MySQL：IN（p1）的功能与IN（p1，p2，...）的功能不同吗？

Question

我刚刚注意到这一点非常奇怪的（至少在我看来），东西...

我已确认的查询不被缓存，但采用的只有一个值查询IN（）功能超级慢......无缘无故我无法想到。我可以把虚值，在那里，但......我真的不想：/

SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST A' 
AND Clues.CultureID IN (1); 

受影响的行：0
时间：3.843ms

SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST B' 
AND Clues.CultureID IN (1,2); 

受影响行：0
时间：0.141ms

SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST C' 
AND Clues.CultureID IN (1); 

影响的行：0
时间：3.815ms

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST D' 
AND Clues.CultureID IN (1,2); 

受影响的行：0
时间：0.142ms

SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST E' 
AND Clues.CultureID IN (1); 

受影响的行：0
时间：3.843ms

点

SELECT Clues.* FROM Clues INNER JOIN SolutionWords 
ON SolutionWords.SolutionWordID = Clues.SolutionWordID 
WHERE SolutionWord COLLATE utf8_general_ci = 'TEST F' 
AND Clues.CultureID IN (1,2); 

受影响的行：0
时间：0.142ms

EDIT：更改参数顺序 - >没有影响

**[SQL] SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE Clues.CultureID IN (1) AND SolutionWord COLLATE utf8_general_ci = 'TEST 1' ; 
Affected rows: 0 
Time: 3.833ms** 

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE SolutionWord COLLATE utf8_general_ci = 'TEST 2' AND Clues.CultureID IN (1,2); 
Affected rows: 0 
Time: 0.141ms 

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE SolutionWord COLLATE utf8_general_ci = 'TEST 3' AND Clues.CultureID IN (1); 
Affected rows: 0 
Time: 3.821ms 

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE SolutionWord COLLATE utf8_general_ci = 'TEST 4' AND Clues.CultureID IN (1,2); 
Affected rows: 0 
Time: 0.141ms 

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE SolutionWord COLLATE utf8_general_ci = 'TEST 5' AND Clues.CultureID IN (1); 
Affected rows: 0 
Time: 3.848ms 

[SQL] 
SELECT Clues.* FROM Clues INNER JOIN SolutionWords ON SolutionWords.SolutionWordID = Clues.SolutionWordID WHERE SolutionWord COLLATE utf8_general_ci = 'TEST 6' AND Clues.CultureID IN (1,2); 
Affected rows: 0 
Time: 0.142ms 


EDIT: EXPLAIN for slow query (actually, it's EXACTLY the same for the fast one as well unless I'm missing something obvious) 

1 SIMPLE Clues ref fk_Clues_Cultures1,fk_Clues_SolutionWords1 fk_Clues_Cultures1 5 const 371462 Using where 
1 SIMPLE SolutionWords eq_ref PRIMARY PRIMARY 4 ClueExplorer.Clues.SolutionWordID 1 Using where 

1 SIMPLE Clues ref fk_Clues_Cultures1,fk_Clues_SolutionWords1 fk_Clues_Cultures1 5 const 371462 Using where 
1 SIMPLE SolutionWords eq_ref PRIMARY PRIMARY 4 ClueExplorer.Clues.SolutionWordID 1 Using where 

编辑：SHOW CREATE TABLE

CREATE TABLE `Clues` (
    `ClueID` int(11) NOT NULL AUTO_INCREMENT, 
    `SolutionWordID` int(11) NOT NULL, 
    `Clue` text CHARACTER SET utf8 COLLATE utf8_bin NOT NULL, 
    `SolutionWordBreakup` varchar(45) DEFAULT NULL, 
    `SpellingID` int(11) DEFAULT NULL, 
    `WordFormID` int(11) DEFAULT NULL, 
    `GrammaticalFormID` int(11) DEFAULT NULL, 
    `ClueCategoryID` int(11) DEFAULT NULL, 
    `ClueFormatID` int(11) DEFAULT NULL, 
    `ClueStyleID` int(11) DEFAULT NULL, 
    `CultureID` int(11) DEFAULT NULL, 
    `Difficulty` int(11) DEFAULT NULL, 
    `Interestingness` int(11) DEFAULT NULL, 
    `ReviewDate` datetime DEFAULT NULL, 
    `Explanation` text, 
    `Citation` text, 
    `RelevantFrom` datetime DEFAULT NULL, 
    `RelevantTo` datetime DEFAULT NULL, 
    PRIMARY KEY (`ClueID`), 
    KEY `fk_Clues_Spellings` (`SpellingID`), 
    KEY `fk_Clues_WordForms1` (`WordFormID`), 
    KEY `fk_Clues_GrammaticalForms1` (`GrammaticalFormID`), 
    KEY `fk_Clues_ClueFormats1` (`ClueFormatID`), 
    KEY `fk_Clues_ClueStyles1` (`ClueStyleID`), 
    KEY `fk_Clues_Cultures1` (`CultureID`), 
    KEY `fk_Clues_ClueCategories1` (`ClueCategoryID`), 
    KEY `fk_Clues_SolutionWords1` (`SolutionWordID`), 
    CONSTRAINT `fk_Clues_ClueCategories1` FOREIGN KEY (`ClueCategoryID`) REFERENCES `ClueCategories` (`ClueCategoryID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_SolutionWords1` FOREIGN KEY (`SolutionWordID`) REFERENCES `SolutionWords` (`SolutionWordID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_ClueFormats1` FOREIGN KEY (`ClueFormatID`) REFERENCES `ClueFormats` (`ClueFormatID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_ClueStyles1` FOREIGN KEY (`ClueStyleID`) REFERENCES `ClueStyles` (`ClueStyleID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_Cultures1` FOREIGN KEY (`CultureID`) REFERENCES `Cultures` (`CultureID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_GrammaticalForms1` FOREIGN KEY (`GrammaticalFormID`) REFERENCES `GrammaticalForms` (`GrammaticalFormID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_Spellings` FOREIGN KEY (`SpellingID`) REFERENCES `Spellings` (`SpellingID`) ON DELETE CASCADE ON UPDATE CASCADE, 
    CONSTRAINT `fk_Clues_WordForms1` FOREIGN KEY (`WordFormID`) REFERENCES `WordForms` (`WordFormID`) ON DELETE CASCADE ON UPDATE CASCADE 
) ENGINE=InnoDB AUTO_INCREMENT=2458300 DEFAULT CHARSET=utf8 

Answer 1

马特，你有没有考虑过你的问题可能与你所在列的顺序有关重新条款？你可以请尝试（作为一个例子）：

...WHERE Clues.CultureID IN (1,2) 
    AND SolutionWord COLLATE utf8_general_ci = 'STRING' 

然后看看它是怎么回事。

Chers，

雷纳托