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Re: Shell Commands executed in perl script

by rinceWind (Monsignor)
on May 22, 2007 at 11:08 UTC ( #616729=note: print w/replies, xml ) Need Help??


in reply to Shell Commands executed in perl script

You have quotes inside your quotes. Try the following:

print $sh->ls( qq(-l ./logs/ |grep -v "^d" |grep -v file1 | awk 'NR>1 +{ print \$9}'));

Or indeed you could use q() as you're not doing any variable substitution inside the string.

Update: yes I know it would be better for the OP to have coded this in perl rather than shelling out 4 subprocesses. Sometimes you need to start with where the person is. Spotting the quote nesting is something glaringly obvious to me - a kind of mistake anyone can make in coding, and worth pointing out.

--
wetware hacker
(Qualified NLP Practitioner and Hypnotherapist)

Replies are listed 'Best First'.
Re^2: Shell Commands executed in perl script
by msk_0984 (Friar) on May 22, 2007 at 11:43 UTC
    Hey rinceWind

    Really thanx for your prompt reply ..... but i am facing another problem after making the changes, when executed it is giving me the error. Even used q() also but no success

    awk: 'NR
    awk: ^ invalid char ''' in expression

    Work Hard Party Harderrr!!
    Sushil Kumar
Re^2: Shell Commands executed in perl script
by Anonymous Monk on May 22, 2007 at 13:24 UTC
    Dude ... Do you really need to use Shell at all ??? Convert this shell to perl code...will probably take less time then debugging this errors one at a time.

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[Eily]: you could tie a variable into not having the same value each time, if you like to make people who try to debug your code facepalm
[Corion]: perl -wle 'package o; use overload q("") => sub {warn "str"; ""}, bool => sub{warn "bool"; 1}; package main; my $o={}; bless $o => o; print "Yay" if ($o && !length($o))'
[Corion]: But people writing such code should document the objects they construct and why it makes sense for an object to be invisible as string while being true in a boolean context
[hippo]: That's equal parts clever and horrendous.
[Eily]: the overload version wouldn't return true with "$x" && !length $x though, I guess
[hippo]: The more I look at this code, the more $x is a plain old scalar and the more this condition will never be true. I'm calling it a bug at this point.
[hippo]: Thanks for your input which has soothed my sanity (a little)
[Corion]: Eily: Sure - if you force both things into stringy things, then you break that magic. But that would also mean that you changed the expression, as now $x = 0.00 will be true instead of false as it were before
[Corion]: Ah no, at least in my feeble experiments that doesn't change the meaning
[Corion]: We sell sanity in small packages ;)

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