regex optional floating decimal

Question

I have a MySQL double(10,2) which I've applied the following regex: /^\d{0,10}\.\d{2}$/

I'm not passing validation for 1234, I'm assuming the decimal is required (I understood ? is the optional char which is absent).

Where do I place the ? char in this regex?

thanks

Why are you treating them as text? Whats the ultimate aim here? — Alex K., Jun 07 '18 at 13:10
@AlexK. to place a number from 0 to 9999999999.99 without requiring a decimal, you can skip the "only numeric" part since I am using a validation package and I just add "numeric|regex: /.../" — erezt, Jun 07 '18 at 13:12
`/^\d{1,10}(\.\d{1,2})?$/`, otherwise you will not accept `123.4`, and also it is good to have at least one digit before the dot. — mik, Jun 07 '18 at 13:29
Yeah, it depends on what the actual requirements are. If 1 digit after `.` is expected, the `{2}` should be replaced with `{1,2}` — Wiktor Stribiżew, Jun 07 '18 at 13:31

score 2 · Accepted Answer · answered Jun 07 '18 at 13:35

To solve the immediate problem you need an optional group around \.\d{2} pattern - (\.\d{2})?:

/^\d{0,10}(\.\d{2})?$/

It matches 0 to 10 digits from the start of the string and then either requires the end of the string, or . and 2 digits.

Next step, you may allow 1 or 2 digits in the fractional part using a {1,2} limiting quantifier:

/^\d{0,10}(\.\d{1,2})?$/

And even use a non-capturing group to get some very tiny efficiency boost:

/^\d{0,10}(?:\.\d{1,2})?$/

1 Answers1