我有一个数据库（MySQL），其中包含有关照片、他们的评分（人们可以投票）和评论的信息。
CREATE TABLE IF NOT EXISTS `Photos` (
`ID` INT NOT NULL AUTO_INCREMENT,
`Description` NVARCHAR(2000),
`Author` INT,
`Contest` INT,
`PhoneModel` NVARCHAR(200),
`PhotoThumbPath` NVARCHAR(2000),
`PhotoPath` NVARCHAR(2000),
`Date` TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
`State` INT DEFAULT 1,
FOREIGN KEY (`Author`) REFERENCES `Users`(`ID`),
FOREIGN KEY (`Contest`) REFERENCES `Contests`(`ID`),
PRIMARY KEY (`ID`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;
CREATE TABLE IF NOT EXISTS `PhotoComments` (
`ID` INT NOT NULL AUTO_INCREMENT,
`Author` INT,
`Photo` INT,
`Comment` NVARCHAR(2000),
`Date` TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
`ReplyTo` INT,
FOREIGN KEY (`Author`) REFERENCES `Users`(`ID`),
FOREIGN KEY (`Photo`) REFERENCES `Contests`(`ID`),
FOREIGN KEY (`ReplyTo`) REFERENCES `PhotoComments`(`ID`),
PRIMARY KEY (`ID`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;
CREATE TABLE IF NOT EXISTS `PhotoComments` (
`ID` INT NOT NULL AUTO_INCREMENT,
`Author` INT,
`Photo` INT,
`Comment` NVARCHAR(2000),
`Date` TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
`ReplyTo` INT,
FOREIGN KEY (`Author`) REFERENCES `Users`(`ID`),
FOREIGN KEY (`Photo`) REFERENCES `Contests`(`ID`),
FOREIGN KEY (`ReplyTo`) REFERENCES `PhotoComments`(`ID`),
PRIMARY KEY (`ID`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;
现在我想向数据库询问某张照片，告诉我它有多少评论，多少评分以及平均评分是多少。
其中哪一种方法会更快：使用子查询 3 进行查询：
SELECT P.PhotoPath, P.Date, P.PhoneModel,
(SELECT COUNT(1) FROM PhotoComments C where C.Photo=P.ID) as CommentsCount,
(SELECT COUNT(1) FROM PhotoRatings R where R.Photo=P.ID) as RatingsCount,
(SELECT COALESCE(AVG(Rating), 0) FROM PhotoRatings R where R.Photo=P.ID) as RatingsAvg
FROM Photos WHERE P.ID=$photoID
或一个带有 2 个左连接的大查询：
SELECT P.PhotoPath, P.Date, P.PhoneModel,
COUNT(DISTINCT C.ID) as CommentsCount,
COUNT(DISTINCT R.ID) as RatingsCount,
COALESCE(AVG(R.Rating), 0) as RatingsAvg
FROM Photos P
LEFT JOIN PhotoComments C ON C.Photo=P.ID
LEFT JOIN PhotoRatings R ON R.Photo=P.ID
WHERE P.ID=$photoID
GROUP BY P.ID