CODEWITHSUNDEEP
regexperl
user2701639
user2701639

Reputation: 921

regular expression what's the meaning of this regular expression s#^.*/##s

what is the meaning of s#^.*/##s because i know that in the pattern '.' denotes that it can represent random letter except the \n.

then '.* 'should represent the random quantity number of random letter .

but in the book it said that this would be delete all the unix type of path.

My question is that, does it means I could substitute random quantity number of random letter by space?

Upvotes: 0

Views: 350

Answers (4)

Dave Cross
Dave Cross

Reputation: 69314

Please excuse a little pedantry. But I keep seeing this and I think it's important to get it right.

s#^.*/##s is not a regular expression.

^.* is a regular expression.

s/// is the substitution operator.

The substitution operator takes two arguments. The first is a regular expression. The second is a replacement string.

The substitution operator (like many other quote-like operators in Perl) allows you you change the delimiter character that you use.

So s### is also a substitution operator (just using # instead of /).

s#^.*/## means "find the text that matches the regular expression ^.*/ and replace it with an empty string. And the s on the end is a option which changes the regex so that the . matches "\n" as well as all other characters.

Upvotes: 0

Keith Thompson
Keith Thompson

Reputation: 263507

Substitutions conventionally use the / character as a delimiter (s/this/that/), but you can use other punctuation characters if it's more convenient. In this case, # is used because the regexp itself contains a / character; if / were used as the delimiter, any / in the pattern would have to be escaped as \/. (# is not the character I would have chosen, but it's perfectly valid.)

^ matches the beginning of the string (or line; see below)

.*/ matches any sequence of characters up to and including a / character. Since * is greedy, it will match all characters up to an including the last / character; any precedng / characters are "eaten" by the .*. (The final / is not, because if .* matched all / characters the final / would fail to match.)

The trailing s modifier treats the string as a single line, i.e., causes . to match any character including a newline. See the m and s modifiers in perldoc perlre for more information.

So this:

s#^.*/##s

replaces everything from the beginning of the string ($_ in this case, since that's the default) up to the last / character by nothing.

If there are no / characters in $_, the match fails and the substitution does nothing.

This might be used to replace all directory components of an absolute or relative path name, for example changing /home/username/dir/file.txt to file.txt.

Upvotes: 4

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89584

s      -> subsitution
#      -> pattern delimiter
^.*    -> all chars 0 or more times from the begining
/      -> literal /
##     -> replace by nothing (2 delimiters)
s      -> single line mode ( the dot can match newline)

Upvotes: 5

Toto
Toto

Reputation: 91498

It will delete all characters, including line breaks because of the s modifier, in a string until the last slash included.

Upvotes: 3

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