Regular expression for accepting integers, decimal values and limiting by maximum value

Question

I need a regular expression that accept only integers or decimals from 1 to 9999.99. I tried this

^(?:[1-9][0-9]{0,4}(?:.d{1,2})?|9999|9999.99)$

but it is also accepting 10000.

Thank you for the reply, but this expression is not accepting all 2 digit, 3 digit, 4 digit numbers. — Rakhi, Dec 30 '21 at 12:28
@Rakhi As per the examples shared in the demo link, it does match 1/2/3/4 digit nos — Gurmanjot Singh, Dec 30 '21 at 12:30
Are you using a specific programming language? Arithmetic is usually a much better solution. — MonkeyZeus, Dec 30 '21 at 13:49
You can [generate any numeric range regex here](https://stackoverflow.com/a/67502416/3832970). Anyway, `d` can never match a digit, `\d` matches a digit. Also, `.` matches any char other than line break chars, a `\.` matches a dot. — Wiktor Stribiżew, Dec 30 '21 at 15:50

The fourth bird · Accepted Answer · 2021-12-30T15:32:27.060

There are a few issues in the pattern:

This part [1-9][0-9]{0,4} can match 1-5 digits as the quantifier is 0-4 times so it can match 10000 as well
You don't need 9999|9999.99 as those can already be matched by 1-9][0-9]{0,4}
This part .d matches any char except a newline followed by a d char. You have to escape both to match a dot literally and a single digit
With those changes, you can omit the outer capture group

The updated pattern looks like:

^[1-9]\d{0,3}(?:\.\d{1,2})?$

This **is** the right answer. I started regex101'ing before reading the answers and my regex turned out identical to yours. — MonkeyZeus, Dec 30 '21 at 13:52

score 2 · Answer 2 · answered Dec 30 '21 at 12:23

2

Try this regex:

^0*(?!\.\d+)[1-9]\d{0,3}(?:\.\d{1,2})?$

Explanation:

^ - matches start of the string
0* - matches 0+ occurrences of digit 0
(?!\.\d+) - zero-length match if the current position is not followed by a . and 1+ digits
[1-9] - matches a digit from 1 to 9
\d{0,3} - matches at-least 0 or at-most 3 occurences of a digit
(?:\.\d{1,2})? - matches the decimal . followed by 1 or 2 digits. A ? at the end to make the fractional part optional
$ - matches the end of string

answered Dec 30 '21 at 12:23

Gurmanjot Singh

Thank you. This regex not supporting 22, 33, 434 etc..supports only when adding 0 before each number. Help is appreciated – Rakhi Dec 30 '21 at 12:35
[Sorry, but it does match the values you have mentioned](https://regex101.com/r/oQYglx/1) – Gurmanjot Singh Dec 30 '21 at 12:36
I'm not sure why you put the `0*(?!\.\d+)`. And if you remove this, it's almost identical with the other answer, but still a bit earlier it seems. – bobble bubble Dec 30 '21 at 15:08
@bobblebubble I had added `0*` to match inputs like `01`,`001`,`00001`. I had added `(?!\.\d+)` so as to exclude inputs like `.99`, `.11` as the OP had asked to include the numbers>=1 – Gurmanjot Singh Dec 30 '21 at 15:16
Thanks for your quick reply! Well, not clear about that, however upvoted for it already. – bobble bubble Dec 30 '21 at 15:23

score 1 · Answer 3 · answered Dec 30 '21 at 15:27

1

Another idea but there is not much difference regarding performance (just for fun).

^(?!0)\d{1,4}(?:\.\d\d?)?$

The looakhead (?!0) prevents starting with zero, so we can further match \d{1,4}.

Does not respect leading zeros. See this demo at regex101.

answered Dec 30 '21 at 15:27

bobble bubble

3 Answers3