Java regular expression for ending with *

Question

I am trying to write regular expression in java to match alphanumeric string which can optionally have *. But if it contains * then it has to be at the end of that string. It can only contain one *

Below inputs should return true

abcd

abcd*

*

Where as below input should return false since it contains * which is not at the end of that string

abc*d
abcd**

I tried writing java program as shown below , but it is not working

public static void main(String[] a){
    String pattern = "[a-zA-Z0-9]+[\\*]$";
    String test = "abcd*";
    System.out.println(test.matches(pattern));
}

It return true for abcd* but not for abcd and *

Please let me know how to fix this regular expression. Thanks

[\\*] is redundant. Either use a character class so * has no special meaning ([*]), or use escape so * has no special meaning (\\*). Doing both works, but is excessive. — Andreas
– Andreas, Commented Sep 11, 2018 at 23:51
@Andreas Yes, you are right. Thanks. It is now working with [a-zA-Z0-9]*[*]? — MLS
– MLS, Commented Sep 11, 2018 at 23:57

Dave Zabriskie · Accepted Answer · 2018-09-11 23:56:53Z

3

You have to specify that the trailing asterisk is optional by using a ?

[a-zA-Z0-9]*[*]?

edited Sep 11, 2018 at 23:56

answered Sep 11, 2018 at 23:36

Dave Zabriskie

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4 Comments

MLS Over a year ago

It worked, thanks. But I also wanted regular expression to return true for *. I have edited my question now

Dave Zabriskie Over a year ago

Updated to check for 0 or more alphanumerics (*) instead of 1 or more (+).

Andreas Over a year ago

Should be just [a-zA-Z0-9]*\\*? (or [a-zA-Z0-9]*[*]?), i.e. no need for both \\ and [], and no need for $.

Dave Zabriskie Over a year ago

Good point here and in question comments. Updated to remove escaping and terminal ($) check. @Andreas

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