Regex for pattern integer,integer or float

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I want to validate entry in textbox using regex. The pattern should be

integer, integer or float

so valid entries can be

1234, 1234
123, 123.45

and invalid entries will be

abd, 123
123, abc
abc, abc

so far I have tried

var entry = "123,123.4" 
var patt = new RegExp("^[0-9]+,(?:[\d+(\.\d+]+,?){0,1}$");
res = patt.test(entry);  

but it returns false, though it should return true

Replace you regExp by this one:

var reg = new RegExp("^[0-9]+,[0-9]+.?[0-9]*");

I think this is what you want.

To accept space after comma:

var reg = new RegExp("^[0-9]+, [0-9]+.?[0-9]*");

For case of 1a it's needed to ad $ in the end

Example: Matching Floating Point Numbers with a Regular Expression, Our regex should also match integers and floating point numbers where the integer part is not given. We will not try to match numbers with an exponent, such as� The Regex will match an integer or float, with a preceding +/- sign. I don't care for E notation, so I ignored it. nhahtdh mentioned "(5.) and (.6)": "5." does not look right to me, the trailing period has no purpose. However, ".6" is indeed valid, as the leading period indicates a positive value < 1 (i.e., it's short for "0.6").

Check if a given string is a valid number (Integer or Floating Point) in , Hence one of the regular expression for an integer number is [+-]?[0-9][0-9]* import java.util.regex.Matcher;. import java.util.regex.Pattern;. This regular expression matches an optional sign, that is eitherfollowed by zero or more digits followed by a dot and one or more digits (a floating point number with optional integer part), or that is followed by one or more digits (an integer). This is a far better definition. Any match must include at least one digit.

First split string by ","

var entry = "123,123.4";
var entryArray = entry.split(',');

then just check entryArray items using following function I got from link

function isFloat(n) {
    return n === +n && n !== (n|0);

function isInteger(n) {
    return n === +n && n === (n|0);

And here is full recipe

function isFloat(n) {
    return n === +n && n !== (n|0);

function isInteger(n) {
    return n === +n && n === (n|0);

function isValidInput(input){
    if (!input) {return false};
    var inputArray = input.split(',');
    inputArray ={
        return item.trim();
    inputArray = inputArray.filter(function(n){
        var number = parseFloat(n);
        return number && (isInteger(parseFloat(n)) || isFloat(parseFloat(n)) )
    return (inputArray.length === 2)

Solved: Regular expression that detects all numbers (integer or float , Solved: hi, i search a regular expression that detects all numbers (integer or float ) in any string ? thank you for your help. Accepts a number and a unit. The number part can be integer or decimal. The unit can be several variations of weeks, days, and hours: e.g., w, wk, week, ws, wks, weeks are all valid.


Will be valid to

1234, 1234
123, 123.45

And Invalid To

abd, 123
123, abc
abc, abc

A regex pattern that matches all forms of integers and decimal , integer ::= decinteger | bininteger | octinteger | hexinteger decinteger exponentfloat ::= (digitpart | pointfloat) exponent digitpart ::= digit To create the regex, simply translate the BNF into the corresponding regex patterns. REGEX_MATCH(string,REGEX pattern) – Returns True if the regex pattern is matched in the string field. REGEX_REPLACE(string,REGEX pattern, replacement) – Replaces the matched pattern with the specified replacement text.

Try this one:


652.16 valid

.12 invalid

511.22 valid

1.23 valid

12.3 valid

12.34 valid

6.10. Floating-Point Numbers, 6.10. Floating-Point Numbers Problem You want to match a floating-point number and specify whether the sign, integer, fraction and exponent parts of the� Below are other short-hands regular expression for an integer number [+-]?[0-9]+ [+-]?\d\d* [+-]?\d+ Check if a given string is a valid floating point number. For floating point number : Below is the regular definition for a floating point number.

JavaScript (and thus also JSON) does not have distinct types for integers and floating-point values. Therefore, JSON Schema can not use type alone to distinguish between integers and non-integers. The JSON Schema specification recommends, but does not require, that validators use the mathematical value to determine whether a number is an

When attempting to build a logical “or” operation using regular expressions, we have a few approaches to follow. Fortunately the grouping and alternation facilities provided by the regex engine are very capable, but when all else fails we can just perform a second match using a separate regular expression – supported by the tool or native language of your choice.

Best practices for regular expressions in .NET. 06/30/2020; 39 minutes to read +10; In this article. The regular expression engine in .NET is a powerful, full-featured tool that processes text based on pattern matches rather than on comparing and matching literal text.

  • The first one intenger (every) and the second one Integer of Float?
  • You should escape your dot
  • What are you trying to do with the '?+' at the end? Did you mean to put a '*'?
  • This works, only can you update it to even accept space please. So if its 123, 123.4 (with a space) it should also pass. Thanks in advance
  • If you need it to accept spaces you should probably add a couple of \s+. But it would make sense to do what Dhaval said and use String.split(',') and then String.trim() each string.
  • This will match on 1, 1a - which isn't really valid.
  • The link used is from 2010, there are much more modern method now, such as Number.isInteger()
  • There's a million ways to verify whether a number is an int or a float, that doesn't make this method less valid.
  • @Andrew It's not less valid but it might be less efficient, and anyway it seems wasteful to write code when there's a native function that already does the same thing.
  • @Aaron Number.isInteger() is not cross browser supported, you have to write Polyfill for that.
  • @Dhaval It's defined in ES6, which is usually good enough. However, according to the MDN it looks like you're right and its support is lagging behind :-/