Honestly, I thought it would be easy to find a regex that would match something like:

1,234.56 OR 1234.56 OR 123,456.789029329 or even -223.233434332 etc.

But it wasn’t. I found some but they didn’t quite work or

Anywhere here is the regex I crafted and settled on:

**^(\d -)?(\d ,\d{3})\.?\d$**

It isn’t perfect, but it is as close as I could manage. Here are some matches:

  • 1322.23
  • 1,322.23
  • 1,322
  • 1132

Here are some numbers that fail to match

  • 1,32
  • 1,32,2
  • 1320.223,232

So, in general this works great but it is possible to get by the regex somewhat because I just am not strong enough with regex’s at the moment to fix it:

Here is an example:

  • 1,2345,232

Notice there are four numbers between the first and the second commas. My rule insists there has to be at least 3 numbers but doesn’t insist on a comma before the next number. The reason for this is the fact that I don’t want commas to be required before the decimal point. My initial effort which insisted on three numbers after the first comma and a second comma before the next three numbers etc required that any number in the thousands or higher had to have a comma if I wanted to use a decimal point. That wasn’t acceptable so the current regex is my best compromise.

If you can offer one better let me know in the comments.



Try this:


It doesn’t allow trailing zeros (“012”) (“0.000x” is allowed). Negative zero is not allowed. A number is valid if it ends with a period, as in “12.”


Sorry that doesn’t work.

I just tested it and with the number
1,234.232.232 it fails in two ways:

first, it selects 234.232.232 so it drops off the 1, and it allows multiple decimal points.


The following is untested, but I think it may be on the right track, unless I’m breaking the regex engine somehow…

Basically, it first tries to match “0”. If it can’t, it tries to match -1 < x < 1. If it still fails, it tries to match ddddd.ddddd with optional commas. No leading 0’s are allowed.

The only problem I see is that -0.0000 is currently allowed, but I think that could be resolved.

(\d{0,2},(\d{3},)\d{3}) |