var str = 'asd-0.testing';
var regex = /asd-(\d)\.\w+/;
str.replace(regex, 1);
That replaces the entire string str with 1. I want it to replace the matched substring instead of the whole string. Is this possible in Javascript?
6 Answers
var str = 'asd-0.testing';
var regex = /(asd-)\d(\.\w+)/;
str = str.replace(regex, "$11$2");
console.log(str);Or if you're sure there won't be any other digits in the string:
var str = 'asd-0.testing';
var regex = /\d/;
str = str.replace(regex, "1");
console.log(str);
3 Comments
Ivan Rave
or using function: 'asd-0.testing'.replace(/(asd-)\d(\.\w+)/, function(mystring, arg1, arg2){return arg1 + 'mynumber' + arg2})
Martin Massera
is there any answer where you DONT know the structure of the regex? here you are basically creating a new regex with two matches
Bas Slagter
It is good to know that you need braces () around the part you want as $1, $2 etc.
using str.replace(regex, $1);:
var str = 'asd-0.testing';
var regex = /(asd-)\d(\.\w+)/;
if (str.match(regex)) {
str = str.replace(regex, "$1" + "1" + "$2");
}
Edit: adaptation regarding the comment
1 Comment
dave
I want to replace the substring with '1' not the entire string with the substring
I would get the part before and after what you want to replace and put them either side.
Like:
var str = 'asd-0.testing';
var regex = /(asd-)\d(\.\w+)/;
var matches = str.match(regex);
var result = matches[1] + "1" + matches[2];
// With ES6:
var result = `${matches[1]}1${matches[2]}`;
2 Comments
eduncan911
+1 I personally like having the collection of matches to frack with.
Travis J
I agree, having the matched set available is more readable in my opinion.
I think the simplest way to achieve your goal is this:
var str = 'asd-0.testing';
var regex = /(asd-)(\d)(\.\w+)/;
var anyNumber = 1;
var res = str.replace(regex, `$1${anyNumber}$3`);
Comments
A One-Liner based on @Greg7000's answer 👍🏼 which uses a replacer method:
'asd-0.testing'
.replace(
/(asd-)\d(\.\w+)/,
(match: string, p1: string, p2: string) => {
return `${p1}1${p2}`;
});
Comments
Personally, I like the liberty provided by the replacer technique (Search for 'Specifying a function as the replacement' section). This is also simplier to debug in details.
function replacer(match, p1, p2, offset, string) {
console.log('match = ' + match);
console.log('p1 = ' + p1);
console.log('p2 = ' + p2);
console.log('offset = ' + offset);
console.log('string = ' + string);
custom_op = p1 + 1 + p2;
return custom_op;
}
var str = 'asd-0.testing';
var regex = /(asd-)\d(\.\w+)/;
matches = str.match(regex)
res = str.replace(regex, replacer);
console.log(res);
// ----OUTPUT----
// match = asd-0.testing
// p1 = asd-
// p2 = .testing
// offset = 0
// string = asd-0.testing
// asd-1.testing
NOTE: To ease the identification of regex groups, consider using https://regex101.com/
str.replace(regex, "$`1$'")see developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…