Perl/regex to remove first 3 lines and last 3 lines of a string

问题

I was looking to build a regex statement to always remove the first 3 lines of the string, and last 3 lines of the string (the middle part could be any n number of lines content). Any clean regex way to acheive this output? (i.e. always strip our first 3 lines and last 3 lines of the string - and preserve the middle part, which could be a variable # of lines)

Thanks.

e.g.

Input String:

"
1
2
3
<1...n # of lines content>
4
5
6
"

To desired output string:

"<1..n # of lines content>"

回答1:

The previously given solutions are really complex! All one needs is

s/^(?:.*\n){1,3}//;
s/(?:.*\n){1,3}\z//;

If you want to accept a lack of trailing newline at the end, you can use

s/\n?\z/\n/;
s/^(?:.*\n){1,3}//;
s/(?:.*\n){1,3}\z//;

or

s/^(?:.*\n){1,3}//;
s/(?:.*\n){0,2}.*\n?\z//;

回答2:

This should do the trick. You may need to replace "\n" with "\r\n" depending on the newline format of your input string. This will work regardless if the last line is terminated with newline or not.

$input = '1
2
3
a
b
c
d
e
4
5
6';

$input =~ /^(?:.*\n){3} ((?:.*\n)*) (?:.*\n){2}(?:.+\n?|\n)$/x;

$output = $1;

print $output

Output:

a
b
c
d
e

回答3:

for($teststring)
{
    s/<.*?>//g;
    $teststring =~ s%^(.*[\n\r]+){3}([.\n\r]).([\n\r]+){3}$%$2%;
    print "Outputstring" . "\n" . $teststring . "\n";
}

You need to test if it's under 6 lines, etc.

来源：https://stackoverflow.com/questions/7047205/perl-regex-to-remove-first-3-lines-and-last-3-lines-of-a-string