SQL Exercises/The warehouse

Relational Schema

Oracle12c

CREATE TABLE Warehouses ( code INTEGER PRIMARY KEY NOT NULL, location VARCHAR2 (255) NOT NULL, capacity INTEGER NOT NULL);CREATE TABLE Boxes ( code VARCHAR2 (255) PRIMARY KEY NOT NULL, contents VARCHAR2 (255) NOT NULL, value REAL NOT NULL, warehouse INTEGER NOT NULL, CONSTRAINT fk_warehouses_code FOREIGN KEY (warehouse) REFERENCES Warehouses(code));

Sample Dataset

 INSERT INTO Warehouses(Code,Location,Capacity) VALUES(1,'Chicago',3); INSERT INTO Warehouses(Code,Location,Capacity) VALUES(2,'Chicago',4); INSERT INTO Warehouses(Code,Location,Capacity) VALUES(3,'New York',7); INSERT INTO Warehouses(Code,Location,Capacity) VALUES(4,'Los Angeles',2); INSERT INTO Warehouses(Code,Location,Capacity) VALUES(5,'San Francisco',8); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('0MN7','Rocks',180,3); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('4H8P','Rocks',250,1); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('4RT3','Scissors',190,4); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('7G3H','Rocks',200,1); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('8JN6','Papers',75,1); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('8Y6U','Papers',50,3); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('9J6F','Papers',175,2); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('LL08','Rocks',140,4); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('P0H6','Scissors',125,1); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('P2T6','Scissors',150,2); INSERT INTO Boxes(Code,Contents,Value,Warehouse) VALUES('TU55','Papers',90,5);

Exercises

1 Select all warehouses.

SELECT * FROM warehouses

2 Select all boxes with a value larger than $150.

SELECT * FROM boxes WHERE VALUE > 150

3 Select all distinct contents in all the boxes.

SELECT DISTINCT contents FROM boxes

4 Select the average value of all the boxes.

SELECT AVG(value) FROM boxes

5 Select the warehouse code and the average value of the boxes in each warehouse.

SELECT warehouse, COUNT(*) FROM boxes GROUP BY warehouse

6 Same as previous exercise, but select only those warehouses where the average value of the boxes is greater than 150.

SELECT warehouse, AVG(value) FROM boxes GROUP BY warehouse HAVING AVG(value) > 150

7 Select the code of each box, along with the name of the city the box is located in.

SELECT B.code, W.location  FROM boxes B  LEFT JOIN warehouses W    ON B.warehouse = W.code

8 Select the warehouse codes, along with the number of boxes in each warehouse. Optionally, take into account that some warehouses are empty (i.e., the box count should show up as zero, instead of omitting the warehouse from the result).

SELECT W.code, count(B.code)  FROM warehouses W  LEFT JOIN boxes B    ON W.code = B.warehouse GROUP By (W.code)

9 Select the codes of all warehouses that are saturated (a warehouse is saturated if the number of boxes in it is larger than the warehouse’s capacity).

SELECT *  FROM warehouses WHERE capacity <       (SELECT count(*) FROM boxes WHERE warehouse = warehouses.code)SELECT warehouses.Code  FROM warehouses JOIN boxes ON warehouses.code = boxes.warehouse  GROUP BY warehouses.code,warehouses.capacity   HAVING Count(boxes.code) > warehouses.capacity-- for postgresqlSELECT warehouses.Code  FROM warehouses JOIN boxes ON warehouses.code = boxes.warehouse  GROUP BY warehouses.code  HAVING Count(boxes.code) > warehouses.capacity

10 Select the codes of all the boxes located in Chicago.

SELECT code  FROM boxes WHERE warehouse IN       (SELECT code FROM warehouses WHERE location = 'Chicago') SELECT Boxes.Code   FROM Warehouses LEFT JOIN Boxes   ON Warehouses.Code = Boxes.Warehouse   WHERE Location = 'Chicago';

11 Create a new warehouse in New York with a capacity for 3 boxes.

INSERT INTO Warehouses (Location, Capacity) VALUES ('New York', 3);

12 Create a new box, with code “H5RT”, containing “Papers” with a value of $200, and located in warehouse 2.

 INSERT INTO Boxes VALUES('H5RT','Papers',200,2);

13 Reduce the value of all boxes by 15%.

 UPDATE Boxes SET Value = Value * 0.85;

14 Apply a 20% value reduction to boxes with a value larger than the average value of all the boxes.

 UPDATE boxes SET value = value * 0.8 WHERE value > (SELECT AVG(value) FROM boxes)

15 Remove all boxes with a value lower than $100.

DELETE FROM Boxes WHERE Value < 100;

16 Remove all boxes from saturated warehouses.

DELETE FROM boxes WHERE warehouse IN       (SELECT code          FROM warehouses         WHERE capacity < (SELECT count(code)                             FROM boxes                            WHERE warehouse = warehouses.code))