To make the . + optional, you could do: \"(?:. +)?
The metacharacter \b is an anchor like the caret and the dollar sign. It matches at a position that is called a “word boundary”. This match is zero-length. There are three different positions that qualify as word boundaries: Before the first character in the string, if the first character is a word character.
Example: The regex "aa\n" tries to match two consecutive "a"s at the end of a line, inclusive the newline character itself. Example: "a\+" matches "a+" and not a series of one or "a"s. ^ the caret is the anchor for the start of the string, or the negation symbol.
(\d+)\s+(\(.*?\))?\s?Z
Note the escaped parentheses, and the ?
(zero or once) quantifiers. Any of the groups you don't want to capture can be (?:
non-capture groups).
I agree about the spaces. \s
is a better option there. I also changed the quantifier to insure there are digits at the beginning. As far as newlines, that would depend on context: if the file is parsed line by line it won't be a problem. Another option is to anchor the start and end of the line (add a ^
at the front and a $
at the end).
This ought to work:
^\d+\s?(\([^\)]+\)\s?)?Z$
Haven't tested it though, but let me give you the breakdown, so if there are any bugs left they should be pretty straightforward to find:
First the beginning:
^ = beginning of string
\d+ = one or more decimal characters
\s? = one optional whitespace
Then this part:
(\([^\)]+\)\s?)?
Is actually:
(.............)?
Which makes the following contents optional, only if it exists fully
\([^\)]+\)\s?
\( = an opening bracket
[^\)]+ = a series of at least one character that is not a closing bracket
\) = followed by a closing bracket
\s? = followed by one optional whitespace
And the end is made up of
Z$
Where
Z = your constant string
$ = the end of the string
You can do this:
([0-9]+) (\([^)]+\))? Z
This will not work with nested parens for Y, however. Nesting requires recursion which isn't strictly regular any more (but context-free). Modern regexp engines can still handle it, albeit with some difficulties (back-references).
Try this:
X (\(Y\))? Z
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