2

我的问题。我想我有一个性能杀伤子查询，但无法证明它。我第一次尝试使用 JOIN 失败了。有人可以提供更高性能的解决方案或确认这确实可以接受吗？

我有两张表，一张包含待办事项列表（joblist），一张跟踪每个用户的进度（userprogress）。工作可以但不能是看视频。（这是一个电子学习网站。）

观看视频后，它们会在枚举字段上自动设置为“已完成”。用户也可以手动跳过视频（status = 'skipped'）。

下面提供了表结构。

要获取用户根本没有观看的第一个视频（用户进度中没有记录）或已经开始观看（状态 = '开始'），我正在使用此查询。

我已经在用于选择或排序的任何字段上设置了索引。但是我不确定它们是否都需要。

SELECT 语句有两个部分

  1. 一个内部子选择，我在其中获取所有看过或跳过的视频
  2. 主要声明，我在其中获取第一个视频，而不是（1）找到的视频

PHP (:email) 有一个命名参数，以避免 SQL 注入。

SELECT jl.where_to_do_it FROM joblist AS jl
INNER JOIN userprogress AS up
ON (jl.joblistID = up.joblistID)
WHERE jl.what_to_do = 'video'
     AND jl.joblistID NOT IN
      (
        SELECT injl.joblistID
        FROM joblist AS injl
        INNER JOIN userprogress AS inup
        ON (injl.joblistID = inup.joblistID)
        WHERE
             (inup.status = 'finished' OR inup.status = 'skipped')
          AND
             inup.email = :email
          AND
             injl.what_to_do = 'video'
      )
ORDER BY jl.joborder ASC
LIMIT 0,1

这是 EXPLAIN 的输出，我需要一些帮助来理解。

id select_type  table     type  possible_keys                 key         key_len  ref          rows   Extra
1  PRIMARY      jl        ref   PRIMARY,what_to_do            what_to_do  602      const        9      Using where; Using filesort
1  PRIMARY      up        ref   joblistID                     joblistID   3        jl.joblistID 1      Using index
2  DEP-SUB      injl  eq_ref    PRIMARY,what_to_do            PRIMARY     3        func         1      Using where
2  DEP-SUB      inup  eq_ref    nodup,email,joblistID,status  nodup       455      const,func   1      Using where

创建表命令：

CREATE TABLE IF NOT EXISTS `joblist` (
  `joblistID` mediumint(10) unsigned NOT NULL AUTO_INCREMENT,
  `what_to_do` varchar(200) COLLATE utf8_swedish_ci NOT NULL,
  `where_to_do_it` varchar(100) COLLATE utf8_swedish_ci NOT NULL,
  `joborder` mediumint(6) NOT NULL,
  `track` enum('fast','slow','bonus') COLLATE utf8_swedish_ci NOT NULL DEFAULT 'slow',
  `chapter` tinyint(11) unsigned NOT NULL COMMENT 'What book chapter it relates to',
  PRIMARY KEY (`joblistID`),
  KEY `nodupjobs` (`joborder`,`chapter`),
  KEY `what_to_do` (`what_to_do`),
  KEY `where_to_do_it` (`where_to_do_it`),
  KEY `joborder` (`joborder`),
  KEY `track` (`track`),
  KEY `chapter` (`chapter`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_swedish_ci COMMENT='Suggested working order';


CREATE TABLE IF NOT EXISTS `userprogress` (
  `upID` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `email` varchar(150) COLLATE utf8_swedish_ci NOT NULL COMMENT 'user id',
  `joblistID` mediumint(9) unsigned NOT NULL COMMENT 'foreign key',
  `progressdata` varchar(300) COLLATE utf8_swedish_ci DEFAULT NULL COMMENT 'JSON object describing progress',
  `percentage_complete` tinyint(3) unsigned DEFAULT NULL,
  `status` enum('begun','skipped','finished') COLLATE utf8_swedish_ci DEFAULT 'begun',
  `lastupdate` datetime NOT NULL,
  `approved` datetime DEFAULT NULL,
  PRIMARY KEY (`upID`),
  UNIQUE KEY `nodup` (`email`,`joblistID`),
  KEY `email` (`email`),
  KEY `joblistID` (`joblistID`),
  KEY `status` (`status`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8 COLLATE=utf8_swedish_ci COMMENT='Keep track of what the user has done';
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2 回答 2

2

是的，你是对的。IN 和 NOT IN 在 mysql 中表现特别差。这是一个修订版：

SELECT jl.where_to_do_it
FROM joblist jl INNER JOIN
     userprogress up
     ON (jl.joblistID = up.joblistID)
 WHERE jl.what_to_do = 'video' and
       not exists (
           (SELECT 1
            FROM joblist injl INNER JOIN
                 userprogress inup
                 ON (injl.joblistID = inup.joblistID)
            WHERE (inup.status = 'finished' OR inup.status = 'skipped') and
                  inup.email = :email and
                  injl.what_to_do = 'video' and
                  ini1.joblistid = j1.joblistid
          )
ORDER BY jl.joborder ASC
LIMIT 0,1
于 2012-08-17T01:09:21.120 回答
1

看起来你在圈子里跑......你的子查询正在寻找状态为已完成或已跳过的视频，然后在外部查询中寻找那些没有该状态的视频，我会为这样的条件更改它

SELECT jl.where_to_do_it FROM joblist AS jl
INNER JOIN userprogress AS up
ON (jl.joblistID = up.joblistID)
WHERE jl.what_to_do = 'video'
 AND up.status <> 'finished' AND inup.status <> 'skipped'
 AND up.email = :email
 AND jl.what_to_do = 'video'

或者也许我理解错了，无论如何问题似乎是 NOT IN （我不会建议永远使用它）而不是尝试更改条件中的子查询并对其进行左连接并添加条件And SQ.joblistID IS NULL，就像这样

 SELECT jl.where_to_do_it FROM joblist AS jl
 INNER JOIN userprogress AS up
 ON (jl.joblistID = up.joblistID)
 LEFT JOIN (
    SELECT injl.joblistID
    FROM joblist AS injl
    INNER JOIN userprogress AS inup
    ON (injl.joblistID = inup.joblistID)
    WHERE
         (inup.status = 'finished' OR inup.status = 'skipped')
      AND
         inup.email = :email
      AND
         injl.what_to_do = 'video'
  ) SQ ON jl.joblistID = SQ.joblistID
 WHERE jl.what_to_do = 'video'
 AND SQ.joblistID IS NULL      
 ORDER BY jl.joborder ASC

但我认为第一个选项会起作用......

希望能帮助到你

于 2012-08-17T01:27:24.197 回答