如何重用 2 个相同子查询检索到的值?

How to reuse the value retrieved by 2 identical subqueries?

我已经为我的问题准备了一个简单的SQL Fiddle

在 PostgreSQL10 中有 2 个表保存用户 ID 和用户社交网络数据(如名字、照片):

CREATE TABLE words_users (
        uid     SERIAL PRIMARY KEY
);

CREATE TABLE words_social (
        sid     text     NOT NULL,
        social  integer  NOT NULL CHECK (0 < social AND social <= 64),
        given   text     NOT NULL CHECK (given ~ '\S'),
        uid     integer  NOT NULL REFERENCES words_users ON DELETE CASCADE,
        PRIMARY KEY(sid, social)
);

还有2张桌子放着游戏和聊天信息:

CREATE TABLE words_games (
        gid SERIAL PRIMARY KEY,
        player1 integer REFERENCES words_users(uid) ON DELETE CASCADE NOT NULL CHECK (player1 <> player2),
        player2 integer REFERENCES words_users(uid) ON DELETE CASCADE
);

CREATE TABLE words_chat (
        cid     BIGSERIAL PRIMARY KEY,
        created timestamptz NOT NULL,
        gid     integer NOT NULL REFERENCES words_games ON DELETE CASCADE,
        uid     integer NOT NULL REFERENCES words_users ON DELETE CASCADE,
        msg     text    NOT NULL
);

这里我用测试数据填表:

SQL:

INSERT INTO words_users (uid) VALUES (1), (2);

INSERT INTO words_games (gid, player1, player2) VALUES (100, 1, 2);

INSERT INTO words_social (sid, social, given, uid) VALUES 
('1111', 10, 'Alice', 1), 
('2222', 20, 'Bob', 2);

INSERT INTO words_chat (created, gid, uid, msg) VALUES 
(now() + interval '1 min', 100, 2, 'Hello, Alice'), 
(now() + interval '2 min', 100, 1, 'Hello, Bob'), 
(now() + interval '3 min', 100, 2, 'Nice to see you, Alice'), 
(now() + interval '4 min', 100, 1, 'Nice to see you too, Bob'), 
(now() + interval '5 min', 100, 2, 'Goodbye, Alice'), 
(now() + interval '6 min', 100, 1, 'Goodbye, Bob'); 

我正在尝试创建一个 PHP 脚本,当给定 sidsocialgid 会 return 游戏 gid 的完整聊天。

我不只是使用 gid 参数来获取和显示聊天,因为我不希望其他用户监视他们不玩的游戏。

我也不能使用 uid 参数来识别调用我的脚本的用户,我必须使用 sidsocial 参数代替(加上一个秘密,我在上面的测试用例中省略了)。

所以这是我的 SQL-获取聊天的语句:

SELECT uid, msg 
FROM words_chat 
WHERE gid=100 
AND EXISTS (select 1 from words_games where 
   (select uid from words_social 
   where sid='1111' and social=10) in (player1, player2)) 
ORDER BY CREATED ASC;

它有效,但我还需要 1 个东西,这使它变得复杂 - 对于每个聊天 msg 我需要一个布尔值(而不是 uid) 来知道这是不是 "Alice" 的消息(这样我就可以用粗体画那条线)。

所以我正在尝试这样做,SQL 语句变得更加丑陋:

SELECT uid=(select uid from words_social 
   where sid='1111' and social=10) AS mine, msg 
FROM words_chat 
WHERE gid=100 
AND EXISTS (select 1 from words_games where 
   (select uid from words_social 
   where sid='1111' and social=10) in (player1, player2)) 
ORDER BY CREATED ASC;

正如您在上面的屏幕截图中看到的那样有效,但如何改进此查询?

在 SQL 中有没有办法(即我不希望在我的 PHP 脚本中保存 uid 值或使用 pl/pgSQL) 在第二个子查询中重用 uid?

考虑使用 joins 就容易多了:

select wc.*, (wc.uid = ws.uid) as mine
from words_chat wc join
     words_games wg
     on wc.gid = wg.gid join
     words_social ws
     on ws.uid in (wg.player1, player2)
where wc.gid = 100 and
      ws.sid = '1111' and
      ws.social = 10
order by created asc;

Here 是 SQL Fiddle.