Sql Making

Database Operations SQL Making Database Operations With Normalization

2016 Augustine Barlow Programming Assignment Help 16-Apr-16

We will use the below database definition to generate the queries: CLIENT (clientno#,name, client_referred_by#) ORDER (orderno#, clientno#, order_date, empid#) ORDER_LINE (orderno#, order line number#, item_number#, no_of_items, item_ cost, shipping_date) ITEM (item_number#, item_type, cost) EMPLOYEE (empid#, emp_type#, deptno, salary, firstname, lastname)

1. Display all the rows and columns in the CLIENT table. Sort by client name in reverse alphabetical order.

SELECT * FROM CLIENT ORDER BY NAME DESC; 2. Display the item number and total cost for each order line (total cost = no of items X item cost). Name the calculated column TOTAL COST. SELECT A.ITEM _NUMBER “ITEM NUMBER”, B. NO_OF_ITEMS *B. ITEM_COST “TOTAL COST”, C. NAME FROM ITEM A, ORDER LINE B, CLIENT C, ORDER D WHERE A.ITEM_NUMBER =B.ITEM_NUMBER AND B.ORDERNO = D.ORDERNO AND C.CLIENTNO=D.CLIENTNO; 3. Display all the client numbers in the ORDER table. Remove duplicates.

SELECT CLIENTNO FROM ORDER GROUP BY CLIENTNO ; 4. Display the order number and client number from the ORDER table. Output the result in the format. Client <clientno> ordered <orderno> SELECT ‘CLIENT’ +’’+ CLIENTNO + ‘ORDERED’+ ‘ ‘+ ORDERNO FROM ORDER ; 5. Display full details from the ORDER_LINE table where the item number is (first condition) between 1 and 200 (no > or < operators) OR the item number is greater than 1000 AND (second condition) the item cost is not in the list 1000, 2000, 3000 OR the order number is not equal to 1000.

SELECT * FROM ORDER_LINE WHERE (ITEM_NUMBER BETWEEN 1 AND 200 OR ITEM_NUMBER >1000 ) AND (ITEM_COST <> ‘1000’,’2000’,’3000’ OR ORDERNO <> 1000); 6. Display the client name and order date for all orders using the traditional method.

SELECT A.NAME “CLIENT NAME”, B.ORDER_DATE “ORDER DATE” FROM CLIENT A, ORDER B WHERE A.CLIENTNO=B.CLIENTNO ; 7. Repeat query (6) but also display all clients who have never ordered anything.

SELECT A.NAME “CLIENT NAME”, B.ORDER_DATE “ORDER DATE” FROM CLIENT A, ORDER B WHERE A.CLIENTNO(+) = B.CLIENTNO ; 8. Display the client name and order date for all orders using the natural join keywords.

SELECT A.NAME, B.ORDER_DATE FROM CLIENT A, ORDER B JOIN CLIENT A ON A.CLIENTNO=B.CLIENTNO;

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9. Display the client name and order date for all orders using the JOIN . . . USING method.

SELECT CLIENT.NAME, ORDER1.DATE FROM TEST.CLIENT A, TEST.ORDER1 B WHERE A.CLIENTNO = B.CLIENTNO;

10. Display the client number, order date and shipping date for all orders where the shipping date is between three and six months after the order date.

SELECT CLIENTNO, ORDER_DATE, SHIPPING_DATE FROM ORDER A, ORDER_LINE B WHERE SHIPPING_DATE > ORDER_DATE+90 AND SHIPPING_DATE <ORDER_DATE+180; 11. Display the client number and name and the client number and name of the person who referred that client.

SELECT CLIENTNO, NAME, B.CLIENT_NAME REFERREDBYCLIENT_NAME, B.CLIENTNO AS REFERREDBY_CLIENTNO FROM CLIENT A, CLIENT B WHERE A.CLIENTNO=B. CLIENT_REFERRED_BY; 12. Display the client name in upper case only and in lower case only.

SELECT UPPER(NAME), LOWER(NAME) FROM CLIENT 13. Display the second to fifth characters in each client name.

SELECT SUBSTR(NAME,2,5) FROM CLIENT ; 14. Display the item_cost and then round it to the nearest hundred, ten, unit, tenth and hundredth.

SELECT COST, COST, ROUND(COST, -2) HUNDRED_ROUND, ROUND(COST, -1) TEN_ROUND, ROUND(COST, 0) UNIT_ROUND, ROUND(COST, 1) TENTH_ROUND, ROUND(COST, 2) HUNDREDTH_ROUND FROM ITEM;

15. Display the item_cost and then truncate it to the nearest hundred, ten, unit, tenth and hundredth.

SELECT COST, COST, TRUNC(COST, -2) HUNDRED_ROUND, TRUNC (COST, -1) TEN_ROUND, TRUNC (COST, 0) UNIT_ROUND, TRUNC (COST, 1) TENTH_ROUND, TRUNC (COST, 2) HUNDREDTH_ROUND FROM ITEM; 16. Display the order number, order line number and the shipping date. If the shipping date is null, display the string <not shipped yet>.

SELECT ORDERNO,B. ORDER_LINE_NUMBER, DECODE(B.SHIPPING_DATE,NULL,'NOT YET SHIPPED', SHIPPING_DATE) FROM ORDER A, ORDER_LINE B WHERE A.ORDERNO=B.ORDERNO; 17. Display the order number and average item cost for each order.

SELECT ORDER_NO,AVG(ITEM_COST) FROM ORDER_LINE GROUP BY ORDER_NO; 18. Display the clientno and total value for all orders placed by that client. Output the result in the following format: Client <clientno> has placed orders to the value of <total value> SELECT 'CLIENT '||NVL(TOCHAR(CLIENT_NO,'UNKNOWN') ||' HAS PLACED ORDERS TO THE VALUE OF '|| NVL(SUM(ORDERS),0) AS ORDER_SUMMARY Programming Assignment Help

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FROM ORDER_LINE GROUP BY CLIENT_NO; 19. Display all clients whose name begins with the letter J or contains the letter M anywhere or contains E as the third letter.

SELECT CLIENT_NAME FROM CLIENT WHERE CLIENT_NAME LIKE 'J%' OR CLIENT_NAME LIKE '%M%' OR CLIENT_NAME LIKE '___E%'; 20. Using a set operator, display the client number of all clients who have never placed an order.

SELECT CLIENTNO FROM CLIENT MINUS SELECT CLIENTNO FROM ORDER; 21. Using a set operator, display the client number of all clients who have ever placed an order and whose name does not contain the string Sm. SELECT CLIENTNO FROM CLIENT WHERE NAME NOT LIKE %SM% UNION SELECT CLIENTNO FROM ORDER; 22. Display the order number, number of lines in the order, total number of items and total value for all orders that have a total value greater than $100 SELECT ORDERNO, ORDER_LINE_NUMBER, NO_OF_ITEMS, SUM(ITEM_COST) FROM ORDER_LINE GROUP BY ORDERNO, ORDER_LINE_NUMBER, NO_OF_ITEMS; 23. Display the client name for all clients who have placed an order where any order line has more than 3 items. Do not use a table join anywhere in your query.

SELECT NAME,COUNT(ITEM_NUMBER) FROM CLIENT WHERE ITEM_NUMBER IN (SELECT ITEM_NUMBER FROM ITEM ) HAVING COUNT(ITEM_NUMBER)>3 ; 24. Display the order number for all orders whose average item cost is greater than the overall average item cost across all orders.

SELECT ORDER_NO,AVG(ITEM_COST) FROM ORDER_LINE GROUP BY ORDER_NO HAVING AVG(ITEM_COST) > (SELECT AVG(AVG(ITEM_COST)) FROM ORDER_LINE GROUP BY ORDER_NO); 25. Display the client number and the value of the highest value order placed by that client. (High challenge question – well above exam standard).

SELECT A.CLIENTNO,(B.NO_OF_ITEMS * ITEM_COST ) HIGH_VALUE_ORDER FROM ORDER A, ORDER_LINE B WHERE A.ORDERNO =B.ORDERNO AND HIGH_VALUE_ORDER=(SELECT MAX(NO_OF_ITEMS * ITEM_COST) FROM ORDER_LINE); 26. Display the earliest shipping date in the format: DD/MON/YYYY SELECT CONVERT(CHAR(12),GETDATE(SHIPPING_DATE),106) FROM ORDER_LINE ORDER BY SHIPPING_DATE ; 27. Display the order number and the number of months since the order was shipped for all orders that have been shipped in the last year (365 days). (Hint: Unshipped orders will have a null value).

SELECT ORDERNO, MONTHS_BETWEEN(SYSDATE-ORDER_DATE) FROM ORDER A, ORDER_LINE B Programming Assignment Help

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WHERE A.ORDERNO=B.ORDERNO AND B.ORDER_SHIPPING_DATE IS NOT NULL AND TRUNC(SYSDATE) - TO_DATE(ORDER_SHIPPING_DATE, 'DD-MMYYYY')>=’365’; 28. Display the item_number, the total number of times that item has been sold and the average cost for that item. Restrict that output to items that have been included in more than 500 orders. (Hint: Assume an item will appear in only one order line of an order).

SELECT B.ITEMNO, SUM(B.NO_OF_ITEMS), AVG(C.COST) FROM ORDER_LINE B, ITEM C WHERE B.ITEM_NUMBER=C.ITEM_NUMBER AND SUM(B.NO_OF_ITEMS) >500;

30. Display the surname for all employees who earn less the average salary of all those employees in the department with the lowest average salary.

SELECT EMPLOYEE_ID,LAST_NAME FROM EMPLOYEES WHERE SALARY< (SELECT AVG(SALARY) FROM EMPLOYEES)

31. Display the client number and the total number of orders the client has placed for all clients who have placed at least one order in 2011. (Hint: Use exists).

SELECT CLIENTNO, COUNT(ORDERNO) FROM ORDER WHERE TO_DATE(ORDER_DATE,’YYYY’) IN ‘2011’;

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