正则表达式大于或小于3位数和字符

Question

This is probably just the usual "Oh no look another person asking for regex" but I really cant figure this one out. What I need to do is find a simple regex that will match anything that's greater than or less than 3 digits. It must also match characters though.

Just a little explanation. I am trying to match anything that isn't the standard area code for a telephone number. so > 3 < including characters. I am using this for business rules and have already matched the positive version of the area code.

only one record is passed through the regex at one time so there are no need for delimiters.

Okay sorry about that here are some examples:

337   : does not match
123   : does not match
12    : does match
1     : does match
asd   : does match
as2   : does match
12as45: does match
1234  : does match

the opposite is really easy and can just be [0-9]{3} or [\\d]{3}.

PS Its in java

Answer 1

Now this is the solution with "az" (because it seems to be so common):

^(?!\d{3})[a-z0-9]{3}$|^[a-z0-9]{1,2}$|^[a-z0-9]{4,}$

... and this is the true solution, which matches everything except three characters that are all digits:

^(?!\d{3}).{3}$|^.{1,2}$|^.{4,}$

http://regexr.com?358u9

Because we are just checking three alternatives, it's pretty self explanatory.

Answer 2

You can do it this way

^(?:(?!(?<!\d)\d{3}(?!\d))[a-zA-Z\d])+$

See it here on Regexr .

Explained

^                                # match the start of the string (not needed with the matches() method)
    (?:                          # start of a non capturing group
        (?!(?<!\d)\d{3}(?!\d))   # combined lookarounds, fails if there are 3 digits following with not a digit before and ahead of those 3 digits
        [a-zA-Z\d]               # match one ASCII letter or digit
    )+                           # repeat this at least once
$                                # match the end of the string (not needed with the matches() method)

Answer 3

使用\\ d {3}使用负面看法： http ： //www.regular-expressions.info/lookaround.html