SQL Select with joins and calculate percentage - sql

Question

I have three tables and I'm trying to make a select statement to give me a result like the one below
Teams:
ID Name
1 A
2 B
3 C
Players:
ID Name TeamID
1 P1 1
2 P2 1
3 P3 2
Goals: (goaltype: H for home, A for away, T for training)
ID PID goaltype
1 1 A
2 1 A
3 1 H
4 2 A
5 2 H
6 3 A
7 3 T
Result will be Like:
Team totalGoals home away trainig percentage[(home/total)*100]
A 5 2 3 0 40%
B 2 0 1 1 0
C 0 0 0 0 0
This is my current query:
select t.name,
count(g.id) as totalGoals,
sum(case when g.GTYPE = 'H' then 1 else 0 end) as home,
sum(case when g.GTYPE = 'A' then 1 else 0 end) as away,
sum(case when g.GTYPE = 'T' then 1 else 0 end) as training,
--(home/totalGoals) as percentage
from teams t
left join players p on p.TeamID = t.id
left join goals g on g.pid = p.id
group by t.name

Answer 1

You can use conditional aggregation to get the results you want:
SELECT t.Name AS Team,
COUNT(g.goaltype) AS totalGoals,
SUM(CASE WHEN g.goaltype = 'H' THEN 1 ELSE 0 END) AS home,
SUM(CASE WHEN g.goaltype = 'A' THEN 1 ELSE 0 END) AS away,
SUM(CASE WHEN g.goaltype = 'T' THEN 1 ELSE 0 END) AS training,
CASE WHEN COUNT(g.goaltype) = 0 THEN 0
ELSE 100.0 * SUM(CASE WHEN g.goaltype = 'H' THEN 1 ELSE 0 END) /
COUNT(g.goaltype)
END AS percentage
FROM Teams t
LEFT JOIN Players p ON p.TeamID = t.ID
LEFT JOIN Goals g ON g.PID = p.ID
GROUP BY t.Name
ORDER BY t.Name
Output:
team totalgoals home away training percentage
A 5 2 3 0 40
B 2 0 1 1 0
C 0 0 0 0 0
Demo on SQLFiddle