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

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

Wiktor Stribiżew
– Wiktor Stribiżew

2017-12-15 20:23:34 +00:00
Commented Dec 15, 2017 at 20:23 — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Dec 15, 2017 at 20:23

Youcef LAIDANI · Accepted Answer · 2017-12-15 20:28:16Z

2

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'));

answered Dec 15, 2017 at 20:28

Youcef LAIDANI

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Comments

Wiktor Stribiżew · Accepted Answer · 2017-12-15 20:41:11Z

1

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

edited Dec 15, 2017 at 20:41

answered Dec 15, 2017 at 20:33

Wiktor Stribiżew

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2 Comments

Beppe Over a year ago

Thanks, the 2nd solution works just fine! Version 1 does not work, since it is possible that the dot is preceeded by letters of functions (eg "ln.05")

Wiktor Stribiżew Over a year ago

@Beppe Glad it worked for you. Note that without ^ alternative, a .5 number at the start of the string would not have been changed to 0.5.

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