如何使用 5 个子查询优化此查询？答案

【问题标题】：How can I optimize this query with 5 subqueries?如何使用 5 个子查询优化此查询？
【发布时间】：2014-11-20 19:09:40
【问题描述】：

SELECT
  (SELECT COALESCE(SUM(t.hours), 0) AS allotted_hours
   FROM tasks AS t
   WHERE t.projects_id = 8
     AND t.complete != 100
     AND t.name LIKE '%Ongoing%'
     AND t.name NOT LIKE '%Placeholder%') AS allotted_hours_notcomplete_ongoing,

  (SELECT COALESCE(SUM(tl.hours), 0) AS hr
   FROM tasks AS t
   INNER JOIN tasklogs AS tl ON (tl.tasks_id = t.id)
   WHERE t.projects_id = 8
     AND t.complete != 100
     AND t.name LIKE '%Ongoing%'
     AND t.name NOT LIKE '%Placeholder%') AS logged_hours_notcomplete_ongoing,

  (SELECT (allotted_hours_notcomplete_ongoing - logged_hours_notcomplete_ongoing) + logged_hours_notcomplete_ongoing) AS difference_notcomplete_ongoing,

  (SELECT COALESCE(SUM(t.hours), 0) AS hr
   FROM tasks AS t
   WHERE t.projects_id = 8
     AND t.complete != 100
     AND t.name NOT LIKE '%Ongoing%'
     AND t.name NOT LIKE 'Placeholder%') AS allotted_hours_notcomplete_regular,

  (SELECT COALESCE(SUM(tl.hours), 0) AS hr
   FROM tasklogs AS tl
   INNER JOIN tasks AS t ON (t.id = tl.tasks_id
                             AND t.projects_id = 8
                             AND t.complete = 100)
   WHERE hourtypes_id IN (1,
                          2)) AS logged_hours_complete,

  (SELECT logged_hours_complete + allotted_hours_notcomplete_regular + difference_notcomplete_ongoing) AS total

我基本上是在尝试优化此查询，以便在将其合并到我的基本 SELECT * FROM projects 查询之前计算给定项目的任务小时数。

这些子查询中的 1 个仅从 tasks 表中选择，其中 2 个从 tasks 和 tasklogs 上的 INNER JOIN 中选择，反之亦然，其中 2 个是相互简单的计算。

有没有更有效的优化方法？我从来没有使用过临时表，但也许我可以先从 tasks 中选择具有给定项目 ID 的所有任务，然后从该临时表中执行我以后的 SELECT 操作？不胜感激。

如果需要，我可以使用 sql fiddle 复制架构，但这需要一些时间。

【问题讨论】：

标签： mysql optimization

【解决方案1】：

您不止一次从同一张桌子经过。如果没有任何执行计划，我会先尝试以下操作。

    SELECT 
    (SELECT SUM(GT.tHours) FROM GT
    WHERE 1 = 1
    AND t.name LIKE '%Ongoing%'
     AND t.name NOT LIKE '%Placeholder%'
    ) AS allotted_hours_notcomplete_ongoing
    , 
    (SELECT SUM(GT.tlHours) FROM GT
    WHERE 1 = 1
    AND t.name LIKE '%Ongoing%'
     AND t.name NOT LIKE '%Placeholder%'
    ) AS logged_hours_notcomplete_ongoing
    (SELECT SUM(GT.tlHours) FROM GT
    WHERE 1 = 1
    AND t.name LIKE '%Ongoing%'
     AND t.name NOT LIKE '%Placeholder%'
     AND hourtypes_id IN (1,
                          2)) 

    ) AS  logged_hours_complete,
    (SELECT (allotted_hours_notcomplete_ongoing - logged_hours_notcomplete_ongoing) + logged_hours_notcomplete_ongoing) AS difference_notcomplete_ongoing,

    (SELECT COALESCE(SUM(thours), 0) AS hr
    FROM GT
    WHERE
     AND t.name NOT LIKE '%Ongoing%'
     AND t.name NOT LIKE 'Placeholder%') AS allotted_hours_notcomplete_regular,


    FROM
    (
    SELECT SUM(t.hours) AS tHours ,SUM(tl.hours) AS tlHours, t.name,hourtypes_id
    FROM tasks AS t
    LEFT JOIN tasklogs AS tl ON (tl.tasks_id = t.id)
    WHERE t.projects_id = 8
    AND t.complete != 100
    GROUP BY t.name,hourtypes_id
    ) GT

我基本上只尝试了一次总和时间。之后，我将使用内部选择和位置进行过滤。您当然需要以下索引。

任务

t.id
t.projects_id
t.complete
t.name
hourtypes_id
t.hours

任务日志

tl.tasks_id
tl.hours

我假设 hourtypes_id 在任务表中。

【讨论】：