如何在Perl中提取此文件名的部分内容？

Question

In my Excel sheet with column Changeset I am getting the changeset like:

C:\ccviews\hgdasdff-9302\dfcsz\ahgrt\kjhssl\ASGHLS@@\main\ajsdkljlat\hahdasdhfk\1\test.txt\sub\hsdaklfl\3

I need to use split function in a Perl script so that there will be two ouput (input as the above string)

• the part before @@ (eg-here C:\\ccviews\\hgdasdff-9302\\dfcsz\\ahgrt\\kjhssl\\ASGHLS )
• the last character of the string (eg-here 3)

Answer 1

对于常规拆分来说这听起来太复杂了，你需要像这样的普通正则表达式：

my ($first, $second) = / ^ (.+?) @@ .* (.) $ /x;

Answer 2

From Regular Expression Mastery by Mark Dominus:

Randal's Rule

• Randal Schwartz (author of Learning Perl [and also a Stack Overflow user ]) says:

Use capturing or m//g when you know what you want to keep.

Use split when you know what you want to throw away.

You know what you want to keep, so use m//g as in Leon Timmermans's answer .

Answer 3

I think this is what you want, or do you need it all in one statement?

 my ($before, $after) = split '@@', $input;
 my $last_char = substr($after, -1, 1);

Answer 4

Adding to the two answers already, you can try the following using only split function:

$s = 'C:\ccviews\hgdasdff-9302\dfcsz\ahgrt\kjhssl\ASGHLS@@\main\ajsdkljlat\hahdasdhfk\1\test.txt\sub\hsdaklfl\3';

@temp = split/@@/,$s;
$part1 = $temp[0]; # C:\ccviews\hgdasdff-9302\dfcsz\ahgrt\kjhssl\ASGHLS

@temp = split//,$s;
$part2 = $temp[-1]; # 3