SQL Query, Average climbed and pair that has climbed the most peaks - sql

Question

My Databases look like so:
PEAK (NAME, ELEV, DIFF, MAP, REGION)
CLIMBER (NAME, SEX)
PARTICIPATED (TRIP_ID, NAME)
CLIMBED (TRIP_ID, PEAK, WHEN)
PEAK gives info about the mountain peaks that the user is interested in. The table lists the name of each peak, it elevation(in ft), its difficulty level(on a scale of 1-5), the map that it is located on, and the region of the Sierra Nevada that it is located in.
CLIMBER lists the members of club, and gives their name and gender.
PARTICIPATED gives the set of climbers who participated in each of the various climbing trips. The number of participants in each trip varies.
CLIMBED tells which peaks were climbed on each climbing trip, along w/ the data that each peak was climbed.
I need help with writing a query for the following:
Compute the average number of peaks scaled by the men in the club and by the women in the club.
Which pair of climbers have climbed the most peaks together, and how many peaks is that?
Who has climbed more than 20 peaks in some 60 day span?
For the first query, so far I have found a way to compute the total number of peaks climbed by either gender, for men:
SELECT SUM(C)
FROM
(SELECT CD.PEAK, COUNT(*) C
FROM CLIMBED CD
WHERE CD.TRIP_ID IN
(SELECT TRIP_ID
FROM PARTICIPATED PA
WHERE PA.NAME IN
(SELECT NAME
FROM CLIMBER
WHERE SEX = 'M'))
GROUP BY CD.PEAK) T;
For the second query, I have the following which I'm fairly sure isn't correct:
SELECT TEMP2.TRIP_ID, COUNT (*)
FROM
(SELECT P1.NAME, P2.NAME, P1.TRIP_ID
FROM PARTICIPATED P1, PARTICIPATED P2
WHERE P1.NAME <> P2.NAME AND
P1.TRIP_ID = P2.TRIP_ID) TEMP1,
(SELECT *
FROM CLIMBED) TEMP2
WHERE TEMP2.TRIP_ID = TEMP1.TRIP_ID
GROUP BY TEMP2.TRIP_ID;

Answer 1

Question 1:
For total number of trips (including every time a peak was climbed)
SELECT t1.sex, AVG(t1.peak_count) AS average
FROM
(SELECT sex, COUNT(trip_id) AS peak_count
FROM climber c LEFT JOIN participated p ON c.name = p.name GROUP BY c.name, c.sex) t1
For each time a UNIQUE peak was climbed:
SELECT t1.sex, AVG(t1.peak_count) AS average
FROM
(SELECT sex, COUNT(trip_id) AS peak_count
FROM climber c LEFT JOIN participated p ON c.name = p.name GROUP BY c.name, c.sex) t1
Question 2:
SELECT P1.Name, P2.Name, COUNT(DISTINCT p1.trip_id) AS trips
FROM participated p1 INNER JOIN participated p2 ON p1.trip_id = p2.trip_id
WHERE p1.name > p2.name -- > instead of <> gets only one of the pairs
GROUP BY P1.Name, P2.Name
HAVING COUNT(DISTINCT p1.trip_id) > 0
ORDER BY trips DESC
Question 3:
SELECT p.name, cl.when AS span_begin_date, DATEADD(day, 60, cl.when) AS span_end_date, count(c2.trip_id) AS peaks
FROM climbed cl LEFT JOIN
climbed c2 ON c2.when BETWEEN cl.when AND DATEADD(day, 60, cl.when)
GROUP BY p.name, cl.when, DATEADD(day, 60, cl.when)
HAVING COUNT(c2.trip_id) > 20
ORDER BY peaks

Answer 2

Here is my solution. If you provide sample data, this can be verified. For question 3, the some 60 day span is not clear. Can you please specify better?
Question 1
select x.sex, avg(x.peaks_escalated) as peaks
from (
select u.name, u.sex, count(distinct c.peak) as peaks_escalated
from t1_climbed c
inner join t1_participated p on c.trip_id = p.trip_id
inner join t1_climber u on p.name = u.name
group by u.name, u.sex ) x
group by x.sex
Question 2
with list1 as (
select u.name as member, c.trip_id, c.peak, c.when
from t1_climbed c
inner join t1_participated p on c.trip_id = p.trip_id
inner join t1_climber u on p.name = u.name
)
select a.member as m1, b.member as m2, count(distinct a.peak) as total
from list1 a inner join list1 b
on a.trip_id = b.trip_id
and a.peak = b.peak
and a.when = b.when
and a.member <> b.member
group by a.member, b.member

Answer 3

Oracle Setup:
CREATE TABLE PEAK (
NAME VARCHAR2(50) PRIMARY KEY,
ELEV INT,
DIFF INT,
MAP VARCHAR2(10),
REGION VARCHAR2(10)
);
CREATE TABLE CLIMBER (
NAME VARCHAR2(50) PRIMARY KEY,
SEX CHAR(1) CHECK ( SEX IN ( 'M', 'F' ) )
);
-- Created this to have a primary key
CREATE TABLE TRIPS (
TRIP_ID INT PRIMARY KEY
);
CREATE TABLE PARTICIPATED (
TRIP_ID INT REFERENCES TRIPS( TRIP_ID ),
NAME VARCHAR2(50) REFERENCES CLIMBER( NAME ),
PRIMARY KEY ( TRIP_ID, NAME )
);
CREATE TABLE CLIMBED (
TRIP_ID INT REFERENCES TRIPS( TRIP_ID ),
PEAK VARCHAR2(50) REFERENCES PEAK ( NAME ),
"WHEN" DATE
);
Question 1
SELECT sex,
AVG( num_peaks ) AS avg_peaks
FROM (
SELECT c.*,
COUNT( DISTINCT l.peak ) num_peaks
FROM CLIMBED l
INNER JOIN
PARTICIPATED p
ON ( p.trip_id = l.trip_id )
RIGHT OUTER JOIN
CLIMBER c
ON ( p.name = c.name )
GROUP BY c.name, c.sex
)
GROUP BY sex;
You need to OUTER JOIN climbers as they could have not participated in any trips (so having climbed 0 peaks) and this needs to be taken into account in the average. It is also possible that a person could have climbed a peak multiple times - when you want the number of peaks climbed by a person you want to exclude multiple climbs on the same peak and will need to use COUNT( DISTINCT ... ) (or another similar technique) - if you want to count multiple climbs then remove the DISTINCT keyword.
Question 2:
SELECT *
FROM (
SELECT name1,
name2,
COUNT( DISTINCT c.peak ) AS num_peaks_climbed
FROM (
SELECT p1.name AS name1,
p2.name AS name2,
p1.trip_id
FROM PARTICIPATED p1
INNER JOIN
PARTICIPATED p2
ON ( p1.trip_id = p2.trip_id AND p1.name < p2.name )
) p
INNER JOIN
climbed c
ON ( p.trip_id = c.trip_id )
GROUP BY name1, name2
ORDER BY num_peaks_climbed DESC
)
WHERE ROWNUM = 1;
Question 3:
SELECT *
FROM (
SELECT p.name,
COUNT( c.peak ) OVER ( PARTITION BY p.name
ORDER BY c."WHEN"
RANGE BETWEEN INTERVAL '-60' DAY PRECEDING
AND CURRENT ROW
) AS num_peaks_in_60_days,
c."WHEN" AS last_date_of_range
FROM PARTICIPATED p
INNER JOIN
climbed c
ON ( p.trip_id = c.trip_id )
)
WHERE num_peaks_in_60_days > 20;