Re: oneliner to get module version

[graff]

in reply to oneliner to get module version

Right -- I have something quite similar (but a little shorter ;) in my .bashrc:

pmversion () 
{ 
    perl -M$1 -le "print $1->VERSION"
}

Note that bash interpolates both instances of "$1" before perl gets to read the script.

Also, since I tend to work in a rather "diversified" network environment, it's useful to be able to spot the path to a given module:

pmpath () 
{ 
    perl -M$1 -le "print \$INC{'$1.pm'}"
}

Here, I have to guard (escape) the "$" in front of INC so the shell leaves it as-is and perl sees it as "$INC"; bash ignores the single-quotes because they're inside "...", and perl does the right thing there, because $1 has already been interpolated by the shell.

UPDATE: The "pmpath" approach above only works for "top-level" modules. In order to get the path for a module like "Foo::Bar::Baz", it's a bit more complicated:

pmpath () 
{ 
    perl -M$1 -le "(\$_='$1')=~s{::}{/}g; print \$INC{\$_.'.pm'}"
}

My apologies for the misleading initial version.

FINAL UPDATE: There's actually a much better method for finding the path to a module, as pointed out by Your Mother below.

[sflitman]

That is shorter, and I like pmpath().

I have always thought I needed to use local for a Bash function since the args might change from line to line, but you're right, that's really for a multiline shell function (or am I just being paranoid?)

SSF

[Your Mother]

Does your pmpath function get you something perldoc -l doesn't?

cow@moo[609]~/bin>perldoc -l CGI
/usr/local/lib/perl5/5.10.0/CGI.pm

[graff]

Does your pmpath function get you something perldoc -l doesn't?

Evidently, perldoc -l does exactly what I intended to do with my "pmpath" shell function, so now I'll just reduce it to a shell alias:

alias pmpath='perldoc -l'

Much better -- thank you!