Renaming files with criteria - regex

Question

Need some advice. I'm trying to do something with regular expressions that might not be possible, and if it is possible it's over my head. I can't get anything to work. I'm trying to create a tagging system for my PDF files. So if I have this file name:
"csharp 8 in a nutshell[studying programming csharp ebooks].pdf"
I would like all the words inside the '[ ]' to have a '#' in from of them. So the above file name would look like this:
"csharp 8 in a nutshell[#studying #programming #csharp #ebooks].pdf"
The problem is keeping the '#' inside the '[ ]'. For example I'd rather the 'csharp' at the very front of the file name not have the '#'.
Also, I'm using a bulk renamer called 'Bulk Rename Utility' to help me.
Can this be done?
If it can, any hints on how?
Thanks.

Answer 1

Bulk Rename Utility does not support replacing multiple matches, you can only match the whole file name and perform replacements using capturing groups/backreferences.
Since you are using Windows, I suggest using Powershell:
cd 'C:\YOUR_FOLDER\HERE'
Get-ChildItem -File | Rename-Item -NewName { $_.Name -replace '(?<=\[[^][]*?)\w+(?=[^][]*])','#$&' }
See this regex demo and the proof it works with .NET regex flavor.
(?<=\[[^][]*?) - right before this location, there must be a [ and then any amount of chars other than [ and ], as few as possible
\w+ - 1+ word chars
(?=[^][]*]) - right after this location, there must be any amount of chars other than [ and ], as many as possible, and then a ] char.
The replacement is # + the whole match value ($&).
Also, you may use
Get-ChildItem -File | Rename-Item -NewName { $_.Name -replace '(\G(?!\A)[^][\w]+|\[)(\w+)','$1#$2' }
See this regex demo and .NET regex test.
(\G(?!\A)[^][\w]+|\[) - Group 1 ($1): either the end of the previous match and 1+ chars other than ], [ and word chars, or a [ char
(\w+) - Group 2 ($2): one or more word chars.
If you only want to rename *.pdf files, replace Get-ChildItem -File with Get-ChildItem *.pdf.

Answer 2

I assume there is at most one bracket-delimited substring.
You can replace zero-length matches of the following regular expression with '#' when using Perl (click "Perl" then check global and case-different options), Ruby, Python's alternative regex engine, R with perl=true or languages that uses the PCRE regex engine, which includes PHP. With the exception of Ruby, the case-different (\i) and general (\g) flags need be set. Ruby only requires the case-indifferent flag.
r = /(?:^.*\[ *|\G(?<!^)|[a-z]+ +)\K(?<=\[| )(?=[a-z][^\[\]]*\])/
If using Ruby, for example, one would execute
str = "csharp 8 in a nutshell[studying programming csharp ebooks].pdf"
str.gsub(r,'#')
#=> "csharp 8 in a nutshell[#studying #programming #csharp #ebooks].pdf"
I believe all of the languages I named above allow one to run a short script from the command line. (I provide a Ruby script below.)
The regex engine performs the following operations.
(?: : begin non-capture group
^.*\[ * : match beginning of string then 0+ characters then '['
then 0+ spaces
| : or
\G : asserts the position at the end of the previous match
or at the start of the string for the first match
(?<!^) : use a negative lookbehind to assert that the current
location is not the start of the string
| : or
[a-z]+ + : match 1+ letters then 1+ spaces
) : end non-capture group
\K : reset beginning of reported match to current location
and discard all previously-matched characters from match
to be returned
(?<= : begin positive lookbehind
\[|[ ] : match '[' or a space
) : end positive lookbehind
(?= : begin positive lookahead
[a-z][^\[\]]*\] : match a letter then 0+ characters other than '[' and ']'
then ']'
) : end positive lookahead
Another possibility (illustrated with Ruby) is to break the string into three pieces, modify the middle one, then rejoin the pieces:
first, mid, last = str.split /(?<=\[)|(?=\])/
#=> ["csharp 8 in a nutshell[",
# "studying programming csharp ebooks",
# "].pdf"]
first + mid.gsub(/(?<=\A| )(?! )/,'#') + last
#=> "csharp 8 in a nutshell[#studying #programming #csharp #ebooks].pdf"
The regex used by split reads, "match a (zero-width) string that is preceded by '[' ((?<=\[) being a positive lookbehind) or is followed by ']' ((?=\]) being a positive lookahead.) By matching zero-width strings split does not remove any characters.
gsub's regex reads, "match a zero-width string that is at the start of the string or is preceded by a space and is followed by a character other than a space ((?! ) being a negative lookahead). It could alternatively be written /(?<![^ ])(?! )/ ((?<![^ ]) being a negative lookbehind).
A variant:
first + mid.split.map { |s| '#' + s }.join(' ') + last
#=> "csharp 8 in a nutshell[#studying #programming #csharp #ebooks].pdf"
I created a file named 'in' that contains the following two lines:
Little [Miss Muffet sat on her] tuffet
eating her [curds and] whey
Here is an example of a (Ruby) script that could be run from the command line to perform the necessary replacements.
ruby -e "File.open('out', 'w') do |fout|
File.foreach('in') do |str|
first, mid, last = str.split(/(?<=\[)|(?=\])/)
fout.puts(first + mid.gsub(/(?<=\A| )(?! )/,'#') + last)
end
end"
This produces a file named 'out' that contains these two lines:
Little [#Miss #Muffet #sat #on #her] tuffet
eating her [#curds #and] whey