JavaScript: Regex - Cannot start/end with punctuation, no repeating punctuation

Question

I'm trying to create a Regex. The rules are:

1) Only letters (both upper and lower), numbers, periods (.) and underscores (_)

2) Must begin and end with letter or number (i.e. no periods/underscores)

3) Cannot have multiple punctuations in a row, of any kind (no .. or ._ or _. or __ or .._. etc)

4) Must contain at least 2 letters (either upper or lower or both, doesn't matter)

5) No less than 3 chars, no more than 25 chars

So far, what I have is:

^((?!.*[\._]{2})[A-Za-z0-9\._]{3,25})$

Which is close-ish. I've got rules 1, 3 and 5, but cannot figure out 2 and 4.

I thought that

(([A-Za-z0-9]?)([A-Za-z0-9\._]+)([A-Za-z0-9]?)){3,25}

would get rule 2, but it just breaks it.

I've been working on this for a couple hours, but none of the solutions I can find, or can come up with, are quite what I'm looking for.

An explanation of the Regex statement you provide would be greatly appreciated, but is not entirely necessary.

EDIT 1

As per noob's suggestion:

^((?!.*[\._]{2})(([\w\d\.]+)){3,25})$

EDIT 2

Valid:

1rockit.man
q3w
e4.45r.d.2
sp_am

(and anything else between 3 and 25 characters, without leading or repeating punctuation)

Invalid:

.ret.
123
12a
ret..wer
super_.duper
_petrat
petrat_

Use `shorthand` operators like `\w` and `\d`. Easier to comprehend. `\w` also include `_` so that's covered. — , Feb 27 '16 at 06:17
You might want to include a few **valid** and **invalid** matches for quick testing. — , Feb 27 '16 at 06:27
`1rockit.man` and `e4.45r.d.2` are **invalid** according to condition 2. _Cannot begin or end with non-letter/number_. You might wanna check again the given condition or the example. — , Feb 27 '16 at 06:37
@noob *cannot* begin with *non-letter* or *non-number*. i.e., *must* begin and end with either letter or number. I'll make it more clear in the post — Birrel, Feb 27 '16 at 06:40

score 7 · Answer 1 · answered Feb 27 '16 at 07:50

Based upon your conditions I came up with following regex. It's only modification over your current regex.

Regex: ^(?=.*[A-Za-z].*[A-Za-z])([A-Za-z0-9]((?!.*[\._]{2})([\w\d\.]{1,23}))[A-Za-z0-9])$

Explanation:

The first part (?=.*[A-Za-z].*[A-Za-z]) checks for whole string to have two characters separated by anything (that's pretty much the case here).
I added checks on the beginning and end of string which should be either letter or number.
Apart from that the length will also vary now as first and last character is counted rest of the length will be only 23 characters.

Regex101 Demo

score 4 · Accepted Answer · edited May 23 '17 at 12:15

Building the regex from the rules (in an order of convenience):

both upper and lower case => use case insensitive flag //i
letters, numbers, underscore and period => [\w.]*
3-25 characters => ^[\w.]{3,25}$
cannot begin or end with period or underscore => (?!^[\W_]|[W_]$)
(notice uppercase \W meaning not \w)
cannot contain 2 punctuation in a row => (?![\W_]{2})
checking the negative lookahead in every position => (?:(?!...)[\w.]){3,25}
(using non-capturing group (?:) instead of () because we don't need to save the group)
at least 2 letters => (?:.*[a-z]){2} assuming i flag
wrap into lookahead (not consuming any characters) so we can check multiple conditions => (?=(?:...))

The final regex literal:

/^(?=(?:.*[a-z]){2})(?:(?!^[\W_]|[\W_]{2}|[\W_]$)[\w.]){3,25}$/i

https://regex101.com/r/yK1cB3/5#javascript

FTR: in languages that support the x flag (ignoring whitespace inside the regex), you could make it more readable (see a workaround for javascript):

^
(?=
  (?:.*[a-z]){2}
)
(?:
   (?!^[\W_]|[\W_]{2}|[\W_]$)
   [\w.]
){3,25}
$

How would I modify this to allow punctuation (just the periods and underscores) at the start and end? I tried changing the middle section to `(?:(?![\W_]{2}$)[\w.])`, which *does* allow beginning and ending punctuation, but it also allows multiple punctuation in a row. — Birrel, Mar 04 '16 at 04:21
Perfect! Thank you. `/^(?=(?:.*[a-z]){2})(?:(?![\W_]{2})[\w.]){3,25}$/i`, for anyone else who might be wondering. — Birrel, Mar 04 '16 at 15:50

score 1 · Answer 3 · answered Feb 27 '16 at 09:23

You can try with:

^(?=.*[a-zA-Z].*[a-zA-Z].*)[a-zA-Z0-9](?!.*[._]{2,}|.*[._]$)[._a-zA-Z0-9]{2,24}$

or, if you prefer make it using character classes like \w \d \s, you could use:

^(?=.*[a-zA-Z].*[a-zA-Z].*)\w(?!.*[._]{2,}|.*[._]$)(\w|[._]){2,24}$

http://regexr.com/3ct34

score 1 · Answer 4 · edited May 23 '17 at 12:15

Takin part in ur riddle challenge. Here's my version. Would have nice performance as well :)

/^(?=(?:.*?[a-z]){2})[a-z\d](?!.*?[._]{2})[\w.]{1,23}[a-z\d]$/i

i flag for caseless matching where [a-z] is equivalent with alpha [a-zA-Z].
\w is a short for word character [A-Za-z0-9_] thus [\w.] will match allowed characters.
(?: opens a non capturing group mostly used for repitition or alternation (if group not needed).
(?= (?! lookarounds are zero-length assertions triggered at a certain position.

Rule 2: Require one alnum [a-z\d] at ^ start and one at $ end
Rule 1 & 5: No less than 3, no more than 25 with start/end substracted [\w.]{1,23}
Rule 3: (?!.*?[._]{2}) no consecutive punctuation checked by negative lookahead.
Rule 4: (?=(?:.*?[a-z]){2}) at least 2 letters checked with positive lookahead.

Check out demo at regex101

JavaScript: Regex - Cannot start/end with punctuation, no repeating punctuation

4 Answers4