Java Regex match space before or after characters

Matt Zdj :

I try to have a regex validating an input field.

This question is the continuation of this question but I made a mistake and the question changed a bit so I created a new one.

Here is my java regex :

^(?:\?*[a-zA-Z\d]){2}[a-zA-Z\d?]*\*?$

Demo

What I'm tying to match is :

  • Minimum 2 alpha-numeric characters (other than '?' and '*')
  • The '*' can only appears one time and at the end of the string
  • The '?' can appears multiple time
  • No WhiteSpace right at the beginning
  • No WhiteSpace before or after '?' or '*'

So for exemple :

  • abcd = OK
  • ?bcd = OK
  • ab?? = OK
  • ab*= OK
  • ab?* = OK
  • ??cd = OK
  • ab cd = OK
  • *ab = NOT OK
  • a ? b =NOT OK
  • ??? = NOT OK
  • ab? cd = NOT OK
  • ab ?d = NOT OK
  • ab * = NOT OK
  • abcd = NOT OK (space at the begining)

As i've asked in the fisrt question, no white space at all are allowed in my regex now but that's not what I want and I'm a bit lost can you help me please?

Wiktor Stribiżew :

You may use

^(?!\s)(?!.*\s[*?])(?!.*[*?]\s)(?:[?\s]*[a-zA-Z0-9]){2}[a-zA-Z0-9?\s]*\*?$

See the regex demo.

Usage note: if you use it with Java's .matches() method, the ^ and $ can be removed from the pattern. Remember to double escape backslashes in the string literal.

Details

  • ^ - start of string
  • (?!\s) - no whitespace is allowed immediately to the right (at the start of the string)
  • (?!.*\s[*?]) - no whitespace is allowed after any 0+ chars, as many as possible, before * or ?
  • (?!.*[*?]\s) - no whitespace is allowed after any 0+ chars, as many as possible, after * or ?
  • (?:[?\s]*[a-zA-Z0-9]){2} - two sequences of
    • [?\s]* - 0 or more ? or/and whitespaces
    • [a-zA-Z0-9] - an alphanumeric char
  • [a-zA-Z0-9?\s]* - 0 or more letters, digits, ? or whitespaces
  • \*? - an optional ? char
  • $ - end of the string.

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