如何使用正则表达式删除/ 1或/ 2？

Question

Regex gurus,

Here is the following line of code I want to parse with regex:

@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0/1

I want to obtain the following:

@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0

I have written the following regex on rubular.com:

(@.* *.)(!?(\/.))

My idea is to use negation to remove /1 by (!?(\\/.)) . However, this produces the entire line?

@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0/1

Why is (?!thisismystring) not removing /1? I googled the fire out of this, but they seemed to suggest similar things I am already trying? I deeply appreciate your help.

Answer 1

I think what you are trying to write is /(\\@.* .*)(?=\\/\\d)/ (you need to escape the at sign @ to prevent Perl from treating it as an array) but you need a positive look-ahead because you want to match everything up until the following characters are a slash followed by a digit.

Here is a program that demonstrates.

use strict;
use warnings;
use 5.010;

my $s = '@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0/1';

$s =~ /(\@.* .*)(?=\/.)/;

print $1, "\n";

But you would be much better off copying the whole string and removing the slash and everything after it, like this

use strict;
use warnings;

my $s = '@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0/1';

(my $fixed = $s) =~ s{/\d+$}{};

print $fixed, "\n";

output

@ERR030882.2595 HWI-BRUNOP16X_0001:3:1:6649:5175#0