Question

我正在尝试使用正则表达式在Java中实现Azure AD cloud user account policy。

我已经定义了模式字符串，并且按如下方式使用string.matches(pattern)：

// ONE OF THESE CHARACTER CLASS COMBINATIONS NEED TO BE MET
// combination of lowercase, uppercase and numbers, 8 to 16 characters in length
// combination of lowercase, uppercase and symbols, 8 to 16 characters in length
// combination of lowercase, numbers and symbols, 8 to 16 characters in length
// combination of uppercase, numbers and symbols 8 to 16 characters in length
private static final String PATTERN_PASSWORD_REGEX = new StringBuilder('^')
        .append("((?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])).{8,16}")
        .append('|') // -or-
        .append("((?=.*[a-z])(?=.*[A-Z])(?=.*[@#$%^&*\\-_!+=\\[\\]{}|\\:‘,.?\\/`~“\\(\\);])).{8,16}")
        .append('|') // -or-
        .append("((?=.*[a-z])(?=.*[0-9])(?=.*[@#$%^&*\\-_!+=\\[\\]{}|\\:‘,.?\\/`~“\\(\\);])).{8,16}")
        .append('|') // -or-
        .append("((?=.*[A-Z])(?=.*[0-9])(?=.*[@#$%^&*\\-_!+=\\[\\]{}|\\:‘,.?\\/`~“\\(\\);])).{8,16}")
        .append('$')
        .toString();

public static boolean validate(String password) {
    return password.matches(PATTERN_PASSWORD_REGEX);
}

以某种方式，即使我在正则表达式中没有，它也将空格视为特殊字符。下面的JUnit测试方法将传递除最后一个以外的所有断言。

@Test
public void testPassword() {

    Assert.assertTrue("password is valid", validate("Abcdef12"));
    Assert.assertTrue("password is valid", validate("abcde$12"));
    Assert.assertTrue("password is valid", validate("ABCDE%12"));
    Assert.assertTrue("password is valid", validate("Abcde~12"));
    Assert.assertFalse("password is invalid", validate("abcdefgh"));
    Assert.assertFalse("password is invalid", validate("12345678"));
    Assert.assertFalse("password is invalid", validate("ABCDEFGH"));
    Assert.assertFalse("password is invalid", validate("ABCDefgh"));
    Assert.assertFalse("password is invalid", validate("!~$:-_@&"));

    // This assertion fails...
    Assert.assertFalse("password is invalid", validate("Abcdef 12"));
}

关于SO的正则表达式问题，我看到了非常优雅的答案，但通常它们不涉及字符类组合。希望有人能告诉我我做错了什么，或者是否有更好的方法可以做到这一点。

Answer 1

您需要指定最后一部分中允许的字符，而不是点。将.{8,16}替换为[-A-Za-z0-9@#$%^&*_!+=\[\]{}|:',.?/`~\"();]{8,16}：

^(?:
 (?=.*[a-z])(?=.*[A-Z])(?=.*\d)
 |
 (?=.*[a-z])(?=.*[A-Z])(?=.*[-@#$%^&*_!+=\[\]{}|:',.?/`~"();])
 |
 (?=.*[a-z])(?=.*\d)(?=.*[-@#$%^&*_!+=\[\]{}|:',.?/`~"();])
 |
 (?=.*[A-Z])(?=.*\d)(?=.*[-@#$%^&*_!+=\[\]{}|:',.?/`~"();])
)
[-A-Za-z0-9@#$%^&*_!+=\[\]{}|:',.?/`~"();]{8,16}
$

请参见regex demo。

private static final String PATTERN_PASSWORD_REGEX = new StringBuilder('^(?:')
    .append("(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z])(?=[^0-9]*[0-9])")
    .append('|') // -or-
    .append("(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z])(?=.*[-@#$%^&*_!+=\\[\\]{}|:‘,.?/`~“();])")
    .append('|') // -or-
    .append("(?=[^a-z]*[a-z])(?=[^0-9]*[0-9])(?=.*[-@#$%^&*_!+=\\[\\]{}|:‘,.?/`~“();])")
    .append('|') // -or-
    .append("(?=[^A-Z]*[A-Z])(?=[^0-9]*[0-9])(?=.*[-@#$%^&*_!+=\\[\\]{}|:‘,.?/`~“();])")
    .append(')[-A-Za-z0-9@#$%^&*_!+=\\[\\]{}|:‘,.?/`~“();]{8,16}$')
    .toString();

适用于Azure AD云用户帐户密码策略的基于Java的正则表达式

1 个答案: