老实说,我不知道如何处理这个问题。我已经到了SELECT name,然后我迷路了(有点尴尬)。
问题:找出按星球分组的人持有的证书数量。这应该有两列,第一列,"name“将是至少有一个认证的行星的名称。第二列应该是"CertCount“,并将是来自那个星球的人持有的认证数量。例如,如果李获得了”毒蛇“和”机械师“的认证,而卡拉获得了”毒蛇“的认证,并且他们都来自卡布里卡,那么卡布里卡的"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发布于 2016-10-26 09:12:31
将它们连接起来,使用group by进行计数,应该可以做到这一点:
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发布于 2016-10-26 09:17:57
SELECT pl.name, count(cert) AS "CertCount"
FROM bsg_planets pl
JOIN bsg_people pe ON pl.id = pe.homeworld
JOIN bsg_cert_people cp ON cp.pid = pe.id
GROUP BY pl.id我想就是这里了。
https://stackoverflow.com/questions/40251841
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