Review Question 8.12: Answer question 8.10, but useEXISTS. Here is the answer to
ID: 3613330 • Letter: R
Question
Review Question 8.12: Answer question 8.10, but useEXISTS. Here is the answer to question to 8.10: ->SELECT A1.AdvisorName, A1.AdvisorPhone FROM ADVISOR A1 WHERE A1.AdvisorName IN (SELECT A2.AdvisorName FROM ADVISOR A2WHERE A1.AdvisorName =A2.AdvisorName AND A1.AdvisorPhone <> A2.AdvisorPhone); Review Question 8.12: Answer question 8.10, but useEXISTS. Here is the answer to question to 8.10: ->SELECT A1.AdvisorName, A1.AdvisorPhone FROM ADVISOR A1 WHERE A1.AdvisorName IN (SELECT A2.AdvisorName FROM ADVISOR A2
WHERE A1.AdvisorName =A2.AdvisorName AND A1.AdvisorPhone <> A2.AdvisorPhone); ->SELECT A1.AdvisorName, A1.AdvisorPhone FROM ADVISOR A1 WHERE A1.AdvisorName IN WHERE A1.AdvisorName IN (SELECT A2.AdvisorName FROM ADVISOR A2
WHERE A1.AdvisorName =A2.AdvisorName AND A1.AdvisorPhone <> A2.AdvisorPhone);