Variable number of matches / back references in Regex?

Question

I spent some time recently converting some SCSS code to Stylus, and one of my regex steps was to strip out the $ character in variable names, while formatting those names to camelCase rather than hyphenated-case.

For example, $long-hyphenated-variable-name: value; became longHyphenatedVariableName = value; but parameter names like text-size did not become textSize.

anyway, my Regex skills are pretty limited, so I could only figure out how to do it as a series of passes, looking first for variable names that might have as much as 6 parts (5 hyphens) and then for shorter ones on subsequent passes and passing each part of the string as a separate backreference so I could capitalize the first letter. But I wondered if there was a better way to do it: i.e. is there a way to get a variable number of sub-string matches within the string and supply back-references in accordance with how many you got? I just couldn't find any information on this.

This is my code, in six steps, which I'd like to condense to one if it's possible (I'm thinking of writing a multi-step regex routine that can convert SCSS to Stylus for future use). In each case the search expression is followed by the replacement expression:

\$(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(.)
$1\u$2\u$3\u$4\u$5\u$6

\$(\S*?)\-(\S*?)\-(\S*?)\-(\S*?)\-(.)
$1\u$2\u$3\u$4\u$5

\$(\S*?)\-(\S*?)\-(\S*?)\-(.)
$1\u$2\u$3\u$4

\$(\S*?)\-(\S*?)\-(.)
$1\u$2\u$3

\$(\S*?)\-(.)
$1\u$2

\$(.)
$1

Just FYI, I added the `boost-regex` tag because that seems to be the flavor Sublime Text uses. — Alan Moore, Jun 28 '12 at 19:27

score 0 · Answer 1 · answered Jun 25 '12 at 11:34

If you use C# you are in luck because it is one of the few languages (if not the only one) that supports subexpression captures

I'm interested in other languages so if your language supports pleas let us know

http://msdn.microsoft.com/en-us/library/system.text.regularexpressions.capture(v=vs.110)

The .NET API can be looked at as follows

 Matches
     Groups (most regex engines stop here)
         Captures (unique for .NET)

Cylian · Answer 2 · 2012-06-28T08:32:16.773

0

while ($subject =~ m/(\$\S*?)-/g) {
    $subject =~ s/(\$\S*?)-/$1/g;
}

UPDATE

you might use this pattern

(?im)(\$(?![a-z-]+\s*:\s*\d+)\S*?)-

s/b replaced with none.

Within editor you could use multiple find-replace to get the desired output or could create a simple recursive macro implementing the logic.

edited Jun 28 '12 at 08:32

answered Jun 25 '12 at 11:36

Cylian

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This may be the solution, but I don't have the know-how to implement it -- I'm just using Regex within the Find and Replace functionality of my text editor. Is there a way to apply this in that context? Or what else could/should I be using? – lunelson Jun 26 '12 at 14:23

score 0 · Answer 3 · answered Jun 25 '12 at 11:42

There are globals replacements, they don't are about the number of matches. They may not provide a general solution to the problem, but in sed you can do:

$ sample='$test-size: 12; $long-hyphenated-variable-name: value;'
$ echo $sample | sed 's/-\(\w\)/\U\1/g; s/\$\(\w\+\):/\1 =/g'
testSize = 12; longHyphenatedVariableName = value;

Variable number of matches / back references in Regex?

3 Answers3