Here's an alternative way. I'm using the oninput event that is triggered on every value change by user (not only key presses). I'm saving the last valid value and restoring it whenever the new value is invalid.

<input type="text" id="test1" oninput="validateNumber(this);" />
<script>
var validNumber = new RegExp(/^\d*\.?\d*$/);
var lastValid = document.getElementById("test1").value;
function validateNumber(elem) {
  if (validNumber.test(elem.value)) {
    lastValid = elem.value;
  } else {
    elem.value = lastValid;
  }
}
</script>

In contrast to most other answers here this works flawlessly with all input techniques like drag'n'drop, copy'n'paste etc. It also supports special control keys like Ctrl+a (for selecting contents), Pos1, End and so on.

This should be work

var regex = /^\d+(\.\d{1,2})?$/i;

Try the following expression: ^\d+\.\d{0,2}$ If you want the decimal places to be optional, you can use the following: ^\d+(\.\d{1,2})?$

EDIT: To test a string match in Javascript use the following snippet:

var regexp = /^\d+\.\d{0,2}$/;

// returns true
regexp.test('10.5')

Use this:

document.getElementById(target).value = newVal.replace(/[^0-9.]/g, '');

Updated answer:

RegExp -

^(?:0|[1-9]\d+|)?(?:.?\d{0,2})?$

Explanation at regex101:

Original answer:

fiddle Demo

RegExp -

^(\d+)?([.]?\d{0,2})?$

Explanation

Assert position at the beginning of the string «^»
Match the regular expression below and capture its match into backreference number 1 «(\d+)?»
   Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Match a single digit 0..9 «\d+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match the regular expression below and capture its match into backreference number 2 «([.]?\d{0,2})?»
   Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Match the character “.” «[.]?»
      Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
   Match a single digit 0..9 «\d{0,2}»
      Between zero and 2 times, as many times as possible, giving back as needed (greedy) «{0,2}»
Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

Check length via lookahead.

               ↓               ↓       CHANGE THESE PARAMETERS
/^(?=(.[.]?){1,7}$)\d*([.]\d{1,2})?$/
  |---------------|                    LOOK-AHEAD TO CHECK LENGTH
                   |-|                 INTEGER PORTION OF NUMBER
                      |-----------|    OPTIONAL DECIMAL PORTION OF NUMBER

This is more easily scalable to variations of the problem such as "up to seven total digits with up to four decimal places" than other solutions are. Just replace the two characters pointed to by "CHANGE THESE PARAMETERS".

need

/^(?:\d{1,3}(?:,\d{3})*|\d+)(?:\.\d+)?$/

See the regex demo

Details

• ^ - start of string
• (?:\d{1,3}(?:,\d{3})*|\d+) - Either of:
  • \d{1,3}(?:,\d{3})* - 1 to 3 digits followed with 0+ sequences of a , and 3 digits
  • | - or
  • \d+ - 1 or more digits
• (?:\.\d+)? - an optional sequence of . and 1+ digits
• $ - end of string.

var strs = [',,,,', '111', '11,111', '11,111.0', '111111'];
var re = /^(?:\d{1,3}(?:,\d{3})*|\d+)(?:\.\d+)?$/;
for (var s of strs) {
  console.log(s + " => " + re.test(s));
}