我曾尝试为我的问题准备SQL Fiddle -
在多人游戏中，活动游戏存储在表words_games：
CREATE TABLE words_games (
gid SERIAL PRIMARY KEY, /* game id */
created timestamptz NOT NULL,
player1 integer REFERENCES words_users(uid) ON DELETE CASCADE NOT NULL,
player2 integer REFERENCES words_users(uid) ON DELETE CASCADE,
played1 timestamptz,
played2 timestamptz,
score1 integer NOT NULL CHECK(score1 >= 0),
score2 integer NOT NULL CHECK(score2 >= 0),
hand1 varchar[7] NOT NULL,
hand2 varchar[7] NOT NULL,
pile varchar[116] NOT NULL,
letters varchar[15][15] NOT NULL,
values integer[15][15] NOT NULL,
bid integer NOT NULL REFERENCES words_boards ON DELETE CASCADE
);
并且很容易选择所有游戏，例如身份1的玩家参与其中：
SELECT * FROM words_games WHERE player1 = 1 OR player2 = 1;
但是现在我还添加了一个表words_moves，它充当玩家行为的日志记录：
CREATE TYPE words_action AS ENUM ('play', 'skip', 'swap', 'resign');
CREATE TABLE words_moves (
mid SERIAL PRIMARY KEY, /* move id */
action words_action NOT NULL,
gid integer NOT NULL REFERENCES words_games ON DELETE CASCADE,
uid integer NOT NULL REFERENCES words_users ON DELETE CASCADE,
played timestamptz NOT NULL,
tiles jsonb NULL,
score integer NULL CHECK(score > 0) /* score awarded in that move */
);
现在，当用户连接到我的游戏服务器时，我不仅要向她发送所有活动游戏，还要为每个游戏发送最新动作（最高mid）。
如何在一个查询中运行这样的连接（或CTE）？
我尝试了以下INNER JOIN，但它返回所有动作，而我只需要在每场比赛中进行最新动作：
SELECT
g.gid,
EXTRACT(EPOCH FROM g.created)::int AS created,
g.player1,
COALESCE(g.player2, 0) AS player2,
COALESCE(EXTRACT(EPOCH FROM g.played1)::int, 0) AS played1,
COALESCE(EXTRACT(EPOCH FROM g.played2)::int, 0) AS played2,
ARRAY_TO_STRING(g.hand1, '') AS hand1,
ARRAY_TO_STRING(g.hand2, '') AS hand2,
-- g.letters,
-- g.values,
m.action,
m.tiles
FROM words_games g INNER JOIN words_moves m
ON g.gid = m.gid
AND ( g.player1 = m.uid OR g.player2 = m.uid )
AND ( g.player1 = 1 OR g.player2 = 1 )
ORDER BY g.gid;
gid | created | player1 | player2 | played1 | played2 | hand1 | hand2 | action | tiles
-----+------------+---------+---------+------------+------------+---------+---------+--------+---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1 | 1471794994 | 1 | 2 | 1471868012 | 1471810486 | ПЕАЯСАС | ЖИОБАЯС | play | [{"col": 7, "row": 10, "value": 1, "letter": "Н"}, {"col": 7, "row": 8, "value": 2, "letter": "К"}, {"col": 7, "row": 9, "value": 1, "letter": "И"}, {"col": 7, "row": 7, "value": 2, "letter": "С"}]
1 | 1471794994 | 1 | 2 | 1471868012 | 1471810486 | ПЕАЯСАС | ЖИОБАЯС | play | [{"col": 7, "row": 14, "value": 2, "letter": "К"}, {"col": 7, "row": 13, "value": 1, "letter": "Н"}, {"col": 7, "row": 11, "value": 3, "letter": "У"}, {"col": 7, "row": 12, "value": 2, "letter": "П"}]
1 | 1471794994 | 1 | 2 | 1471868012 | 1471810486 | ПЕАЯСАС | ЖИОБАЯС | play | [{"col": 6, "row": 2, "value": 2, "letter": "П"}, {"col": 6, "row": 3, "value": 1, "letter": "О"}, {"col": 6, "row": 4, "value": 1, "letter": "Е"}, {"col": 6, "row": 5, "value": 5, "letter": "Ж"}, {"col": 6, "row": 6, "value": 5, "letter": "Ы"}, {"col": 6, "row": 7, "value": 2, "letter": "П"}, {"col": 6, "row": 8, "value": 5, "letter": "Ы"}]
2 | 1471795037 | 1 | 2 | 1471806484 | 1471865696 | КЙВГКСМ | ЯРХЖИМН | swap | "А"
2 | 1471795037 | 1 | 2 | 1471806484 | 1471865696 | КЙВГКСМ | ЯРХЖИМН | play | [{"col": 7, "row": 10, "value": 5, "letter": "Ы"}, {"col": 7, "row": 9, "value": 2, "letter": "Д"}, {"col": 7, "row": 8, "value": 1, "letter": "А"}, {"col": 7, "row": 7, "value": 2, "letter": "Л"}]
(5 rows)
更新
实际上我需要LEFT JOIN，因为可以有没有任何玩家移动的游戏......
答案 0 :(得分：1)
好的，让我们建立一个sql。首先，我们需要弄清楚所有游戏的最新举措。有很多方法可以做到这一点，但让我们试试这个：
SELECT *
FROM words_moves wm1
WHERE
played = (SELECT max(played)
FROM words_moves wm2
WHERE wm1.gid = wm2.gid);
这不是最快的方式，但它更容易理解 - 从时间戳最近的words_moves获取每一步。
现在我们有了这个，我们可以用它来构建一个查询来获得游戏和动作：
WITH last_moves AS (
SELECT *
FROM words_moves wm1
WHERE
played = (SELECT max(played)
FROM words_moves wm2
WHERE wm1.gid = wm2.gid))
SELECT *
FROM words_games wg
LEFT JOIN last_moves lm
ON (wg.gid = lm.gid)
WHERE
player1 = 1 OR
player2 = 1;
如果您不熟悉，WITH表示common table expression这是一种非常方便的子查询。除此之外，它意味着如果你最终使用不同的方法来获得每场比赛的最新动作（this question有一套很好的替代尝试），那么它也很容易交换很麻烦。
希望有所帮助！
答案 1 :(得分：1)
SELECT g.gid
, EXTRACT(EPOCH FROM g.created)::int AS created
, g.player1
, COALESCE(g.player2, 0) AS player2
, COALESCE(EXTRACT(EPOCH FROM g.played1)::int, 0) AS played1
, COALESCE(EXTRACT(EPOCH FROM g.played2)::int, 0) AS played2
, ARRAY_TO_STRING(g.hand1, '') AS hand1
, ARRAY_TO_STRING(g.hand2, '') AS hand2
, m.action
, m.tiles
FROM words_games g
LEFT JOIN words_moves m
ON g.gid = m.gid
-- this is redundant: m.gid is a FK
-- AND (g.player1 = m.uid OR g.player2 = m.uid)
AND NOT EXISTS ( -- suppress all-but-the-last
SELECT * FROM words_moves nx
WHERE nx.gid = g.gid -- Same game
-- AND nx.mid > m.mid -- but a higher moveid
-- (assuming ascending move_ids)
-- or: you could use m.played, if that is ascending
AND nx.played > m.played
)
WHERE (g.player1 = 1 OR g.player2 = 1)
ORDER BY g.gid;