SQL優化：使用任何子查詢查找第一個未查看視頻

Question

我的問題。我認爲我有一個性能查殺子查詢，但無法證明它。我第一次嘗試使用JOIN失敗。 有人可以提供更高性能的解決方案，或確認這確實可以接受，因爲它是？

我有兩個表，一個包含todo-list（joblist）和一個跟蹤每個用戶進度（userprogress）的表。工作可以，但不能觀看視頻。 （這是一個電子學習網站。）

當觀看視頻時，它們會自動設置爲enum字段上的「已完成」。用戶也可以手動跳過視頻（狀態=「跳過」）。

表結構如下。

要獲取用戶根本沒有看過的第一個視頻（userprogress中沒有記錄）或已開始觀看（status ='started'），我正在使用此查詢。

我已經設置了用於選擇或排序的字段上的索引。但是我不確定他們是否都需要。

SELECT語句有兩個部分

1. 的內部子查詢，我在那裏取都見過或跳過視頻
2. 主要發言，我在那裏取第一視頻不中（1）

有一個用於PHP（：email）的命名參數，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 

這是EXPLAIN輸出，這我需要一些幫助理解。

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，你是正確的。 IN和NOT IN在mysql中表現特別差。下面是一個修訂版：

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

你看上去手忙腳亂......你的子查詢尋找與地位影片完成或跳過，然後在outter查詢找誰不具備的那些這種地位，我會改變，對於這樣的

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，也許我理解錯了的情況，反正這個問題似乎是NOT IN（我不會建議曾經使用過這一點），而不是嘗試改變子查詢條件並做一個左參加它，並添加一個條件And SQ.joblistID IS NULL，這樣的事情

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 

但我認爲第一個選項將工作...

希望它有幫助