Javascript Regex: negative lookbehind

Question

I am trying to replace in a formula all floating numbers that miss the preceding zero. Eg:

"4+.5" should become: "4+0.5"

Now I read look behinds are not supported in JavaScript, so how could I achieve that? The following code also replaces, when a digit is preceding:

 var regex = /(\\.\\d*)/, formula1 = '4+1.5', formula2 = '4+.5'; console.log(formula1.replace(regex, '0$1')); //4+10.5 console.log(formula2.replace(regex, '0$1')); //4+0.5 

Answer 1

Try this regex (\\D)(\\.\\d*)

 var regex = /(\\D)(\\.\\d*)/, formula1 = '4+1.5', formula2 = '4+.5'; console.log(formula1.replace(regex, '$10$2')); console.log(formula2.replace(regex, '$10$2')); 

Answer 2

You may use

s = s.replace(/\B\.\d/g, '0$&')

See the regex demo .

Details

• \\B\\. - matches a . that is either at the start of the string or is not preceded with a word char (letter, digit or _ )
• \\d - a digit.

The 0$& replacement string is adding a 0 right in front of the whole match ( $& ).

JS demo:

 var s = "4+1.5\\n4+.5"; console.log(s.replace(/\\B\\.\\d/g, '0$&')); 

Another idea is by using an alternation group that matches either the start of the string or a non-digit char, capturing it and then using a backreference:

 var s = ".4+1.5\\n4+.5"; console.log(s.replace(/(^|\\D)(\\.\\d)/g, '$10$2')); 

The pattern will match

• (^|\\D) - Group 1 (referred to with $1 from the replacement pattern): start of string ( ^ ) or any non-digit char
• (\\.\\d) - Group 2 (referred to with $2 from the replacement pattern): a . and then a digit