Regular expression to validate that a number has a comma as a decimal symbol

Try this expression

^[0-9]+([,][0-9]+)?$

Debugx Demo

this should be your expression

\d+(,\d+)?

remember that if you pass it to a function as a string you must escape the slashes, like so:

"\\d+(,\\d+)?"

This expression accepts negatives, uses \dinstead of [0-9], without creating unnecessary groups, or escaping unnecessary characters:

^-?\d+(?:,\d+)?$

Alternatively, if you want to accept scientific notation (eg: 1.2e+05):

^-?\d+(?:,\d+)?(?:[Ee][-+]?\d+)?$

Demo en regex101.com

Some notes on the differences:

Negatives . It is a common mistake to forget that a number can also be negative and start with a -. I hope it helps to avoid headaches when the code is already in production.

Using\d . In some languages, \dit matches digits from other alphabets . Among the big ones: Python3, those of .Net, Perl, C/C++ (with boost), R, TCL or Oracle. If you want to avoid that detail, there we would replace all the \dby [0-9]. But for the rest, \dit is exactly the same as [0-9]and visually it is better.

Without creating unnecessary groups . When you use parentheses, you're not just grouping, you're also capturing the text that is matched by that group. But there are groups without capturing , whose syntax is , to avoid using unnecessary memory. Also, it's easier for whoever is reading your regex, because it's clear that you're grouping but you don't care about getting that part back. _{- I know, at first it is difficult to read those and more, which is not a quantifier, but I assure you that in a short time it is read in a row and understood automatically .}(?:subpatrón)
?:

Without escaping what is not necessary . The comma ( ,) is not a metacharacter in regex. As such, it doesn't need to be escaped (neither \,, nor [,]), just the ,in the regular expression is taken as a literal.

Java has no option for string literals ☹. It's the only language (that I know of) where you have to escape each slash to use this regex:

• Code:

  Matcher matcher = Pattern.compile("^-?\\d+(?:,\\d+)?(?:[Ee][-+]?\\d+)?$").matcher(texto);
  if (matcher.find()) {
      //Número válido
  }
  

With this regular expression you validate that the numbers are separated by commas for the thousands without decimals.

[0-9]{1,3}(\,[0-9]{3})

Example:

• 100,000 -> Correct
• 100 -> Correct
• 10.00 -> Wrong
• 10,000 -> Correct

You can validate it at http://regexr.com/

 var patron= /^[0-9]+([.])?([0-9]+)?$/;

This is better because it allows you to put only integers if necessary, and if you put a point it only accepts it, or with decimals, I leave you with some case scenarios

123
123.
123.00
123.0215

Debuggex Demo https://www.debuggex.com/r/X43uNbM0EERtm8NF

With this it validates that it has 1 or 2 characters after the decimal point.

it is not mandatory to place decimals.

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

https://www.debuggex.com/r/qTPyvu3WB9xuUCaj

If I understand your description correctly, the following string would be a valid number:

22,

For that string to be accepted as well, the regular expression would be:

    ^[0-9]+([,][0-9]*)?$

Debugx Demo

^It is important to include the chain start and chain end anchors $.
Without the anchor $the chain 232.33dfdfdas correct when it is not.

If 22,it is not a correct string then the regular expression to use is:

^[0-9]+([,][0-9]+)?$

Debugx Demo