Java regex for validating a user id.

Question

My regex is a little rusty and I could use some help. I am writing a regex to use for validating a user id for a java application.

The two rules that the id has to meet:

Must begin with a -, _, 0-9, a-z or A-Z.
Not allowed to contain a %, backslash, +, /, #, ::, single quote, or double quote

The part I am having trouble with is the double colon. The id can contain a single colon but cannot contain one back to back.

This is what I have come up with but it does not seem to work.

^[a-zA-Z0-9\-_]([^%\+\\\/\#'\"]|^(?!::))+$

Any advice would be great.

it does not seem to work is not a problem statement. You need to elaborate on what exactly that means. — Joe C
– Joe C, Commented Nov 20, 2017 at 22:43
It can't work because ^(?!::) only fails the start of the string if it is followed with ::. Everywhere else, it does not match anything. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 20, 2017 at 22:48

Wiktor Stribiżew · Accepted Answer · 2017-11-20 22:44:37Z

1

The (?!::) is a negative lookahead that only fails the match when there are two colons at the start of the string.

You may use a (?!.*::) negative lookahead right after matching the first char:

String rx = "^[a-zA-Z0-9_-](?!.*::)[^%+\\\\/#'\"]+$";

When using with matches(), you may omit the ^ and $ anchors.

See the regex demo

Details

^ - string start (implicit in matches())
[a-zA-Z0-9_-] - a word or - char (thus, can be reduced to [\w-])
(?!.*::) - no :: after any 0+ chars other than line break chars
[^%+\\\\/#'\"]+ - 1 or more chars other than %, +, \, /, #, ' and "
$ - end of string (implicit in matches())

answered Nov 20, 2017 at 22:44

Wiktor Stribiżew

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3 Comments

Andreas Over a year ago

For potential better performance, bullets 3 and 4 could be combined using | between valid chars excluding :, and a check for single :, e.g. (?:[^%+\\\\/#'\":]|:(?!:))+. This eliminates the backtracking that would otherwise be triggered by .*.

Wiktor Stribiżew Over a year ago

@Andreas I would unroll it then like "^[a-zA-Z0-9_-][^%+\\\\/#'\":]*(?::(?!:)[^%+\\\\/#'\":]*)*$" to eliminate the redundant backtracking caused by |.

dkelley Over a year ago

Thanks this is exactly what I needed!

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