Whats the correct decoding of this regex: ^(\\s\\[abc.?)(\\])

Question

Here is the regex I'm looking at. I I've breaken it down into parts & am trying to understand what the below means .

^(\\s*\\[abc.*?)(\\])

This regex:

^- start of line
\\ --> matches \
s* --> matches character s ( 0 to unlimited times)
\\--> matches \

[abc.*?)(\\] --- I'm not sure what does this match exactly?

Did you try going through some regex tutorial? You'll find `\\s` combinedly matches whitespace. — Rohit Jain, Apr 10 '14 at 17:19
yeah I went through tutorial. Also tried regex101. It says the same there — user3520163, Apr 10 '14 at 17:24

score 1 · Answer 1 · answered Apr 10 '14 at 17:20

Double \\ mess is for Java. Remove one from there and try again explaining.

^(\s*\[abc.*?)(\])

Now the explanation again:

NODE                     EXPLANATION
--------------------------------------------------------------------------------
  ^                        the beginning of the string
--------------------------------------------------------------------------------
  (                        group and capture to \1:
--------------------------------------------------------------------------------
    \s*                      whitespace (\n, \r, \t, \f, and " ") (0
                             or more times (matching the most amount
                             possible))
--------------------------------------------------------------------------------
    \[                       '['
--------------------------------------------------------------------------------
    abc                      'abc'
--------------------------------------------------------------------------------
    .*?                      any character except \n (0 or more times
                             (matching the least amount possible))
--------------------------------------------------------------------------------
  )                        end of \1
--------------------------------------------------------------------------------
  (                        group and capture to \2:
--------------------------------------------------------------------------------
    \]                       ']'
--------------------------------------------------------------------------------
  )                        end of \2

score 0 · Answer 2 · edited May 23 '17 at 12:13

^(\\s*\\[abc.*?)(\\])

is a regular expression that is escaped for use in a Java string. After unescaping (after it's compiled by the java.util.regex.Pattern), it's

^(\s*\[abc.*?)(\])

Each piece:

is a ^:start of line/input anchor

is the start of a capture group

\s*

is *:zero or more \s:whitespace characters

\[abc

is the literal string [abc. The open-square-bracket is literal because it's escaped. Were it not escaped, it would be a special character indicating the start of a character class.

.*?

is a reluctantly zero or more of any character

is the end of the capture group

(\])

is another capture group with a single literal close-square-bracket (literal because it's escaped).

All links in this answer are from the Stack Overflow Regular Expressions FAQ.

score 0 · Answer 3 · answered Apr 10 '14 at 17:33

0

Pictorial representation from debuggex beta

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answered Apr 10 '14 at 17:33

Braj

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Whats the correct decoding of this regex: ^(\\s*\\[abc.*?)(\\])

3 Answers3

Whats the correct decoding of this regex: ^(\\s\\[abc.?)(\\])