I've been searching but haven't found a solution.
I have a regexp:
var reg = /^((\d){1,5}(\.|,)(\d{2}))/;
And a string:
var str = '12323.098765421';
I'm trying to keep part that match regexp and delete all the rest.
What I'm doing:
str.replace(/^((\d){1,5}(\.|,)(\d{2}))/, '$1 // can't understand what should I put here to replace first part');
Let's imagine you have a string that starts with a number that conforms to the pattern:
1 to 5 digits + decimal separator + 2 digits
This pattern may be followed with any character, even including a newline. Then, in JS, you can use the following replacement:
.replace(/^(\d{1,5}[.,]\d{2})[\s\S]*/, "$1")
Where
^ - matches the start of the string(\d{1,5}[.,]\d{2}) - matches and captures our pattern (with , or . as decimal separators)[\s\S]* - matches any 0+ characters, even including a newline.var re = /^(\d{1,5}[.,]\d{2})[\s\S]*/;
var str = '12323.098765421';
var result = str.replace(re, "$1");
console.log(result);The $1 is a backreference to the value captured with Group 1 (formed with the help of a pair of unescaped parentheses).
See more information on capturing groups and backreferences.
The regex will match the required output, since you want to remove the rest then you need to update it like this.
var str = '12323.098765421';
console.log(
str.replace(/^(\d{1,5}[.,]\d{2})\d+/, '$1')
);Or you need to use String#match method to get the matched string
var str = '12323.098765421';
console.log(
str.match(/^\d{1,5}[.,]\d{2}/)[0]
);If the intent is to round to two places, you can use the following regex:
'12323.09876542'.replace(/(\.[0-9][0-9])\d+/, '$1');
Or, you can use the toFixed() Number function:
Number('12323.09876542').toFixed(2)
12323.09? See regex101.com/r/jF5tV6/1. But why remove? You can useRegExp#match:"12323.098765421".match(/^\d{1,5}[.,]\d{2}/)[0].can't understand what should I put here to replace first partpart.