Customer and Loan Table Having One To Many Relationship. Consider the Following Entities and Their Relationship. Customer(cno,cname,city) Loan(lno,loan_amt,no_of_years) Constraints: Primary Key loan_amt should be>0
SQL> create table cust 2 (cno number primary key, 3 cname varchar(20), 4 city varchar(20)); Table created. SQL> create table loan 2 (lno number primary key, 3 loan_amt number check(loan_amt>0), 4 no_of_years number, 5 cno number references cust); Table created. SQL> select * from cust; CNO CNAME CITY ---------- -------------------- ----------------- 1 Raj Mumbai 2 Ramesh Mumbai 3 Ganesh Pune 4 Harsh Delhi 5 Gagan Delhi 6 Satish Mumbai 6 rows selected. SQL> select * from loan; LNO LOAN_AMT NO_OF_YEARS CNO ---------- ---------- ----------- ---------------- 101 200000 4 1 102 100000 2 1 103 500000 6 2 104 150000 1 3 105 300000 8 4 106 100000 1 4 107 250000 2 5 108 800000 6 6 8 rows selected. a) Create or Replace a PL/SQL Function to Find Total Loan Amout From 'Mumbai' City. SQL> set serveroutput on; SQL> create or replace function f12 2 return number 3 as 4 loanamt number; 5 begin 6 select sum(loan_amt) into loanamt from loan,cust 7 where cust.cno=loan.cno 8 and city='Mumbai'; 9 if SQL%Found Then 10 return loanamt; 11 else 12 return 0; 13 end if; 14 end f12; 15 / Function created. ***CALL PROGRAM*** SQL> declare 2 a number; 3 begin 4 a:=f12; 5 if(a=0)then 6 dbms_output.put_line('No Record Found'); 7 else 8 dbms_output.put_line('The Total Loan Amt From Mumbai City is' ||' '|| a); 9 end if; 10 end; 11 / The Total Loan Amt From Mumbai City is 1600000 PL/SQL procedure successfully completed. b) Write a Cursor To display Details of all Customers who have Taken Loan For More Than 3 Years. SQL> declare 2 cursor c12 is 3 select cust.cno,cname,city,loan_amt,no_of_years from cust,loan 4 where cust.cno=loan.cno 5 and no_of_years>3; 6 y c12%rowtype; 7 begin 8 dbms_output.put_line('CNO CNAME CITY LOANAMT NO_OF_YEARS'); 9 dbms_output.put_line('-------------------------------------------'); 10 for y in c12 11 loop 12 dbms_output.put_line(y.cno ||' '|| y.cname ||' '|| y.city ||' '|| y.loan_amt ||' '|| y.no_of_years); 13 end loop; 14 end; 15 / CNO CNAME CITY LOANAMT NO_OF_YEARS ------------------------------------------- 1 Raj Mumbai 200000 4 2 Ramesh Mumbai 500000 6 4 Harsh Delhi 300000 8 6 Satish Mumbai 800000 6 PL/SQL procedure successfully completed.
Customer and Loan Table Having One To Many Relationship. Consider the Following Entities and Their Relationship. Customer(cno,cname,city) Loan(lno,loan_amt,no_of_years) Constraints: Primary Key loan_amt should be>0
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November 17, 2013
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