mysql - 我的奇怪子选择，需要 LEFT JOIN 改进

Question

以下是 SQL 转储示例:https://gist.github.com/JREAM/99287d033320b2978728

• 我有一个可以获取大量用户的 SELECT。
• 然后，我执行一个 foreach 循环，将所有关联的 tree_processes 附加到该用户。
• 所以我最终做了 X 查询:用户 * 树。

将两者一起获取不是效率更高吗？

• 我曾考虑过执行 LEFT JOIN Subselect，但很难正确执行。
• 下面我已经完成了一个查询，以在 SELECT 中选择正确的数据，但是我必须对所有 15 行执行此操作，这似乎是一种可怕的内存浪费。

这是我肮脏的尝试:

-

    SELECT
            s.id,
            s.firstname,
            s.lastname,
            s.email,
            (
                SELECT tp.id FROM tree_processes AS tp
                JOIN tree AS t ON (
                    t.id = tp.tree_id
                )
                WHERE subscribers_id = s.id
                ORDER BY tp.id DESC
                LIMIT 1
            ) AS newest_tree_id,
            #
            # Don't want to have to do this below for every row
            (
                SELECT t.type FROM tree_processes AS tp
                JOIN tree AS t ON (
                    t.id = tp.tree_id
                )
                WHERE subscribers_id = s.id
                ORDER BY tp.id DESC
                LIMIT 1
            ) AS tree_type


    FROM subscribers AS s
    INNER JOIN scenario_subscriptions AS ss ON (
        ss.subscribers_id = s.id
    )

    WHERE ss.scenarios_id = 1
    AND ss.completed != 1
    AND ss.purchased_exit != 1
    AND deleted != 1

    GROUP BY s.id
    LIMIT 0, 100

这是我的LEFT JOIN尝试，但我在获取 SELECT 值时遇到问题

SELECT
        s.id,
        s.firstname,
        s.lastname,
        s.email,
        freshness.id,
        # freshness.subscribers_id < -- Cant get multiples out of the LEFT join


FROM subscribers AS s
INNER JOIN scenario_subscriptions AS ss ON (
    ss.subscribers_id = s.id
)


LEFT JOIN ( SELECT tp.id, tp.subscribers_id AS tp FROM tree_processes AS tp
            JOIN tree AS t ON (
                t.id = tp.tree_id
            )
            ORDER BY tp.id DESC
            LIMIT 1 ) AS freshness 
ON (
    s.id = subscribers_id
)

WHERE ss.scenarios_id = 1
AND ss.completed != 1
AND ss.purchased_exit != 1
AND deleted != 1

GROUP BY s.id
LIMIT 0, 100

Answer 1

在 LEFT JOIN 中，您使用“freshness”作为表别名。在您选择时，您需要另外说明您想要从中获得哪些列。由于只有一列(id)，您需要添加:

freshness.id

到 select 子句。

您的左连接 ON 子句看起来也很危险。也许 freshness.id = ss.subscribers_id？

干杯-