将嵌套元素与通配符匹配

Question

I am trying to delete parts of a string with regex. I want to delete all parts, which look like this:

[[File: bla bla bla]]

so I used the following regular expression (note: [\\x5B] is '[' and [\\x5D] is ']'):

@"[\x5B][\x5B]File(.*?)[\x5D][\x5D]"
// Translates to @"[[File(.*?)]]"

However the text in the file tag can contain further tags, ie

[[File: bla bla [[foo]] bla]]

The above expression would leave

bla]]

because it detects the closing parentheses as the end of the match (note, I am searching non-greedy).

I therefore came up with this:

@"[\x5B][\x5B]File(.*?)[\x5B][\x5B](.*?)[\x5D][\x5D](.*?)[\x5D][\x5D]"
// Translates to @"[[File(.*?)[[(.*?)]](.*?)]]" 

which deletes all File tags, containing one inner tag. First calling this regex and then the simpler one above would delete all with one and zero inner tags.

However, the File tag can contain an arbitrary number of inner tags and obviously, my approach isn't very nice. I am just getting started with regex and any help is much appreciated.

Answer 1

I think this is how it is in the books.
@"\\[(?>(?:(?!\\[|\\]).)+|\\[(?<Depth>)|\\](?<-Depth>))*(?(Depth)(?!))\\]"

 \[                            # Match opening [
 (?>                           # Then either match (possessively):
      (?:                           #  the following group which matches
           (?! \[ | \] )                 #  (but only if we're not at the start of [ or ] )
           .                             #  any character
      )+                            #  once or more
   |                              # or
      \[                            #  [ (and increase the braces counter)
      (?<Depth> )
   |                              # or
      \]                            #  ] (and decrease the braces counter).
      (?<-Depth> )
 )*                            # Repeat as needed.
 (?(Depth)                     # Assert that the braces counter is at zero.
      (?!)                          # Fail this part if depth > 0
 )
 \]                            # Then match a closing ].