I have this line:
val = val.replace(/[^0-9\.]/g, '')
and it replaces anything that is not a number, but I need a regular expression that limits val to be 2 numbers, period and then 2 numbers, like this:
- 11.33
- 12.34
- 54.65
I've already tried something like this but it didn't work:
val = val.replace(/^[^0-9\.]{1,2}/g, '')
Normally with replace you scan the entire string and keep the middle part. So start with beginning (^), scan some stuff you don't care about (.), then scan your number ([0-9]{1,2}(?:.[0-9]{0-2})?), then scan the rest which you don't care about (.), then you're at the end ($).
Then you replace with the middle capture group.
val.replace(/^(.*)([0-9]{1,2}(?:\.[0-9]{0-2})?)(.*)$/gm,'\2');
Use the m flag to process line by line.
.* in first group it will consume anything up to the last digit, rather use lazy .*? 2.) The {0-2} quantifier should be {0,2} 3.) The replacement is rather $2 than \2 (working demo).Sometimes it is easier to use multiple regexes instead of one. Here is a solution that uses a first regex to strip the string from anything but number digits, then a second regex that reduces the number string to the proper length:
const regex1 = /[^\d\.]/g;
const regex2 = /^.*(\d\d\.\d\d).*$/;
[
'aaa11.33zzz',
'aaa123.456zzz',
'-54.65',
'12xtra.34'
].forEach(str => {
let result = str.replace(regex1, '').replace(regex2, '$1');
console.log(str, '==>', result);
});Output:
aaa11.33zzz ==> 11.33
aaa123.456zzz ==> 23.45
-54.65 ==> 54.65
12xtra.34 ==> 12.34
Explanation of regex1:
[^\d\.] -- anything not 0-9 and .g -- flag to replace pattern multiple timesExplanation of regex2:
^ -- anchor at start of string.* -- greedy scan(\d\d\.\d\d) -- expect 2 digits, a dot, and 2 digits.* -- greedy scan$ -- anchor at end of stringYou did not specify what should happen if the input string has less than the 2digits, dot, 2 digits pattern. So this solution does not address that case.
123.456?