Verifying password policy on existing users

This may not be popular among your users, but I believe the only way you can know for sure is to force a password change for every SQL login with CHECK_POLICY = ON. This will generate a set of ALTER LOGIN commands with blank passwords, you can update the query giving them all a common password or manually update each one with individual passwords - just make sure they meet your policy. Of course you need to be sure that the password policy is as complex as you expect, and that it is enabled (Control Panel > Administrative Tools > Local Security Policy > Account Policies > Password Policy > Password must meet complexity requirements).

SELECT N'ALTER LOGIN ' + QUOTENAME(name) 
  + N' WITH PASSWORD = N'''' MUST_CHANGE, CHECK_POLICY = ON;' 
  FROM sys.sql_logins 
  --WHERE is_policy_checked = 0;

Steve Jones wrote about this a while back. Note that - due to what I uncovered below - you can't rely on is_policy_checked = 1 to mean that the password actually meets your current policy, since the login could have been created with a hashed password (in which case the plain text password can't be checked) or while the local complexity policy was disabled (which still leads to is_policy_checked = 1).

Another approach I thought would work would be to try to create a copy of every login with their current password_hash and with CHECK_POLICY = ON, and make a note of every one that fails. However, this can not work - even with CHECK_POLICY = ON, it doesn't perform any validation of an already-hashed password. I'll include the code for posterity - but, by design, the policy simply can't be checked.

SELECT N'BEGIN TRY
  CREATE LOGIN ' + QUOTENAME(N'copy_of_' + name) 
    + N' WITH PASSWORD = ' 
    + CONVERT(NVARCHAR(255), password_hash, 1)
    + ' HASHED, CHECK_POLICY = ON;
  DROP LOGIN ' + QUOTENAME(N'copy_of_' + name) + ';
END TRY
BEGIN CATCH
  IF ERROR_NUMBER() = 15118
    PRINT N''' + REPLACE(name, '''', '''''') 
      + N' was not complex enough.'';
END CATCH'
FROM sys.sql_logins;

Personally, I think this is a bug. If the syntax allows me to create a login using a hashed password, and I can stipulate that that password must meet my complexity policy, it should generate an error or warning that the policy was not, in fact, checked.

UPDATE: I filed a bug against this behavior.

• CHECK_POLICY is a no-op with CREATE LOGIN ... PASSWORD ... HASHED

There is no way that you are going to get this 100% accurate. Though you can use PWDCOMPARE to check against a list of weak passwords (you can add to the list of weak passwords and do a comparison).

I have written a similar script that does the comparison and gives you the results. I have posted it on github.

EDIT:

Now you can have a list of weak passwords in a csv and then use dbatools Test-DbaLoginPassword with -Dictionary switch (Specifies a list of passwords to include in the test for weak passwords.)