SQL优化：使用任何子查询查找第一个未查看视频

Question

我的问题。我认为我有一个性能查杀子查询，但无法证明它。我第一次尝试使用JOIN失败。 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，因为它是？

我有两个表，一个包含todo-list（joblist）和一个跟踪每个用户进度（userprogress）的表。工作可以，但不能观看视频。 （这是一个电子学习网站。）

当观看视频时，它们会自动设置为enum字段上的“已完成”。用户也可以手动跳过视频（状态=“跳过”）。

表结构如下。

要获取用户根本没有看过的第一个视频（userprogress中没有记录）或已开始观看（status ='started'），我正在使用此查询。

我已经设置了用于选择或排序的字段上的索引。但是我不确定他们是否都需要。

SELECT语句有两个部分

1. 的内部子查询，我在那里取都见过或跳过视频
2. 主要发言，我在那里取第一视频不中（1）

有一个用于PHP（：email）的命名参数，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输出，这我需要一些帮助理解。

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 

在CREATE TABLE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'; 

Answer 1

是的，你是正确的。 IN和NOT IN在mysql中表现特别差。下面是一个修订版：

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 

Answer 2

你看上去手忙脚乱......你的子查询寻找与地位影片完成或跳过，然后在outter查询找谁不具备的那些这种地位，我会改变，对于这样的

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，也许我理解错了的情况，反正这个问题似乎是NOT IN（我不会建议曾经使用过这一点），而不是尝试改变子查询条件并做一个左参加它，并添加一个条件And SQ.joblistID IS NULL，这样的事情

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将工作...

希望它有帮助