Java Regex在字符之前或之后匹配空格

Question

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?

Answer 1

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.