- 10.1 Introduction
- 10.2 Grouping on One Column
- 10.3 Grouping on Two or More Columns
- 10.4 Grouping on Expressions
- 10.5 Grouping of NULL Values
- 10.6 General Rules for the GROUP BY Clause
- 10.7 Complex Examples with GROUP BY
- 10.8 Grouping with WITH ROLLUP
- 10.9 Grouping with WITH CUBE
- 10.10 Grouping Sets
- 10.11 Grouping with ROLLUP and CUBE
- 10.12 Combining Grouping Sets
- 10.13 Answers
This chapter is from the book
This chapter is from the book
This chapter is from the book
10.7 Complex Examples with GROUP BY
Here are several other examples to illustrate the extensive possibilities of the GROUP BY clause.
Example 10.12. What is the average total amount of penalties for players who live in Stratford and Inglewood?
SELECT AVG(TOTAL)
FROM (SELECT PLAYERNO, SUM(AMOUNT) AS TOTAL
FROM PENALTIES
GROUP BY PLAYERNO) AS TOTALS
WHERE PLAYERNO IN
(SELECT PLAYERNO
FROM PLAYERS
WHERE TOWN = 'Stratford' OR TOWN = 'Inglewood')
The result is:
AVG(TOTAL) ---------- 85
Explanation: The intermediate result of the subquery in the FROM clause is a table consisting of two columns, called PLAYERNO and TOTAL, and contains five rows (players 6, 8, 27, 44, and 104). This table is passed on to the WHERE clause, where a subquery selects players from Stratford and Inglewood (players 6, 8, and 44). Finally, the average is calculated in the SELECT clause of the column TOTAL.
Example 10.13. For each player who incurred penalties and is captain, get the player number, the name, the number of penalties that he or she incurred, and the number of teams that he or she captains.
SELECT PLAYERS.PLAYERNO, NAME, NUMBER_OF_PENALTIES,
NUMBER_OF_TEAMS
FROM PLAYERS,
(SELECT PLAYERNO, COUNT(*) AS NUMBER_OF_PENALTIES
FROM PENALTIES
GROUP BY PLAYERNO) AS NUMBER_PENALTIES,
(SELECT PLAYERNO, COUNT(*) AS NUMBER_OF_TEAMS
FROM TEAMS
GROUP BY PLAYERNO) AS NUMBER_TEAMS
WHERE PLAYERS.PLAYERNO = NUMBER_PENALTIES.PLAYERNO
AND PLAYERS.PLAYERNO = NUMBER_TEAMS.PLAYERNO
The result is:
PLAYERNO NAME NUMBER_OF_PENALTIES NUMBER_OF_TEAMS
-------- --------- ------------------- ---------------
6 Parmenter 1 1
27 Collins 2 1
Explanation: The FROM clause contains two subqueries that both have a GROUP BY clause.
The previous statement could have been formulated more easily by including subqueries in the SELECT clause, which makes GROUP BY clauses no longer required; see the next example. Now, the only difference is that all players appear in the result.
SELECT PLAYERS.PLAYERNO, NAME,
(SELECT COUNT(*)
FROM PENALTIES
WHERE PLAYERS.PLAYERNO =
PENALTIES.PLAYERNO) AS NUMBER_OF_PENALTIES,
(SELECT COUNT(*)
FROM TEAMS
WHERE PLAYERS.PLAYERNO =
TEAMS.PLAYERNO) AS NUMBER_OF_TEAMS
FROM PLAYERS
Example 10.14. Get the player number and the total number of penalties for each player who played a match.
SELECT DISTINCT M.PLAYERNO, NUMBERP
FROM MATCHES AS M LEFT OUTER JOIN
(SELECT PLAYERNO, COUNT(*) AS NUMBERP
FROM PENALTIES
GROUP BY PLAYERNO) AS NP
ON M.PLAYERNO = NP.PLAYERNO
Explanation: In this statement, the subquery creates the following intermediate result (this is the NP table):
PLAYERNO NUMBERP
-------- -------
6 1
8 1
27 2
44 3
104 1
Next, this table is joined with the MATCHES table. We execute a left outer join, so no players disappear from this table. The final result is:
PLAYERNO NUMBERP
-------- -------
2 ?
6 1
8 1
27 2
44 3
57 ?
83 ?
104 1
112 ?
Example 10.15. Group the penalties on the basis of payment date. Group 1 should contain all penalties between January 1, 1980, and June 30, 1982; group 2 should contain all penalties between July 1, 1981, and December 31, 1982; and group 3 should contain all penalties between January 1, 1983, and December 31, 1984. Get for each group the sum of all penalties.
SELECT GROUPS.PGROUP, SUM(P.AMOUNT)
FROM PENALTIES AS P,
(SELECT 1 AS PGROUP, '1980-01-01' AS START,
'1981-06-30' AS END
UNION
SELECT 2, '1981-07-01', '1982-12-31'
UNION
SELECT 3, '1983-01-01', '1984-12-31') AS GROUPS
WHERE P.PAYMENT_DATE BETWEEN START AND END
GROUP BY GROUPS.PGROUP
ORDER BY 1
The result is:
GROUP SUM(P.AMOUNT)
----- -------------
1 225.00
2 30.00
3 225.00
Explanation: In the FROM clause, a new (virtual) table is created in which the three groups have been defined. This GROUPS table is joined with the PENALTIES table. A BETWEEN operator is used to join the two tables. If there are penalties that fall outside these groups with respect to payment date, they will not be included in the result.
Example 10.16. For each penalty, get the penalty amount plus the sum of that amount and the amounts of all penalties with a lower payment number (cumulative value).
SELECT P1.PAYMENTNO, P1.AMOUNT, SUM(P2.AMOUNT) FROM PENALTIES AS P1, PENALTIES AS P2 WHERE P1.PAYMENTNO >= P2. PAYMENTNO GROUP BY P1. PAYMENTNO, P1.AMOUNT ORDER BY P1. PAYMENTNO
For convenience, we assume that the PENALTIES table consists of the following three rows only (you can create this, too, by temporarily removing all penalties with a number greater than 3):
PAYMENTNO PLAYERNO PAYMENT_DATE AMOUNT
--------- -------- ------------ ------
1 6 1980-12-08 100
2 44 1981-05-05 75
3 27 1983-09-10 100
The desired result is:
PAYMENTNO AMOUNT SUM
--------- ------ ---
1 100 100
2 75 175
3 100 275
The intermediate result of the FROM clause (we show only the columns PAYMENTNO and AMOUNT):
P1.PAYNO P1.AMOUNT P2.PAYNO P2.AMOUNT
-------- --------- -------- ---------
1 100 1 100
1 100 2 75
1 100 3 100
2 75 1 100
2 75 2 75
2 75 3 100
3 100 1 100
3 100 2 75
3 100 3 100
The intermediate result of the WHERE clause:
P1.PAYNO P1.AMOUNT P2.PAYNO P2.AMOUNT
-------- --------- -------- ---------
1 100 1 100
2 75 1 100
2 75 2 75
3 100 1 100
3 100 2 75
3 100 3 100
The intermediate result of the GROUP BY clause:
P1.PAYNO P1.AMOUNT P2.PAYNO P2.AMOUNT
-------- --------- -------- --------------
1 100 {1} {100}
2 75 {1, 2} {100, 75}
3 100 {1, 2, 3} {100, 75, 100}
The intermediate result of the SELECT clause:
P1.PAYNO P1.AMOUNT SUM(P2.AMOUNT)
-------- --------- --------------
1 100 100
2 75 175
3 100 275
This final result is equal to the desired table.
Most joins in this book and in reality are equi joins. Non-equi joins are rare. The previous statement is an example that shows that non-equi joins can be useful and that powerful statements can be formulated with them.
Example 10.17. For each penalty, get the payment number, the penalty amount, and the percentage that the amount forms of the sum of all amounts (again, we use the same PENALTIES table as in the previous example).
SELECT P1.PAYMENTNO, P1.AMOUNT,
(P1.AMOUNT * 100) / SUM(P2.AMOUNT)
FROM PENALTIES AS P1, PENALTIES AS P2
GROUP BY P1.PAYMENTNO, P1.AMOUNT
ORDER BY P1.PAYMENTNO
The intermediate result of the FROM clause is equal to that of the previous example. However, the intermediate result of the GROUP BY clause differs:
P1.PAYNO P1.AMOUNT P2.PAYNO P2.AMOUNT
-------- --------- -------- --------------
1 100 {1, 2, 3} {100, 75, 100}
2 75 {1, 2, 3} {100, 75, 100}
3 100 {1, 2, 3} {100, 75, 100}
The intermediate result of the SELECT clause:
P1.PAYNO P1.AMOUNT (P1.AMOUNT * 100) / SUM(P2.AMOUNT)
-------- --------- ----------------------------------
1 100 36.36
2 75 27.27
3 100 36.36
Find out whether this is the final result as well.
Exercise 10.14: How many players live in a town, on average?
Exercise 10.15: For each team, get the team number, the division, and the number of players that played matches for that team.
Exercise 10.16: For each player, get the player number, the name, the sum of all penalties that he or she incurred, and the number of teams from the first division that he or she captains.
Exercise 10.17: For each team captained by a player who lives in Stratford, get the team number and the number of players who have won at least one match for that team.
Exercise 10.18: For each player, get the player number, the name, and the difference between the year in which he or she joined the club and the average year of joining the club.
Exercise 10.19: For each player, get the player number, the name, and the difference between the year in which he or she joined the club and the average year in which players who live in the same town joined the club.