Regex: matching for digits excluding some digits at a position

user_2444661 0
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I have a string which has a pattern, say "xxx.xxx".

So first three are digits, then a . and then 3 more digits.

I want to check if first digit is always 3, and next 2 can be any digits except 30 and 40, rest can be anything.

My regex attempt is:

"3[^34][0-9]."

But it doesn't seem to work. Which part here is wrong?

regex
3 Answers

Try this: ^3(?!30|40)\d{2}\.\d{3}$

Test regex here: https://regex101.com/r/iCtKnB/1

^          matches the start of string
3          matches the number 3
(?!30|40)  checks if the next two numbers are not either 30 or 40
\d{2}      matches two digits if they are not 30 or 40
\.         matches period
\d{3}      matches the next three digits
$

Basically this will match any number of your pattern which begins with 3 not followed by 30 or 40 then a dot, followed by another three digits.

In your regex: "3[^34][0-9]."

  • 3 > This checks the number starts with 3
  • [^34] > then the Next digit is not 3 or 4
  • [0-9] > followed by one digits which can be 0 to 9
  • . > followed by any other char that is not white spaces

So your requirements were not being checked in this regex.

user_73698090

In case lookaheads are not supported you can use:

^3(?:[0-25-9][0-9]|[34][1-9])\.[0-9]{3}$

RegEx Demo

user_5482250

You may try this:

^3(?!30|40)\d{2}\.\d{3}$

Explanation:

  1. ^ asserts position at start of a line
  2. 3 starts with 3
  3. (?!30|40) negative look ahead to see the following two characters do not match 30 or 40 if found successful then \d{2} ensures any two digits.
  4. \. means a dot
  5. \d{3} means 3 digits
  6. $ asserts position at the end of a line

Demo

user_7611320
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