Exam 1z0-082 topic 1 question 150 discussion

A: select * from customers c1 where exists ( select 1 from customers c2 where c1.custname=c2.custname and c1.CUSTADDRESS=c2.CUSTADDRESS and c1.CUST_CREDIT_LIMIT=c2.CUST_CREDIT_LIMIT and C1.CUSTINO<>c2.CUSTINO); B: select * from customers c1 join customers c2 on c1.custname=c2.custname and c1.CUSTADDRESS=c2.CUSTADDRESS and c1.CUST_CREDIT_LIMIT=c2.CUST_CREDIT_LIMIT and C1.CUSTINO<>c2.custino;

A, B are correct answers. See example: A query: SELECT custino,custname,custaddress,cust_credit_limit FROM customers WHERE custname IN ( SELECT custname FROM customers GROUP BY custname HAVING COUNT(custname)>1 ) AND custaddress IN ( SELECT custaddress FROM customers GROUP BY custaddress HAVING COUNT(custaddress)>1 ) AND cust_credit_limit IN ( SELECT cust_credit_limit FROM customers GROUP BY cust_credit_limit HAVING COUNT(cust_credit_limit)>1 ) ORDER BY custname; B query: SELECT DISTINCT(c.custino),c.custname,c.custaddress,c.cust_credit_limit FROM customers c, customers u WHERE c.custname = u.custname AND c.custaddress = u.custaddress AND c.cust_credit_limit = u.cust_credit_limit AND c.custino <> u.custino ORDER BY c.custname, c.custino;

Provided answer are correct.