Re^2: word with some number in between

[Anonymous Monk]

hey i did this , but there is some issue with the command , the first $ is not a part of regex as per perl . with this regex what i tried doing is that some $ before num and some $ after it . please tell what is the mistake

if ($array_A[$i] =~ /^$+[0-9]+$/)

[hippo]

The /^$+ part makes no sense. If you you just want to test for the presence of digits then a simple /[0-9]/ will work splendidly.

[Happy-the-monk]

Continuing hippo's advice, you should continue to the grep part next.

Cheers, Sören

[Anonymous Monk]

but since /0-9/ can be anywhere in the word , not necessarily in beginning or end

[hippo]

Precisely. eg.

$ perl -e 'print "Matched.\n" if "foo8baz" =~ /[0-9]/;'
Matched.
$ perl -e 'print "Matched.\n" if "foobaz" =~ /[0-9]/;'

[ww]

" ... what i tried doing is that some $ before num and some $ after it "

Do you, by any chance, mean you tried to deal with data that included a "$" sign... such as:  abc12$ or  $98abc or  $$zyx5abc or something similar?

If so, you'll have to escape the "$" sign (with a preceding backslash, "\") in your regex.

perl -E "my $s='$abc123'; if ($s =~ /\$/ ) { say 'Found a dollar-sign';}

which outputs:

Found a dollar-sign

Otherwise, (slightly oversimplified), the regex engine uses the "$" sign is only to mark an EOL.

Updated by quoting a possibly mis-understood part of OP's latest question.