哪个查询更有效？内部联接与子查询？总和案例与拥有

Question

我正在使用 PostgreSql 9。我有一个简单的问题。 哪个查询效率更高？

SELECT users_sessions.session_id, users_sessions.series 
FROM users_sessions 
WHERE users_sessions.user_id = 8 
AND users_sessions.session_id IN (
    SELECT session_id 
    FROM sessions_history 
    GROUP BY sessions_history.session_id 
    HAVING SUM(CASE WHEN sessions_history.action = 2 THEN 1 END) = 0
) 

VS.

SELECT US8.session_id,Us8.series

FROM

( SELECT us.session_id as S_ID, us.series 
        FROM users_sessions as US
        WHERE US.user_id = 8 ) AS US8
INNER JOIN
(SELECT SH.session_id as SH_ID
        FROM session_history as SH
        WHERE SH.action <> 2) AS SH2    
ON US8.session_id = SH2.session_id 

Answer 1

另一种解决方案使用 NOT EXISTS，这是您似乎要问的内容的精确翻译：

给我不存在操作 2 的会话：

SELECT session_id, series 
FROM users_sessions as us
WHERE users_sessions.user_id = 8 
AND NOT EXISTS
 (  SELECT * 
    FROM sessions_history as sh
    WHERE action = 2
    AND us.session_id = sh.session_id 
 ) 

Answer 2

您的第二个查询 is 在语法上无效。您不能在HAVING 子句中引用SELECT 列表（“输出列”）中的列别名。

不管怎样，这两个查询都不好。如果您确实想找到不存在的(user_id, action) 组合，请尝试：

SELECT t.*, 0 AS s
FROM  (SELECT 8 AS user_id, 2 AS action) t
LEFT  JOIN (
            users_sessions  us
       JOIN session_history sh USING (session_id)
           ) USING (user_id, action)
WHERE  sh.session_id IS NULL;

我将子查询t 和LEFT JOIN 中的单行派生表引入到两个基表的组合中。如果组合 不 存在，则只返回一行。 假设列名 user_id 和 action 在两个表中只出现一次。否则，对第二个连接使用更明确的条件：

ON t.user_id = us.user_id AND t.action = sh.action

详情：

• Select rows which are not present in other table

另类

可能更快，但是：

SELECT t.*, 0 AS s
FROM  (SELECT 8 AS user_id, 2 AS action) t
LEFT   JOIN users_sessions  us USING (user_id)
LEFT   JOIN session_history sh USING (action, session_id)
WHERE  sh.session_id IS NULL;

USING 在本例中无论哪种方式都是安全的。

Answer 3

我的 tuppence 值；

我可以用两种不同的方式解读你的意图，并且对每一种都有不同的解决方案（我首选的解决方案在每种情况下都是第一个）；

具有至少一个历史记录的会话 2
（而不关心有多少历史记录在其中 action = 2）

SELECT
    us.session_id,
    us.series
FROM
    users_sessions   AS us
INNER JOIN
    session_history  AS sh
        ON  sh.session_id  = us.session_id
        AND sh.action     <> 2
WHERE
    us.user_id = 8
GROUP BY
    us.session_id,
    us.series

或

SELECT
    us.session_id,
    us.series
FROM
    users_sessions   AS us
WHERE
        us.user_id = 8
    AND EXISTS (SELECT *
                  FROM session_history  AS sh
                 WHERE sh.session_id  = us.session_id
                   AND sh.action     <> 2
               )

或

SELECT
    us.session_id,
    us.series
FROM
    users_sessions   AS us
INNER JOIN
(
    SELECT
        session_id
    FROM
        session_history
    WHERE
        sh.action <> 2
    GROUP BY
        session_id
)
    AS sh
        ON  sh.session_id  = us.session_id
WHERE
    us.user_id = 8

没有历史记录且 action = 2 的会话
（但可以有其他历史记录）

SELECT
    us.session_id,
    us.series
FROM
    users_sessions   AS us
LEFT JOIN
    session_history  AS sh
        ON  sh.session_id = us.session_id
        AND sh.action     = 2
WHERE
        us.user_id     = 8
    AND sh.session_id IS NULL

-- No GROUP BY needed this time

或

SELECT
    us.session_id,
    us.series
FROM
    users_sessions   AS us
WHERE
        us.user_id = 8
    AND NOT EXISTS (SELECT *
                     FROM session_history  AS sh
                     WHERE sh.session_id = us.session_id
                       AND sh.action     = 2
                   )

Answer 4

您对此有何看法？

SELECT US8.session_id,Us8.series

FROM

( SELECT us.session_id as S_ID, us.series 
        FROM users_sessions as US
        WHERE US.user_id = 8 ) AS US8
INNER JOIN
(SELECT SH.session_id as SH_ID
        FROM session_history as SH
        WHERE SH.action <> 2) AS SH2    
ON US8.session_id = SH2.session_id