perl - using a scaler variable to represent an R.E.

[Mr Peter Kempner]

I am trying to write a perl script and I am confused. I want to use a scalar variable to hold an RE against which to pattern match. It doesnt work. eg

$pattern='/humpty|jack/'
.
.
if ($variable_x =~ $pattern)
{
print 'Match found for humpty or jack'
}

Does not work.

If instead of $pattern I use /$var_a|$var_b/ where $var_a = humpty and $var_b = jack then it works fine! What am I not getting???

[harry d brown jr]

Change $pattern to

$pattern=/humpty|jack/;

get rid of the quotes

live free or die
harry

[Rodney Hills]

The "/" is part of the syntax for the matching operation.

Therefore do the following-

$pattern='humpty|jack'
.
.
if ($variable_x =~ /$pattern/)
{
print 'Match found for humpty or jack'
}

Should work.

Note- If $pattern does not change, you can use the "o" option (ie /$pattern/o) to have perl generate more effecient code for execution.

-- Rod Hills

[harry d brown jr]

Rodney,

I like your solution much better!

live free or die
harry

[H.Merijn Brand (procura]

You got it almost right the first time, but you forgot to read the docs (use the qr// operator):

$pattern = qr/humpty|jack/;
.
.
if ($variable_x =~ $pattern)
{
print 'Match found for humpty or jack'
}

Harry's solution is plain wrong: it'd asign the result of matching $_ against the pattern. Rodney's solution is technically correct, but still suboptimal and error-prone.

[Ralph Grothe]

Sorry, a wee "correction" to Rodney's answer.

In deed you can use almost any character pair to embrace search and substituion patterns when using the operators m, s, y, tr

At the shell type

perldoc perlop
perldoc -f m
perldoc -f y
perldoc -f s
perldoc -f qr


This comes handy when you have to deal with paths, as subdirectories are delimited by /, which you otherwise would have to escape by
instead of writing

$path =~ /\/usr/\/local/\/bin/

you can say

$path =~ m|/usr/local/bin/|

or

$path =~ m(/usr/local/bin)

Of course you can also reference variables in patterns

@users = qw(Humpty Jack);

print "found match\n" if ($name =~ /$users[0]|$users[1]/)

You can even use the qr operator if you insist

$pattern = qr(Humpty|Jack);
print "found match" if $name =~ $pattern;

You can inverse logic by using '!~' instead of '=~'
But this should be familiar from awk.