如何计算同一个表中的两个单独的列并将它们相加成一个新列

Question

我有两张表：playersnames和ma​​tches。

SELECT * FROM playernames;
 id |       name       
----+------------------
 38 | Abe Lincoln
 39 | Richard Nixon
 40 | Ronald Reagan
(3 rows)

SELECT * FROM matches;
 match_id | winner | loser 
----------+--------+-------
        6 |     38 |    39
        8 |     38 |    39
        9 |     39 |    38
       10 |     38 |    39
       11 |     40 |    39
(5 rows)

我需要创建一个返回四列的查询：id、name、wins、matches。但即使有连接和子查询，我似乎也无法在一个查询中得到所有这些。我最接近的是执行两个单独的查询。这是计算每位玩家总获胜次数的查询：

SELECT playernames.id, name, COUNT(*) AS wins 
FROM matches, playernames 
WHERE winner = playernames.id 
GROUP BY playernames.id, winner;
 id |       name       | wins 
----+------------------+------
 38 | Abe Lincoln      |    3
 39 | Richard Nixon    |    1
 40 | Ronald Reagan    |    1
(3 rows)

但如果我想正确计算每个玩家的比赛总数，我必须发出单独的查询：

SELECT playernames.id, name, COUNT(*) 
FROM matches, playernames 
WHERE playernames.id = winner 
OR playernames.id = loser 
GROUP BY playernames.id;
 id |       name       | count 
----+------------------+-------
 40 | Ronald Reagan    |     1
 38 | Abe Lincoln      |     4
 39 | Richard Nixon    |     5
(3 rows)

我正在避免将这个问题与我在将这些组合成一个查询时所做的许多错误尝试搞混。谁能告诉我如何正确地将这两个查询合并为一个？

Answer 1

我会使用一次连接来获取玩家参加的所有比赛，然后计算 case 表达式以仅提取他赢得的比赛：

SELECT    playernames.id, name, 
          COUNT(CASE playernames.id WHEN winner THEN 1 ELSE NULL END) AS wins, 
          COUNT(match_id) AS matches
FROM      playernames 
LEFT JOIN matches ON playernames.id IN (winner, loser)
GROUP BY  playernames.id, name;