Hi guys I trying to check if password contains special caracters and at least 2 characters (digit or letter) So like this:
("&#€$&&÷") ====> false
("&#&'&*#5") ====> false
(">~<<`<•5t") ====> true
("{\><>\tt") =====> true
("65%#^$*@") ====> true
("7373673") ====> false
("7267373~") ====> true
I've tried this regular expression but It seems not working:
/^((?=.*\d{2})|(?=.*?[a-zA-Z]{2}))/
CodePudding user response：
You can assert the 2 occurrences of a character or digit in the same lookahead, and then match at least a single "special" character.
Using a case insensitive pattern:
^(?=(?:[^a-z\d\n]*[a-z\d]){2})[a-z\d]*[~!@#$%^&*()_ <>•`{}\\][~!@#$%^&*()_ <>•`{}\\a-z\d]*$
The pattern matches:
^Start of string(?:[^a-z\d\n]*[a-z\d]){2}Assert 2 occurrences of either a char a-z or a digit. The[^a-z\d\n]*part negates the character class using[^to prevent unnecessary backtracking[a-z\d]*Match optional chars a-z or a digit[~!@#$%^&*()_ <>•`{}\]Match a special character[~!@#$%^&()_ <>•`{}\a-z\d]Match optional allowed chars$End of string