学生表名为：student, 单科成绩的字段为：subject。学生名字为：name
查询单科成绩前十：mysql语句为：select * from student order by subject desc limit 10
查询总分成绩前十：mysql语句为：select sum(subject) subject,name from student group by name order by subject desc limit 10
注：
select sum(subject) subject,name
这句的意思是：sum(subject) subject 把单科成绩加总用subject 来命名（就是总成绩），name就是一个字段，这里只有两个字段。
group by name order by subject ： group by name 的意思按照名字这一栏来分组，当然，学生成绩表名字有可能是一样的，按照学号是最准确的，这里只是举个例子。
order by subject 这句的意思是按照总分成绩排序，这里的subject 是前面重命名总分的意思。
select sum(subject) as countsubject,name from student group by name order by countsubject desc limit 10
扩展资料：学生成绩表常用sql
1. 在表中插入符合主键
[sql]
/*成绩表*/
CREATE TABLE SC
(
Sid INT REFERENCES Student(Sid), /*学生学号*/
Cid INT REFERENCES Course(Cid), /*课程编号*/
Score INT NOT NULL, /*课程分数*/
PRIMARY KEY(Sid,Cid) /*将学生学号和课程编号设为复合主键*/
)
2. 查询各科成绩最高分，最低分以及平均分
[sql]
SELECT c.Cname, MAX(s.Score) AS Max, MIN(s.Score) AS Min, AVG(s.Score) AS Average
FROM Course c JOIN SC s ON c.Cid = s.Cid
GROUP BY c.Cname
/*此处应注意，若不按照c.Cname进行分组，SQL语句会报错，c.Cname在SELECT语句中不合法，因为它并未出现在聚合函数中也没有出现在GROUP BY语句中*/
3. 查询平均成绩大于80分的学生姓名以及平均成绩
[sql]
SELECT Sname, AVG(Score) AS Average FROM Student JOIN SC
ON Student.Sid=SC.Sid
GROUP BY Sname
HAVING AVG(Score)>80
/*以聚合函数为条件进行删选只能在HAVING语句中进行，WHERE语句不支持聚合函数*/
4. 按总分为学生排名，总分相同名次相同
[sql]
SELECT RANK() OVER (ORDER BY SUM(ss.Score) DESC) AS Rank, s.Sname,
ISNULL(SUM(ss.Score),0)
FROM Student s LEFT JOIN SC ss
ON s.Sid = ss.Sid
GROUP BY s.Sname
ORDER BY SUM(ss.Score) DESC
/*RANK()是SQL Server的一个built-in函数，语法为
RANK() OVER ( [ partition_by_clause ] order_by_clause ).*/
5. 查询总分在100至200之间的学生姓名及总分
[sql]
SELECT s.Sname,SUM(ss.Score) FROM Student s JOIN SC ss ON s.Sid=ss.Sid
GROUP BY s.Sname HAVING SUM(ss.Score) BETWEEN 100 AND 200
借题主答复另一个问题的答案：
表一（考试名称表）：CREATE TABLE `zm_exam` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT COMMENT '主键id，考试表',
`title` varchar(120) NOT NULL DEFAULT '' COMMENT '考试名称',
`t_id` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '外键，老师表',
`cg_id` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '外键，班级表',
`exam_time` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '考试时间',
`status` tinyint(2) unsigned NOT NULL DEFAULT '1' COMMENT '1正常，2禁用',
`type` tinyint(2) unsigned NOT NULL DEFAULT '1' COMMENT '考试类别，1单科，2全部',
`remark` varchar(255) NOT NULL DEFAULT '' COMMENT '考试说明',
`createdtime` datetime DEFAULT '0000-00-00 00:00:00',
`changedtime` datetime DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=3 DEFAULT CHARSET=utf8 表二（考试科目记录表）：
CREATE TABLE `zm_score` (
`id` int(11) unsigned NOT NULL AUTO_INCREMENT COMMENT '主键id，成绩表',
`e_id` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '外键，考试表',
`c_id` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '外键，科目表',
`stu_id` int(11) unsigned NOT NULL DEFAULT '0' COMMENT '外键，学生表',
`grade` float unsigned NOT NULL DEFAULT '0' COMMENT '成绩分数',
`status` tinyint(2) unsigned NOT NULL DEFAULT '1' COMMENT '1正常，2禁用',
`remark` varchar(255) NOT NULL DEFAULT '' COMMENT '说明',
`createdtime` datetime NOT NULL DEFAULT '0000-00-00 00:00:00',
`changedtime` datetime NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=47 DEFAULT CHARSET=utf8
问题：查出每次考试中，该学生在此次考试中的分数排名：
-- a表、b表同为一模一样的临时表-- 通过比对a、b两表的考试记录id&考试分数个数（ WHERE a.id = b.id AND a.grade < b.grade）
SELECT *,
(SELECT COUNT(*)+1 FROM
(SELECT e.`id`,e.`title`,e.`remark`,s.`grade`,s.`stu_id`,e.`exam_time` FROM zm_student stu,zm_exam e,zm_score s
WHERE stu.`id` = 132 AND stu.`status` = 1 AND stu.`c_g_id` = e.`cg_id` AND e.`type` = 1 AND e.`id` = s.`e_id` AND s.`status` = 1 ORDER BY s.`grade` DESC) AS b WHERE a.id = b.id AND a.grade < b.grade) AS paiming -- 名次
FROM
(SELECT e.`id`,e.`title`,e.`remark`,s.`grade`,s.`stu_id`,e.`exam_time` FROM zm_student stu,zm_exam e,zm_score s
WHERE stu.`id` = 132 AND stu.`status` = 1 AND stu.`c_g_id` = e.`cg_id` AND e.`type` = 1 AND e.`id` = s.`e_id` AND s.`status` = 1 ORDER BY s.`grade` DESC) AS a WHERE a.stu_id = 132 ORDER BY id DESC
欢迎分享，转载请注明来源：内存溢出
微信扫一扫
支付宝扫一扫
评论列表（0条）