基於Oracle SQL中的計數的查找率

Question

我試圖列出每場比賽中比賽的總比賽失敗比例。

我已經試過了，但它似乎以下不工作：

select teams.name as Name, 
Lost 
from teams 
join matches on teams.name in (matches.home, matches.away) 
group by teams.name 
having (1 - (matches.winner/(count (matches.home) + count (matches.away)))) as Lost; 

是與我寫在上線的功能方式的問題？參考

關係圖：

Answer 1

HAVING該子句用於篩選結果集的任何分組已發生後，不以限定的功能。 您需要編寫在SELECT子句中需要返回的列的所有邏輯。

SELECT T.NAME 
     ,COUNT(M.*) AS MATCHES_PLAYED 
     ,SUM(CASE WHEN M.WINNER = T.NAME THEN 1 ELSE 0 END) AS WINS 
     ,SUM(CASE WHEN M.WINNER = T.NAME THEN 0 ELSE 1 END) AS LOSSES 
     ,SUM(CASE WHEN M.WINNER = T.NAME THEN 1 ELSE 0 END)/COUNT(M.*) AS WIN_PERCENT 
     ,SUM(CASE WHEN M.WINNER = T.NAME THEN 0 ELSE 1 END)/COUNT(M.*) AS LOSS_PERCENT 
FROM TEAMS T 
INNER JOIN 
     MATCHES M 
ON  T.NAME = M.HOME 
OR  T.NAME = M.AWAY   
GROUP BY 
     T.NAME