I need regexp to remove all not digits and not dash ('-'). new RegExp('[^0-9-]') seems works but new RegExp('[\\D-]') remove dash also. Why is this different?
I need regexp to remove all not digits and not dash ('-'). new RegExp('[^0-9-]') seems works but new RegExp('[\\D-]') remove dash also. Why is this different?
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2
Because
[\D-]is 'match any non-digit symbol OR dash', while[^0-9-]is 'match any symbol that's not digit or dash'. – raina77ow Commented Feb 5, 2013 at 20:40 -
1
[\D-]is equivalent to[\D], as the-is already matched by\D. – Blender Commented Feb 5, 2013 at 20:41 -
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Guys, it's JS: clearly
\\Dis how it's written in the string literal, not the regex one (it wouldn't match digits at all otherwise). The difference is worth to be mentioned, though. ) – raina77ow Commented Feb 5, 2013 at 20:46 -
@raina77ow: I don't normally use
new RegExp, I use regex literals, so I didn't think of that. – gen_Eric Commented Feb 5, 2013 at 20:53
5 Answers
Reset to default 13[^0-9-] is "anything that is NOT a digit, or is NOT a dash
[\D-] is "anythign that is NOT a digit, or IS a dash
the ^ inverts the entire [] character class, so on your \D version, there's no inversion, so a - is a legitimate match.
Because there is no negation in front of the dash in the second one. The \D (there should only be one backslash really) means 'all not-digits', the dash means a dash.
Because you have interpreted the meaning of your second regex wrong.
The ^ at the beginning of the character class [^0-9-] matches all characters specified in it, specifically anything thats not 0-9 and not -. Whereas [\D-] matches anything thats not a digit and is a -.
When you use the ^ in a character class ([]), it means "NOT anything in this class". \D is just a special sequence that means "anything that's not a digit".
So:
[^0-9-] matches anything that isn't a digit or a -.
[\D-] means "any non-digit character (or a -, which is already a non-digit). There is no need for the [] here, this is the same as just \D by itself.
\D is simple shorthand for ^0-9