Regular expression " symbol is not working

Question

• Password must not match or contain last name.
• Password must contain at least 1 special characters.
• Password must contain at least 1 numeric characters.
• Password must contain at least 2 alphabetic characters.
• Password must contain at least 1 uppercase letters.
• Password must not match or contain user ID.
• Password must not match or contain first name.
• Password must not contain the following characters: !
• Password must not be longer than 25 characters.
• Password must be at least 8 characters long.
• Password must contain at least 1 lowercase letters.

These are the symbols should contain in !"#$%&'()*+-./:;<=>?@[\]^_{|}~`

but i am trying to put " this special character it is giving error[compile time error]

private static final String PASSWORD_PATTERN = "((?=.*[a-z])(?=.*\\d)(?=.*[A-Z])(?=.*["@#$%!%^&*()_+=?/[],.<>|~`'-]).{8,32})";

Can an one hep , thanks in advance

Answer 1

I recommend to use https://www.debuggex.com/ when you are trying to work with regular expressions. It's easier and directly tells you if your regex is incorrect.

Answer 2

Of course you can't simply put a quote in a string literal, that ends the string. And it's not related to regular expression, that would be the same whatever you do later with the string.

Simply escape it : replace " with \"

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Addendum regarding the new question in comments : if you put [ and ] in a character class (that is between [ and ]), then you must escape them for the regular expression. And as you do that in a string literal, that makes a double escaping because you must escape the \. And you must also escape the - in a character class.

So change

["@#$%!%^&*+=?/[],.<>|~`'-:/<>]

to

["@#$%!%^&*+=?/\\[\\],.<>|~`'\\-:/<>]