Chris Wagner wrote:
At 01:44 PM 1/12/2006 -0800, Lyle Kopnicky wrote:
If I put a relative path in @INC, I guess it would be relative to the
current directory. Which is whatever the directory the console is in
when it runs the script. Not very useful for finding modules. Instead
I want to find a path relative to where the script is located.
(Actually, I'm interested in finding the parent directory of the one
where the script is located.) As far as I can tell, there is no
function that tells me that. That is why I have to calculate it.
The current directory is automatically in @INC. If the script and module
are in the same dir then u have to do nothing.
Wrong. The script is not necessarily in the current directory. Let's
say you type at the prompt:
C:\> \devel\perl\project2\main.pl
Then the current directory is 'C:\', but the script is in
'C:\devel\perl\project2\main.pl'. If your modules are all in
'C:\devel\perl\project2\', then you want that path in @INC. But I can't
use the literal path, it must be computed, lest I move the project to
'C:\test\project2' or something like that.
If the the path is always
something known reletive to the current dir u can try use lib "../dir"; or
first find the current dir with use Cwd; and then chopping off the last dir
component and putting the result into @INC.
That's the kind of stuff done by the code I attached. Except it first
has to figure out what the script's dir is. What we do know is that $0
is the path to the script, relative to the current dir. Hence the call
to File::Spec->rel2abs().
What is the most common way that people handle this? That is, say your
modules are in the same folder as the main script. (Which I would think
. is always in @INC.
But, as I said, '.' is not necessarily where the script is. That's why
in my script, I computed where the script is, went up one directory, and
did a chdir.
--
Lyle Kopnicky
Software Project Engineer
Veicon Technology, Inc.
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