Question

Use Workbench/Command Line to create the commands that will run the following queries/problem scenarios.

Use MySQL and the Colonial Adventure Tours database to complete the following exercises.

1. List the last name of each guide that does not live in Massachusetts (MA).

2. List the trip name of each trip that has the type Biking.

3. List the trip name of each trip that has the season Summer.

4. List the trip name of each trip that has the type Hiking and that has a distance longer than 10 miles.

5. List the customer number, customer last name, and customer first name of each customer that lives in New Jersey (NJ), New York (NY) or Pennsylvania (PA). Use the IN operator in your command.

6. Repeat Exercise 5 and sort the records by state in descending order and then by customer last name in ascending order.

7. How many trips are in the states of Maine (ME) or Massachusetts (MA)?

8. How many trips originate in each state?

9. How many reservations include a trip price that is greater than $20 but less than $75?

10. How many trips of each type are there?

11. Colonial Adventure Tours calculates the total price of a trip by adding the trip price plus other fees and multiplying the result by the number of persons included in the reservation. List the reservation ID, trip ID, customer number, and total price for all reservations where the number of persons is greater than four. Use the column name TOTAL_PRICE for the calculated field.

12. Find the name of each trip containing the word “Pond.”

13. List the guide’s last name and guide’s first name for all guides that were hired before June 10, 2013.

14. What is the average distance and the average maximum group size for each type of trip?

15. Display the different seasons in which trips are offered. List each season only once.

16. List the reservation IDs for reservations that are for a paddling trip. (Hint: Use a subquery.)

17. What is the longest distance for a biking trip?

18. For each trip in the RESERVATION table that has more than one reservation, group by trip ID and sum the trip price. (Hint: Use the COUNT function and a HAVING clause.)

19. How many current reservations does Colonial Adventure Tours have and what is the total number of persons for all reservations?

Answer 1

Solution

1)
SELECT LAST_NAME FROM GUIDE
WHERE STATE <> 'MA';

2)
SELECT TRIP_NAME FROM TRIP
WHERE TYPE='Biking';

3)
SELECT TRIP_NAME FROM TRIP
WHERE SEASON='Summer';

4)
SELECT TRIP_NAME FROM TRIP
WHERE TYPE='Hiking' AND DISTANCE > 10;

5)
SELECT CUSTOMER_NUM, LAST_NAME, FIRST_NAME FROM CUSTOMER
WHERE STATE IN('NJ','NY','PA');

6)
SELECT CUSTOMER_NUM, LAST_NAME, FIRST_NAME FROM CUSTOMER
WHERE STATE IN('NJ','NY','PA')
ORDER BY STATE DESC,LAST_NAME ASC;

7)
SELECT COUNT(*) FROM TRIP
WHERE STATE IN('ME','MA');

8)
SELECT STATE, COUNT(TRIP_ID) FROM TRIP
GROUP BY STATE
ORDER BY STATE;

9)
SELECT COUNT(*) FROM RESERVATION
WHERE TRIP_PRICE>20 AND TRIP_PRICE<75;

10)
SELECT TYPE, COUNT(TRIP_ID) FROM TRIP
GROUP BY TYPE
ORDER BY TYPE;

11)
SELECT RESERVATION_ID, TRIP_ID, CUSTOMER_NUM,
(TRIP_PRICE+OTHER_FEES)*NUM_PERSONS AS TOTAL_PRICE
FROM RESERVATION
WHERE NUM_PERSONS>4;

12)
SELECT TRIP_NAME FROM TRIP
WHERE TRIP_NAME LIKE '%Pond%';

13)
SELECT LAST_NAME, FIRST_NAME FROM GUIDE
WHERE HIRE_DATE<20130610;

14)
SELECT TYPE, AVG(DISTANCE), AVG(MAX_GRP_SIZE) FROM TRIP
GROUP BY TYPE;

15)
SELECT DISTINCT SEASON FROM TRIP;

16)
SELECT RESERVATION_ID FROM RESERVATION
WHERE TRIP_ID IN
(SELECT TRIP_ID FROM TRIP WHERE TYPE='Paddling');

17)
SELECT MAX(DISTANCE) FROM TRIP
WHERE TYPE='Biking';

18)
SELECT TRIP_ID,SUM(TRIP_PRICE) FROM RESERVATION
GROUP BY TRIP_ID
HAVING COUNT(*)>1;

19)
SELECT COUNT(*), SUM(NUM_PERSONS) FROM RESERVATION;

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