] [error] 51806: ModPerl::Registry:
/prog/www/wood/perl/menu.css not found or unable to stat [Tue Feb 03
15:08:58 2004] [error] 41882: ModPerl::Registry:
/prog/www/wood/perl/images not found or unable to stat
If I put the perl files in the root directory (/wood/) and setup httpd
conf this way : <Directory /prog/www/wood/>
SetHandler perl-script
PerlResponseHandler ModPerl::Registry
PerlOptions +ParseHeaders
Options +ExecCGI
</Directory>
I make it look under /images/ but it tries to read the .jpg byte per
byte like if it would be a perl file!
The best solution for me is to have all perl and html files in the root
directory, and have the images under the /images/ subdirectory, and
maybe the stylesheets in the /style/ subdirectory.
What would be the best httpd.conf configuration to do that?
Thanks a lot for your help
Best regards,
Steve Hemond
Programmeur Analyste / Analyst Programmer
Smurfit-Stone, Ressources Forestieres
La Tuque, P.Q.
Tel.: (819) 676-8100 X2833
[EMAIL PROTECTED]
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