Create preg_match for password validation allowing (!@#$%)

Question

I would like to create a preg_match function to validate my passowrds, but I'm not sure how to write it to allow the following special characters to be used: !@#$%.

if(!preg_match(?????)$/', $password))

Here are my password rules that I want to work into the regex:

• May contain letter and numbers
• Must contain at least 1 number and 1 letter
• May contain any of these characters: !@#$%
• Must be 8-12 characters

Thank you for any help you can offer.

Answer 1

I think this should look like that:

if(!preg_match('/^(?=.*\d)(?=.*[A-Za-z])[0-9A-Za-z!@#$%]{8,12}$/', $password)) {
    echo 'the password does not meet the requirements!';
}

Between start -> ^
And end -> $
of the string there has to be at least one number -> (?=.*\d)
and at least one letter -> (?=.*[A-Za-z])
and it has to be a number, a letter or one of the following: !@#$% -> [0-9A-Za-z!@#$%]
and there have to be 8-12 characters -> {8,12}

As user557846 commented to your question, I would also suggest you to allow more characters, I usually (if i use a maximum) take at least 50 :)

btw, you might want to take a look at this regex tutorial

Answer 2

preg_match('/^(?=.*\d)(?=.*[@#\-_$%^&+=§!\?])(?=.*[a-z])(?=.*[A-Z])[0-9A-Za-z@#\-_$%^&+=§!\?]{8,20}$/',$password)

• at least one lowercase char
• at least one uppercase char
• at least one digit
• at least one special sign of @#-_$%^&+=§!?