Regard Regex and pattern compilation in java

Question

Please give me the code in regex for password validation in java which should consist of one Caps character,one integer ,one following symbols( @,#,$,%,^,&,+,=) and small characters.

I have been trying this with different separate regular expressions and one combined regular expression.

Actually i am already having a single regex that evaluates all the conditions in javascript. I am not able to use it in Java back end. I tried by escaping \. Its also not working.

Here is my code:

    Pattern pattern = Pattern.compile("/.*(?=.{6,})(?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).*$/");
    Matcher matcher = pattern.matcher("Aa@1");
    if(matcher.matches()){
        System.out.println("Matched");
    }
    else{
        System.out.println("No mat");
    }

The original javascript regex is

/.*(?=.{6,})(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).*$/

In that the \d gave me error due to the escaping character. So, i added another \ before that in the Java Version.

I am not able to understand what is going wrong.

Thanks in Advance.

Answer 1

You were nearly there, but you missed a few details:

First, the starting point is bad - that JavaScript regex is ugly. Instead of

/.*(?=.{6,})(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).*$/

use this:

/^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).{6,}$/

Then, to translate the regex to Java, you need to remove the delimiters (and use quotes instead, not additionally like you did) and double the backslashes (like you already did):

Pattern pattern = Pattern.compile("^(?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).{6,}$");

Now it should work.

Answer 2

You don't need the /'s in Java. It will actually match a slash. Also, the leading .* is useless (although it won't affect the result).

Pattern pattern = Pattern.compile("(?=.{6,})(?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&+=]).*$");