instruction: Drive-by users with one [identification] question.
response: SELECT Users.Id AS "user_link", Posts.Id AS "post_link" FROM Users INNER JOIN Posts ON Posts.OwnerUserId = Users.Id INNER JOIN PostTags ON Posts.Id = PostTags.PostId INNER JOIN Tags ON Tags.Id = PostTags.TagId WHERE Tags.TagName LIKE 'identification' AND Posts.ParentId IS NULL AND Users.Id IN (SELECT Users.Id FROM Users INNER JOIN Posts ON Posts.OwnerUserId = Users.Id WHERE Posts.ParentId IS NULL GROUP BY Users.Id HAVING COUNT(Posts.Id) = 1) ORDER BY Users.Id DESC