$value = $list[1] ~ s/\D//g;
syntax error at try1.pl line 53, near "] ~"
Execution of try1.pl aborted due to compilation errors.
I am trying to extract the digits from the second element of @list, and store it into $value.
views:
80answers:
4
+1
A:
maybe you wanted the =~ operator?
P.S. note that $value will not get assigned the resulting string (the string itself is changed in place). $value will get assigned the number of substitutions that were made
newacct
2010-08-27 05:06:51
@newacct: Then how come it gets assigned in [this case](http://stackoverflow.com/questions/3577087/how-to-read-from-a-file-and-do-a-grep-in-perl/3577114#3577114)?
Lazer
2010-08-27 05:13:57
@values = ($list[1] =~ /\d/g);
hlynur
2010-08-27 05:22:44
@newacct: thanks!
Lazer
2010-08-27 05:48:29
A:
And wanted \digits not non-\Digits. And have a superfluous s/ubstitute operator where a match makes more sense.
if ($list[1] =~ /(\d+)/) {
$value = $1;
}
msw
2010-08-27 05:08:18
your match only works if the digits are contiguous. the search-and-replace will work on something like `$100,000` for example
plusplus
2010-08-27 09:55:01
+5
A:
You mean =~, not ~. ~ is a unary bitwise negation operator.
A couple of ways to do this:
($value) = $list[1] =~ /(\d+)/;
Both sets of parens are important; only if there are capturing parentheses does the match operation return actual content instead of just an indication of success, and then only in list context (provided by the list-assign operator ()=).
Or the common idiom of copy and then modify:
($value = $list[1]) =~ s/\D//;
ysth
2010-08-27 05:19:21
Actually, it's a lot like C. The operator you were using doesn't do in C what you were trying to do either. :)
brian d foy
2010-08-27 08:27:08
A:
You said in a comment that are trying to get rid of non-digits. It looks like you are trying to preserve the old value and get the modified value in a new variable. The Perl idiom for that is:
( my $new = $old ) =~ s/\D//g;
brian d foy
2010-08-27 08:28:38