在现有正则表达式中添加几个特殊字符

Question

I use a regex pattern in a change password feature of my app. The requirements for the password are:

• at least 1 small letter
• at least 1 upper case letter
• minimum of 8 characters
• at least 1 number
• at least 1 special character

I want to add these characters to my pattern, but I couldn't seem to add the "\\" and the single and double quotes. How do I escape them? I'm using Eclipse (Java) so escaping needs 2 backslashes

Right now, my pattern looks like this and it's giving me an error:

String customPattern = "((?=.*\\d)(?=.*[a-z])(?=.*[A-Z])
(?=.*[!@#$%^&*()\\-_+=/?.,>~\\\\|\\[];:\\"'{}])(?=[\\S]+$).{8,})";

Answer 1

把一个反斜杠双引号之前，也难逃]目前焦炭类中。

String customPattern = "^(?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[!@#$%^&*()\\-_+=/?.,>~\\\\|\\[\\];:\"'{}])(?=\\S+$).{8,}";

Answer 2

You did not escape the ] inside the character class [!@#$%^&*()\\\\-_+=/?.,>~\\\\\\\\|\\\\[];:\\\\"'{}] (right after \\\\[ ). Here is a working regex (with ^ / $ anchors that will really enforce length check of the whole string if not used with matches() ):

^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[!@#$%^&*()\-_+=\/?.,>~\\|\[\];:"'{}])(?=\S+$).{8,}$
                                                                ^^

See demo

In Java, you can use it as

String customPattern = "^(?=.*\\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[!@#$%^&*()\\-_+=/?.,>~\\\\|\\[\\];:\"'{}])(?=\\S+$).{8,}$";