老实说,我不知道如何解决这个问题。我已经SELECT name
了,然后我就迷路了(有点尴尬)。
问题:找出按星球分组的人们持有的认证数量。这应该有两列,第一列,“名称”将是至少具有一个认证的行星的名称。第二列应该是“CertCount”,并且将是来自该星球的人持有的认证数量,例如,如果 Lee 获得了“Viper”和“Mechanic”认证,而 Kara 获得了“Viper”认证,并且他们都来自 Caprica,那么 caprica 的“CertCount”应该是 3:
CREATE TABLE `bsg_cert` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB
CREATE TABLE `bsg_cert_people` (
`cid` int(11) NOT NULL DEFAULT '0',
`pid` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`cid`,`pid`),
KEY `pid` (`pid`),
CONSTRAINT `bsg_cert_people_ibfk_1` FOREIGN KEY (`cid`) REFERENCES `bsg_cert` (`id`),
CONSTRAINT `bsg_cert_people_ibfk_2` FOREIGN KEY (`pid`) REFERENCES `bsg_people` (`id`)
) ENGINE=InnoDB
CREATE TABLE `bsg_people` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`fname` varchar(255) NOT NULL,
`lname` varchar(255) DEFAULT NULL,
`homeworld` int(11) DEFAULT NULL,
`age` int(11) DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `homeworld` (`homeworld`),
CONSTRAINT `bsg_people_ibfk_1` FOREIGN KEY (`homeworld`) REFERENCES `bsg_planets` (`id`) ON DELETE SET NULL ON UPDATE CASCADE
) ENGINE=InnoDB
CREATE TABLE `bsg_planets` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`population` bigint(20) DEFAULT NULL,
`language` varchar(255) DEFAULT NULL,
`capital` varchar(255) DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `name` (`name`)
) ENGINE=InnoDB
最佳答案
将它们全部加入并使用 group by 进行计数应该可以解决问题:
SELECT planet.name ,
COUNT(*) AS cert_count
FROM bsg_cert_people people_cert
JOIN bsg_people people ON people.id = people_cert.pid
JOIN bsg_planet planet ON people.homeworld = planet.id
GROUP BY planet.name
关于mysql - SQL - 获取地球上人们的认证总数(3 个单独的表),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40251841/