Example strings :
2222
333333
12345
111
123456789
12345678
Expected result:
2@222
333@333
12@345
111
123@456@789
12@345@678
i.e. '@' should be inserted at the 4th,8th,12th etc last position from the end of the string.
I believe this can be done using replace and some other methods in JavaScript.
for validation of output string i have made the regex :
^(\d{1,3})(\.\d{3})*?$
You can use this regular expression:
/(\d)(\d{3})$/
this will match and group the first digit \d and group the last three \d{3} which are then grouped in their own group. Using the matched groups, you can then reference them in your replacement string using $1 and $2.
See example below:
const transform = str => str.replace(/(\d)(\d{3})$/, '$1@$2');
console.log(transform("2222")); // 2@222
console.log(transform("333333")); // 333@333
console.log(transform("12345")); // 12@345
console.log(transform("111")); // 111For larger strings of size N, you could use other methods such as .match() and reverse the string like so:
const reverse = str => Array.from(str).reverse().join('');
const transform = str => {
return reverse(reverse(str).match(/(\d{1,3})/g).join('@'));
}
console.log(transform("2222")); // 2@222
console.log(transform("333333")); // 333@333
console.log(transform("12345")); // 12@345
console.log(transform("111")); // 111
console.log(transform("123456789")); // 123@456@789
console.log(transform("12345678")); // 12@345@678
(\d)(\d{3})$ will be enough
var test = [
'111',
'2222',
'333333',
'12345',
'123456789',
'1234567890123456'
];
console.log(test.map(function (a) {
return a.replace(/(?=(?:\B\d{3})+$)/g, '@');
}));
You could match all the digits. In the replacement insert an @ after every third digit from the right using a positive lookahead.
(?=(?:\B\d{3})+$)
(?= Positive lookahead, what is on the right is
(?:\B\d{3})+ Repeat 1+ times not a word boundary and 3 digits$ Assert end of string) Close lookaheadconst regex = /^\d+$/;
["2222",
"333333",
"12345",
"111",
"123456789",
"12345678"
].forEach(s => console.log(
s.replace(/(?=(?:\B\d{3})+$)/g, "@")
));
.in it ? can you please explain ?