I have a requirement to build a regex pattern to validate a String in Java. Hence I build a pattern
[A-Z][a-z]*\s?[A-Z]?[a-z]*$ for the conditions:
Pattern.matches("[A-Z][a-z]*\s?[A-Z]?[a-z]*$","Joe V") returns false for me in java.
But the same pattern returns true for the data "Joe V" in regexr.com.
What might be the cause?
Javascript has native support for regex while Java doesn't. Since Java uses \ for special signs in strings (like \n) you have to escape the \ to actually be a \ sign. That's done with another \. So any \ you use in Java should be written as \\.
Thus your regex / code should be:
Pattern.matches("[A-Z][a-z]*\\s?[A-Z]?[a-z]*$", "Joe V")
which returns true
P.s. \s is interpreted as a Space in any Java-String
\s being interpreted as a space is a recent addition (Java 15, when text blocks where released)You can use
Pattern.matches("[A-Z][a-z]*(?:\\s[A-Z][a-z]*)*","Joe V")
Pattern.matches("\\p{Lu}\\p{Ll}*(?:\\s\\p{Lu}\\p{Ll}*)*","Joe V")
See the regex demo #1 and regex demo #2.
Note that .matches requires a full string match, hence the use of ^ and $ anchors on the testing site and their absence in the code.
Details:
^ - start of string (implied in .matches)[A-Z] / \p{Lu} - an (Unicode) uppercase letter[a-z]* / \p{Ll}* - zero or more (Unicode) lowercase letters(?:\s[A-Z][a-z]*)* / (?:\s\p{Lu}\p{Ll}*)* - zero or more sequences of
\s - one whitespace[A-Z][a-z]* /\p{Lu}\p{Ll}* - an uppercase (Unicode) letter and then zero or more (Unicode) lowercase letters.$ - end of string (implied in .matches)
s?, it seems you expect that to match a space..., but that would need to be\s?, and in a string literal with escaped backslash...