Using local perl variables in shell command

Question

I am trying to run a remote find command but I have to pass it local variables whose value I get from command line arguments from user. But, I am getting an "undefined variable" error. I am doing something like :

my $ssh = Net::OpenSSH->new($host);
#I am getting a part of the $path and $pattern from command line
#I have to look for $pattern at the $path
my @files = $ssh->capture(q(find $path -name $pattern) );

This command runs fine when I give the exact path and pattern but I get error when I replace it with variables.

Schwern · Accepted Answer · 2017-04-12 03:44:18Z

2

q(find $path -name $pattern)

This is the literal string find $path -name $pattern, dollar signs and all.

Perl has several different kinds of quotes. "" will interpolate variables (ie. replace them with their values), '' will not. q() is another way to write ''. You want qq().

See Quote and Quote-Like Operators in perlop for more info.

answered Apr 12, 2017 at 3:44

Schwern

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3 Comments

emma Over a year ago

yes thank you for your reply. Now, it is not giving me an error but it is unable to get the matched patterns. That is, I am getting a "find:no match" when I store the same pattern in a variable but is running when I use value of pattern directly in command. Thanks again :)

emma Over a year ago

my $pattern = "40786"; #This runs when I write my @files = $ssh->capture2(qq(find /u1/temp -name "40786") ); #but returns not found when I use variable $pattern Sorry if its a silly question. I am new to perl and couldn't find an explanation elsewhere.

Schwern Over a year ago

@emma That should work. Without the surrounding code, I can't tell you what's wrong. I'd suggest you ask a new question and provide a complete program that doesn't work.

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