Hot answers tagged

2

Enlightenment. Don't be impressed by a pattern. Format it to something readable: [^.!?\s] [^.!?\n]* (?: [.!?] (?! ['"]? \s | $ ) [^.!?]* )* [.!?]? ['"]? (?= \s | $ ) There are three simple parts. No madness here, this pattern isn't complicated, but before assuming parts are useless you have to fully understand what they do. The first ...


1

You don't need to use Regex for this, and your regex doesn't cover most of cases anyway. You can simply splitting the input and try parse them into decimals. If they contains any invalid inputs then return three spaces (" ") otherwise print it in "00.000/00.000" format. Function StringFormatting(input As String) As String Dim parts = ...


1

/\b[^\We]+\b/g \W means NOT a "word" character. ^\W means a "word" character. [^\We] means a "word" character, but not an "e". see it in action: word without e "and" Operator for Regular Expressions BTW, I think this pattern can be used as an "and" operator for regular expressions. In general, if: ...


Only top voted, non community-wiki answers of a minimum length are eligible