P
PerlFAQ Server
This is an excerpt from the latest version perlfaq4.pod, which
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
to review and update the answers. The latest version of the complete
perlfaq is at http://faq.perl.org .
--------------------------------------------------------------------
4.32: How do I strip blank space from the beginning/end of a string?
(contributed by brian d foy)
A substitution can do this for you. For a single line, you want to
replace all the leading or trailing whitespace with nothing. You can do
that with a pair of substitutions:
s/^\s+//;
s/\s+$//;
You can also write that as a single substitution, although it turns out
the combined statement is slower than the separate ones. That might not
matter to you, though:
s/^\s+|\s+$//g;
In this regular expression, the alternation matches either at the
beginning or the end of the string since the anchors have a lower
precedence than the alternation. With the "/g" flag, the substitution
makes all possible matches, so it gets both. Remember, the trailing
newline matches the "\s+", and the "$" anchor can match to the absolute
end of the string, so the newline disappears too. Just add the newline
to the output, which has the added benefit of preserving "blank"
(consisting entirely of whitespace) lines which the "^\s+" would remove
all by itself:
while( <> ) {
s/^\s+|\s+$//g;
print "$_\n";
}
For a multi-line string, you can apply the regular expression to each
logical line in the string by adding the "/m" flag (for "multi-line").
With the "/m" flag, the "$" matches *before* an embedded newline, so it
doesn't remove it. This pattern still removes the newline at the end of
the string:
$string =~ s/^\s+|\s+$//gm;
Remember that lines consisting entirely of whitespace will disappear,
since the first part of the alternation can match the entire string and
replace it with nothing. If you need to keep embedded blank lines, you
have to do a little more work. Instead of matching any whitespace (since
that includes a newline), just match the other whitespace:
$string =~ s/^[\t\f ]+|[\t\f ]+$//mg;
--------------------------------------------------------------------
The perlfaq-workers, a group of volunteers, maintain the perlfaq. They
are not necessarily experts in every domain where Perl might show up,
so please include as much information as possible and relevant in any
corrections. The perlfaq-workers also don't have access to every
operating system or platform, so please include relevant details for
corrections to examples that do not work on particular platforms.
Working code is greatly appreciated.
If you'd like to help maintain the perlfaq, see the details in
perlfaq.pod.
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
to review and update the answers. The latest version of the complete
perlfaq is at http://faq.perl.org .
--------------------------------------------------------------------
4.32: How do I strip blank space from the beginning/end of a string?
(contributed by brian d foy)
A substitution can do this for you. For a single line, you want to
replace all the leading or trailing whitespace with nothing. You can do
that with a pair of substitutions:
s/^\s+//;
s/\s+$//;
You can also write that as a single substitution, although it turns out
the combined statement is slower than the separate ones. That might not
matter to you, though:
s/^\s+|\s+$//g;
In this regular expression, the alternation matches either at the
beginning or the end of the string since the anchors have a lower
precedence than the alternation. With the "/g" flag, the substitution
makes all possible matches, so it gets both. Remember, the trailing
newline matches the "\s+", and the "$" anchor can match to the absolute
end of the string, so the newline disappears too. Just add the newline
to the output, which has the added benefit of preserving "blank"
(consisting entirely of whitespace) lines which the "^\s+" would remove
all by itself:
while( <> ) {
s/^\s+|\s+$//g;
print "$_\n";
}
For a multi-line string, you can apply the regular expression to each
logical line in the string by adding the "/m" flag (for "multi-line").
With the "/m" flag, the "$" matches *before* an embedded newline, so it
doesn't remove it. This pattern still removes the newline at the end of
the string:
$string =~ s/^\s+|\s+$//gm;
Remember that lines consisting entirely of whitespace will disappear,
since the first part of the alternation can match the entire string and
replace it with nothing. If you need to keep embedded blank lines, you
have to do a little more work. Instead of matching any whitespace (since
that includes a newline), just match the other whitespace:
$string =~ s/^[\t\f ]+|[\t\f ]+$//mg;
--------------------------------------------------------------------
The perlfaq-workers, a group of volunteers, maintain the perlfaq. They
are not necessarily experts in every domain where Perl might show up,
so please include as much information as possible and relevant in any
corrections. The perlfaq-workers also don't have access to every
operating system or platform, so please include relevant details for
corrections to examples that do not work on particular platforms.
Working code is greatly appreciated.
If you'd like to help maintain the perlfaq, see the details in
perlfaq.pod.