Regex for decimal number with only single dot use replace() method

You could count the dots and replace onlc the firts with a dot, the rest remove the dot.

function singleDot(s) {
    return s.replace(/\./g, (c => _ => c++ ? '' : '.')(0));
}

console.log(['0.0', '0....0', '0.0...1'].map(singleDot));

You might use split and the dot as a separator:

const strings = [
  "10.01.1",
  "10.00............1",
  "100",
  ""
];
strings.forEach((s) => {
  const [head, ...rest] = s.split('.');
  console.log(rest.length ? head + '.' + rest.join('') : head);
});

Try /[^\d.\s]|(?<=\.\d*)\./g;. In short, replace every character except digits 0-9, ".", and white-spaces OR any "." that is preceded by a "." and zero or more digits 0-9. The positive look-behind (?<=) ensures that any "." that has a "." before it will be replaced which means that the first "." of every group will not be replaced. Note, that every group separated by a space or new line is considered a separate number due to not replacing any \s.

[^\d.\s]

|

(?<=\.\d*)\.

RegEx101

const rgx = /[^\d.\s]|(?<=\.\d*)\./g;
const str = `.08.2  5.880..
0.00..5.3
223...9.`;
const res = str.replace(rgx, "");

console.log(res);

If you want to do it via regex only, you can search for an alternation that captures your first period but simply matches all others and then replace all matches with the captured group. Since the captured string will be empty for the matches that aren't captured, only the first period group is replaced by itself while the other periods are deleted. See regex demo.

Search for (^\d*\.)|\. and replace with $1.

A simple solution that uses three replace operations, without regular expressions.

Probably not the fastest but definitely the easiest to read and understand.

function repair(nb) {
  return nb.replaceAll('.', '#').replace('#', '.').replaceAll('#', '');
}

console.log(['10.01.1', '10.00............1', '100', ''].map((i) => repair(i)));

String.replaceAll() has been introduced in 2020. It does not work in older browser. However, there is a solution that works with older browsers too and uses regexps!

function repair(nb) {
  return nb.replace(/\./g, '#').replace('#', '.').replace(/#/g, '');
}

console.log(['10.01.1', '10.00............1', '100', ''].map((i) => repair(i)));