PostgreSQL - 带有错误计数输出的左连接答案

【问题标题】：PostgreSQL - Left Join with Bad Count OutputPostgreSQL - 带有错误计数输出的左连接
【发布时间】：2015-02-12 12:53:32
【问题描述】：

我正在尝试设置一组简单的表格来显示比赛结果 - 我有以下结构：

CREATE TABLE players(
    id SERIAL PRIMARY KEY,
    name TEXT);

CREATE TABLE matches(
    id SERIAL PRIMARY KEY,
    player_one_id INTEGER REFERENCES players,
    player_two_id INTEGER REFERENCES players,
    winner_id INTEGER REFERENCES players);

并且我已经输入了一些测试数据，如下：

INSERT INTO players (name) VALUES ('Mike Jones');
INSERT INTO players (name) VALUES ('Albert Awesome');
INSERT INTO players (name) VALUES ('Sad Sally');
INSERT INTO players (name) VALUES ('Lonely Lenny');

INSERT INTO matches (player_one_id, player_two_id, winner_id) VALUES (1,2,1);
INSERT INTO matches (player_one_id, player_two_id, winner_id) VALUES (3,4,4);

我正在尝试执行一个查询，为每个玩家提供以下结果：

id、姓名、matched_won、matches_played。

到目前为止，我有以下查询：

SELECT players.id, players.name, count(matches.winner_id) as matches_won
                               , count(matches.id) as matches_played
    FROM players left join matches
    ON players.id = matches.winner_id
GROUP BY players.id
ORDER BY matches_won DESC

不幸的是，我得到的输出不正确（每个玩家应该有 1 个matches_played）：

 id |      name      | matches_won | matches_played 
----+----------------+-------------+----------------
  4 | Lonely Lenny   |           1 |              1
  1 | Mike Jones     |           1 |              1
  2 | Albert Awesome |           0 |              0
  3 | Sad Sally      |           0 |              0
(4 rows)

现在，我知道这个错误输出的原因是因为加入了 player.id = matches.winner_id，但是，我的问题是：

是否有可能只使用一个左连接查询来获得这些结果？如果是这样，怎么做？如果可能，我想避免进行多次查询。

【问题讨论】：

标签： sql postgresql aggregate-filter

【解决方案1】：

是的。首先，您需要了解count() 只是计算具有非 NULL 值的行数，因此您的两个计数应该相同。

要获得胜利者，请使用条件聚合：

SELECT p.id, p.name,
       sum(case when m.winner_id = p.id then 1 else 0 end) as matches_won,
       count(m.id) as matches_played
FROM players p left join
     matches m
     ON p.id in (m.player_one_id, m.player_two_id)
GROUP BY p.id
ORDER BY matches_won DESC;

您还需要修复join 条件。您不能仅仅加入获胜者并期望获得所有比赛的计数。

【讨论】：

【解决方案2】：

分选解决方案：

SELECT players.id, players.name,
       (select count(*)
        from matches
        where matches.winner_id = players.id) as matches_won,
       (select count(*)
        from matches
        where players.id in (player_one_id, player_two_id)) as matches_played
FROM players
ORDER BY matches_won DESC

【讨论】：

【解决方案3】：

除了 Gordon 的回答，您还可以使用 COUNT() 代替 SUM()，使用 NULLIF 或 FILTER（从 PostgreSQL 9.4 开始过滤）：

使用 NULLIF() 因为使用列名时 NULL 不计算在内：

SELECT p.id, p.name,
       count(nullif(m.winner_id <> p.id, false)) as matches_won,
       count(m.id) as matches_played
FROM players p 
    left join matches m ON p.id in (m.player_one_id, m.player_two_id)
GROUP BY p.id
ORDER BY 
    matches_won DESC;

并使用过滤器：

SELECT p.id, p.name,
       count(*) filter (WHERE m.winner_id = p.id) as matches_won,
       count(m.id) as matches_played
FROM players p 
    left join matches m ON p.id in (m.player_one_id, m.player_two_id)
GROUP BY p.id
ORDER BY 
    matches_won DESC;

【讨论】：

【解决方案4】：

SELECT p.name,COUNT(m.player_one_id)+ COUNT(m1.player_two_id) AS num_of_matches_played
,COUNT(m2.winner_id) AS num_of_matches_won FROM players p 
LEFT OUTER JOIN matches m ON p.id = m.player_one_id
LEFT OUTER JOIN matches m1 ON p.id = m1.player_two_id
LEFT OUTER JOIN matches m2 ON p.id = m2.winner_id
GROUP BY p.name

【讨论】：