I need to write a regex that validates the input for a city. Except for characters only spaces are allowed and * acts as a wildcard character, so it can be either in the end or in the beginning of the string. * character should be allowed to be the only input as well.

Accepted strings:

• *city
• *cit *
• '*' (only asterisk)
• *city one

Not accepted strings:

• **
• *!@
• %^&*

I have written this, but it doesn't allow only '*" as input. Any ideas? '^.*[A-Za-zÀ-ÖØ-öø-ž ]{1,34}.*$'

CodePudding user response：

You can use

^(?:\*|\*?\p{L} (?:\h \p{L}*)* \*?)$

See the regex demo.

Details:

• ^ - start of string
• (?:\*|\*?\p{L} (?:\h \p{L}*)* \*?) - either of the two patterns:
  • \* - an asterisk
  • | - or
  • \*? - an optional asterisk
  • \p{L} - one or more letters
  • (?:\h \p{L}*)* - one or more repetitions of one or more horizontal whitespaces followed with zero or more letters
  • \*? - an optional asterisk
• $ - end of string.

In Java:

bool matched = text.matches("\\*|\\*?\\p{L} (?:\\h \\p{L}*)* \\*?");

CodePudding user response：

Here is a regex that might be shorter and efficient:

^(?:\*?[\p{L}\h] \*?|\*)$

RegEx Demo

RegEx Breakup:

• ^: Start
• (?:: Start non-capture group
  • \*?: Match an optional *
  • [\p{L}\h] : Match 1 of any unicode letter or horizontal whitespace
  • \*?: Match an optional *
  • |: OR
  • \*: Match a single *
• ): End non-capture group
• $: End

Java Code:

final String regex = "^(?:\\*?[\\p{L}\\h] \\*?|\\*)$";
String s = "* cit *";
System.out.println( s.matches(regex) );

CodePudding user response：

Welcome to StackOverflow!

Escape the asterisk with a backslash so that it isn't interpreted as a greedy repetition quantifier: \*.

Use beginning (^) and end ($) anchors to ensure that your asterisk can only occur at the start or end of the string.

Lastly, use a negative lookahead ((?!...)) to ensure that the match can't only be two asterisks.

Putting it all together, you get:

(?!^\*\*$)^\*?[\p{L} ]*\*?$