Sql assignment #5

Naman1109 · 2022-06-08T05:21:57Z

SQL Assignment Added

ashishvijaywargiya · 2022-06-09T06:11:36Z

SQL_Assignment.txt

+
+3)	Find the largest order taken by each salesperson on each date.
+
+Ans :- mysql> select salespeople.snum, salespeople.sname, max(amt) as MAX_ORDER from orders inner join salespeople on orders.snum=salespeople.snum group by snum;


Grouping should be done on the OrderDate. This is a wrong query.

mysql> select salespeople.snum, salespeople.sname, orders.odate, max(amt) as MAX_ORDER from orders inner join salespeople on orders.snum=salespeople.snum group by odate;
+------+---------+----------+-----------+
| snum | sname | odate | MAX_ORDER |
+------+---------+----------+-----------+
| 1007 | rifkin | 10/03/90 | 767.19 |
| 1003 | AlexRod | 10/04/90 | 75.75 |
| 1001 | peel | 10/05/90 | 4723.00 |
| 1002 | seeres | 10/06/90 | 9891.88 |
+------+---------+----------+-----------+

ashishvijaywargiya · 2022-06-09T06:12:52Z

SQL_Assignment.txt

+
+5)	Find which salespeople currently have orders in the order table.
+
+Ans :- mysql> select salespeople.sname from salespeople join orders on (salespeople.snum = orders.snum);


Please display only distinct sales person name.

ashishvijaywargiya · 2022-06-09T07:28:31Z

SQL_Assignment.txt

+
+10)	Match salespeople to customers according to what city they live in.
+
+Ans :-mysql> select salespeople.sname,customers.city,customers.cname from salespeople join customers on (customers.city = salespeople.city );


This is a wrong query

ashishvijaywargiya · 2022-06-09T10:25:10Z

SQL_Assignment.txt

+
+16)	Select all orders that had amounts that were greater than at least one of the orders from October 6.
+
+Ans :- mysql> select * from orders where amt > (select min(amt) from orders where odate='10/06/90');


This is a wrong query

ashishvijaywargiya · 2022-06-13T06:16:56Z

SQL_Assignment.txt

+
+20)	Find the largest order taken by each salesperson on each date, eliminating those Maximum orders, which are less than 3000.
+
+Ans :- mysql> select sp.SNAME,od.AMT,od.ODate from salespeople as sp,orders AS od WHERE sp.SNUM = od.SNUM AND od.AMT > 1000 group by ODATE;


The conditional check will be for 3000 instead of 1000.

ashishvijaywargiya · 2022-06-13T06:19:42Z

SQL_Assignment.txt

+
+27) Find salespeople with customers located in their own cities.
+
+Ans :-mysql>  select sp.SNAME,cust.CNAME,cust.CITY from salespeople as sp Inner join customers as cust on sp.CITY=cust.CITY order by cust.CITY;


This is a wrong query

ashishvijaywargiya · 2022-06-13T06:23:42Z

SQL_Assignment.txt

+------+-------+----------+------+------+
+1 row in set (0.00 sec)
+
+Ans :- mysql>  select sp.SNAME,max(od.AMT) as LargestOrder  from orders as od inner join  salespeople as sp on sp.SNAME in ('Serres','Rifkin' ) And od.SNUM=sp.SNUM group by sp.SNUM;


The largest order taken by Rifkin is 1098.16, why your result is showing 10000?

ashishvijaywargiya · 2022-06-15T13:21:03Z

SQL_Assignment.txt

+
+38) Find all orders with amounts smaller than any amount for a customer in SanJose.
+
+Ans :-mysql> select od.* from orders as od left join customers as cust on cust.city='sanjose' and cust.CNUM=od.CNUM;


This is a wrong query

ashishvijaywargiya · 2022-06-15T13:23:45Z

SQL_Assignment.txt

+
+39) Find all orders with above average amounts for their customers.
+
+Ans :- mysql> select * from orders natural join customers where AMT > (select AVG(AMT) from orders);


This is a wrong query

ashishvijaywargiya · 2022-06-15T13:30:25Z

SQL_Assignment.txt

+
+49) Write two queries that will produce all orders taken on October 3 or October 4.
+
+Ans :- mysql> select * from orders where ODATE = '10,03,90' or odate = '10,04,90' ;


This is a wrong query

ashishvijaywargiya · 2022-06-15T13:35:56Z

SQL_Assignment.txt

+
+55)	Produce all combinations of salespeople and customer names such that the former precedes the latter alphabetically, and the latter has a rating of less than 200.
+
+Ans :- mysql> SELECT Customers.cname, salespeople.sname , customers.rating FROM Customers CROSS JOIN salespeople WHERE customers.rating<200 order by Customers.cname, salespeople.sname;


This is a wrong query

ashishvijaywargiya · 2022-06-15T13:42:24Z

SQL_Assignment.txt

+
+64)	List the commissions of all salespeople servicing customers in London.
+
+Ans :- mysql> select salespeople.comm,salespeople.sname , customers.cname, customers.city from salespeople inner join customers where customers.city = 'london';


This is a wrong query

ashishvijaywargiya · 2022-06-15T13:48:30Z

SQL_Assignment.txt

+
+67)	Write a query that selects all customers serviced by Peel or Motika. (Hint: The snum field relates the 2 tables to one another.)
+
+Ans :-mysql> select customers.cname, salespeople.sname from customers inner join salespeople on salespeople.sname='motika' or salespeople.sname='peel' order by salespeople.sname ;


This is a wrong query

ashishvijaywargiya · 2022-06-16T04:51:12Z

SQL_Assignment.txt

+
+71)	Find all salespeople who have customers with more than one current order.
+
+Ans :- mysql> select sp.sname,count(cust.cnum) from salespeople sp  Inner join customers cust on sp.snum=cust.cnum group by cust.snum having count(cust.snum)>1;


This is a wrong query

ashishvijaywargiya · 2022-06-16T04:53:21Z

SQL_Assignment.txt

+
+72)	Write a query that extracts from the customer’s table every customer assigned to a salesperson, who is currently having at least one another customer(besides the customer being selected) with orders in the Orders Table.
+
+Ans:- mysql> select (select sname from salespeople where snum=cust.snum ) as salespeoplesHavingmoreCustomers from customers as cust  group by snum having count(snum)>1;


This is a wrong query

ashishvijaywargiya · 2022-06-16T05:13:49Z

SQL_Assignment.txt

+
+80)	 Write a query that selects each customer’s smallest order.
+
+Ans:-mysql> select ord.*,min(ord.AMT) as SmallestOrderOfCustommer from customers as cust natural join orders as ord  group by cust.CName ;


Please display the name of the customer as well in the query.

ashishvijaywargiya · 2022-06-16T05:28:09Z

SQL_Assignment.txt

+
+104) Select all salespeople by name and number who have customers in their city whom they don’t service. 
+
+ Ans :- mysql> select sname, snum from salespeople where snum in ( select snum from customers group by snum having count(snum)>1);


This is a wrong query.

Naman added 2 commits June 8, 2022 10:45

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Sql assignment #5

Sql assignment #5

Naman1109 commented Jun 8, 2022

ashishvijaywargiya Jun 9, 2022

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		3) Find the largest order taken by each salesperson on each date.

		Ans :- mysql> select salespeople.snum, salespeople.sname, max(amt) as MAX_ORDER from orders inner join salespeople on orders.snum=salespeople.snum group by snum;


		5) Find which salespeople currently have orders in the order table.

		Ans :- mysql> select salespeople.sname from salespeople join orders on (salespeople.snum = orders.snum);


		10) Match salespeople to customers according to what city they live in.

		Ans :-mysql> select salespeople.sname,customers.city,customers.cname from salespeople join customers on (customers.city = salespeople.city );


		16) Select all orders that had amounts that were greater than at least one of the orders from October 6.

		Ans :- mysql> select * from orders where amt > (select min(amt) from orders where odate='10/06/90');


		20) Find the largest order taken by each salesperson on each date, eliminating those Maximum orders, which are less than 3000.

		Ans :- mysql> select sp.SNAME,od.AMT,od.ODate from salespeople as sp,orders AS od WHERE sp.SNUM = od.SNUM AND od.AMT > 1000 group by ODATE;


		27) Find salespeople with customers located in their own cities.

		Ans :-mysql> select sp.SNAME,cust.CNAME,cust.CITY from salespeople as sp Inner join customers as cust on sp.CITY=cust.CITY order by cust.CITY;


		38) Find all orders with amounts smaller than any amount for a customer in SanJose.

		Ans :-mysql> select od.* from orders as od left join customers as cust on cust.city='sanjose' and cust.CNUM=od.CNUM;


		39) Find all orders with above average amounts for their customers.

		Ans :- mysql> select * from orders natural join customers where AMT > (select AVG(AMT) from orders);


		49) Write two queries that will produce all orders taken on October 3 or October 4.

		Ans :- mysql> select * from orders where ODATE = '10,03,90' or odate = '10,04,90' ;


		55) Produce all combinations of salespeople and customer names such that the former precedes the latter alphabetically, and the latter has a rating of less than 200.

		Ans :- mysql> SELECT Customers.cname, salespeople.sname , customers.rating FROM Customers CROSS JOIN salespeople WHERE customers.rating<200 order by Customers.cname, salespeople.sname;


		64) List the commissions of all salespeople servicing customers in London.

		Ans :- mysql> select salespeople.comm,salespeople.sname , customers.cname, customers.city from salespeople inner join customers where customers.city = 'london';


		67) Write a query that selects all customers serviced by Peel or Motika. (Hint: The snum field relates the 2 tables to one another.)

		Ans :-mysql> select customers.cname, salespeople.sname from customers inner join salespeople on salespeople.sname='motika' or salespeople.sname='peel' order by salespeople.sname ;


		71) Find all salespeople who have customers with more than one current order.

		Ans :- mysql> select sp.sname,count(cust.cnum) from salespeople sp Inner join customers cust on sp.snum=cust.cnum group by cust.snum having count(cust.snum)>1;


		72) Write a query that extracts from the customer’s table every customer assigned to a salesperson, who is currently having at least one another customer(besides the customer being selected) with orders in the Orders Table.

		Ans:- mysql> select (select sname from salespeople where snum=cust.snum ) as salespeoplesHavingmoreCustomers from customers as cust group by snum having count(snum)>1;


		80) Write a query that selects each customer’s smallest order.

		Ans:-mysql> select ord.*,min(ord.AMT) as SmallestOrderOfCustommer from customers as cust natural join orders as ord group by cust.CName ;


		104) Select all salespeople by name and number who have customers in their city whom they don’t service.

		Ans :- mysql> select sname, snum from salespeople where snum in ( select snum from customers group by snum having count(snum)>1);

Sql assignment #5

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Sql assignment #5

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