FAQ 4.32 How do I strip blank space from the beginning/end of a string?

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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 physical
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. It 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 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;

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