javascript - Look behind replace all occurrences -


i want replace occurences of .digit 0.digit.

i'm new regular expressions far understand use behind this. js not support that, i'd know if knows solution. show problem wrote following code. 

str = "0.11blabla.22bla0.33bla.33" allow = "\\.\\d*" str.match(new regexp(allow,"g")) [".11", ".22", ".33", ".33"] deny = "0\\.\\d*" str.match(new regexp(deny,"g")) ["0.11", "0.33"] diffreg= new regexp("(?!"+deny+")"+allow,"g") // translates to: /(?!0\.\d*)\.\d*/g str.match(diffreg) [".11", ".22", ".33", ".33"]

obviously allow matches decimal values whereas deny matches values preceding 0. result should of course set difference between two: [".33", ".33"].

i think looking regex instead

  [0]?(\.\d*)

so in code have:

intersectionreg = new regexp("[0]?("+allow+")","g")

thanks @richard, edited


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