How to allow underscore in this 'Password Complexity' regex

Here it is:

/(?=^.{8,}$)((?=.开发者_开发技巧*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$/

It only passes if the password contains upper case AND lower case letters, and also either 1 digit or 1 special character, however I want underscore _ to count as a special character as well and it currently does not, how can modify this regex so that it will allow underscore to count as a special character?

EDIT: here is the context...

jQuery.validator.addMethod("complexity", function(value, element) {
    return this.optional(element) || /(?=^.{8,}$)((?=.*\d)|(?=.*\W+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$/.test(value);
}, "password is not complex, see requirements above");

/(?=^.{8,}$)((?=.*\d)|(?=.*[\W_]+))(?![.\n])(?=.*[A-Z])(?=.*[a-z]).*$/

aioobe was close replace \W with [\W_]

Just so you know this considers a space a special character.

Also I don't see where you are checking for numbers. EDIT: nevermind I found it. (man complex regexes are like a wheres waldo.)

Here is a simplifed version that does not allow spaces (and it is easyier to maintain)

(?=^.{8,}$)(?=.*[a-z])(?=.*[A-Z])(?=.*[\W_])(?=^.*[^\s].*$).*$
^          ^          ^          ^            ^
|          |          |          |            L--does not contain a whitespace
|          |          |          L--at least one non word character(a-zA-Z0-9_) or _ or 0-9
|          |          L--at least one upper case letter
|          L--at least one lowercase Letter
L--Number of charaters

These are your building blocks

(?=.*[a-z]) // Whatever is inside the [] meens the string contains at least one charter inside that set.
            // If you wanted a minimum of three lowercase letters you can chain the inner block like so 
               (?=(.*[a-z]){3,})

Any particular reason you want to do all of this in one regex? Seems to me this would be a lot more readable if you just broke it down:

 if (password.length() < 8) return "Password too short!";
 if (!password.matches("\d")) return "Password must contain at least one digit!";
 if (!password.matches("[A-Z]")) return "Password must contain at least one uppercase leter!";
 if (!password.matches("[a-z]")) return "Password must contain at least one lowercase leter!";

 // Password is valid, save it, etc.

I suppose your \W denotes a special character (It's usually defined as "A non-word character")

So, if you would like to add _ to that set of characters, you should be able to write [\W_].

(Updated according to Scott Chamberlains comment.)

I'm not sure why you have (?![.\n]). Do you want to prevent newlines in the password?

Otherwise, this looks like a simplified regexp:

String regexp = "(?=^.{8,}$)(?=.*[A-Z])(?=.*[a-z])(?=.*[\\W_\\d]).*$";
for (String s : new String[] {
        "abcdefg",
        "abcdefgh",
        "abCdefgh",
        "abc1efgh",
        "abC1efgh",
        "abC1e@gh",
        "abC1e_gh"}) {

    System.out.println(s + " is " + (s.matches(regexp) ? "ok" : "not ok"));

}

Prints

abcdefg is not ok
abcdefgh is not ok
abCdefgh is not ok
abc1efgh is not ok
abC1efgh is ok
abC1e@gh is ok
abC1e_gh is ok

Do you consider this an acceptable password?

Aa8 ½ý_

Because your regex matches it. You might want to change that first lookahead from (?=^.{8,}$) to (?=^[!-~]{8,}$), limiting the possibilities to printing ASCII characters (i.e., not whitespace or control characters). If you want to allow non-ASCII characters it gets a lot more complicated; JavaScript regexes don't do Unicode very well.

Here's how I would write the regex:

/^(?=[!-~]{8,}$)(?=.*[A-Z])(?=.*[a-z])(?=.*[^A-Za-z]).*$/

Like Scott, I don't see the point of this bit: (?![.\n]). All it does is make sure the first character is not a period or a linefeed. That can't be what you (or the original author) meant, but I can't even guess what the actual intent was.

For future searchers, here's my implementation that includes requiring a special character AND a digit:

/(?=^[!-~]{8,}$)(?=.*[a-z])(?=.*[A-Z])(?=.*[^A-Za-z0-9])(?=^.*[^\s].*$)(?=.*[\d]).*$/

Heres an example