How to assign the match of string to another variable in perl?

Posted on 2012-12-21
Last Modified: 2012-12-21
I try to assign the matched result to another variable but instead the other variable gets the value 1 which means that there is match.

Here is the example:

my ($clientRoot) = $infoArray[3] =~ /Client root:\s.*/;

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$infoArray[3]  is something like this: Client root: some/path/in/here

Expected result for $clientRoot:


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Question by:Tolgar

Accepted Solution

brendonfeeley earned 200 total points
ID: 38714243
This should work:

my $clientRoot= $1 if $infoArray[3] =~ /Client root:\s(.*)/;

Author Comment

ID: 38714262
I normally use my other pattern but why didn't it work in here? Is it because I have an array instead of a single variable?
LVL 26

Assisted Solution

wilcoxon earned 200 total points
ID: 38714270
Or, more verbosely:
if ($infoArray[3] =~ m{Client root:\s+(.*)}) {
    $clientRoot = $1;

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Or, more compactly:
my ($clientRoot) = ($infoArray[3] =~ m{Client root:\s+(.*)});

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Expert Comment

ID: 38714271
You can do it this way:

my ($clientRoot) = $infoArray[3] =~ /Client root:\s(.*)/;

Only difference is the capturing group in the regex.
LVL 26

Expert Comment

ID: 38714275
Your original pattern did not work because you forgot the capturing parens (see my second option for the correct equivalent of your original idea (which I missed when reading the question)).  The parens around the match may not be required in mine but I usually include them just to make precedence explicit.
my ($clientRoot) = ($infoArray[3] =~ m{Client root:\s+(.*)});

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LVL 75

Assisted Solution

by:käµfm³d 👽
käµfm³d   👽 earned 100 total points
ID: 38714283
It did work, but the only thing output is the entire match. This is because you didn't specify any capture groups (parentheses, like brendonfeeley demonstrated). When you capture the result of a regex in list context in Perl, the capture groups are what are output to the variable. Since you didn't specify any groups, the output reverted to the overall match.

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