Javascript Match Function For Special Characters

I am working on this code and using 'match' function to detect strength of password. how can I detect if string has special characters in it? if(password.match(/[a-z]+/)) score++;

Solution 1:

If you mean !@#$% and ë as special character you can use:

/[^a-zA-Z ]+/

The ^ means if it is not something like a-z or A-Z or a space.

And if you mean only things like !@$&$ use:

/\W+/

\w matches word characters, \W matching not word characters.

Solution 2:

You'll have to whitelist them individually, like so:

if(password.match(/[`~!@#\$%\^&\*\(\)\-=_+\\\[\]{}/\?,\.\<\> ...

and so on. Note that you'll have to escape regex control characters with a \.

While less elegant than /[^A-Za-z0-9]+/, this will avoid internationalization issues (e.g., will not automatically whitelist Far Eastern Language characters such as Chinese or Japanese).

Solution 3:

you can always negate the character class:

if(password.match(/[^a-z\d]+/i)) {
    // password contains characters that are *not*
    // a-z, A-Z or 0-9
}

However, I'd suggest using a ready-made script. With the code above, you could just type a bunch of spaces, and get a better score.

Solution 4:

Just do what you did above, but create a group for !@#$%^&*() etc. Just be sure to escape characters that have meaning in regex, like ^ and ( etc....

EDIT -- I just found this which lists characters that have meaning in regex.

Solution 5:

if(password.match(/[^\w\s]/)) score++;

This will match anything that is not alphanumeric or blank space. If whitespaces should match too, just use /[^\w]/.