mysql练习

    技术2022-07-10  151

    #创建0630mysql练习题数据库 CREATE DATABASE IF NOT EXISTS 0630_practice; USE 0630_practice; #创建学生信息表 CREATE TABLE IF NOT EXISTS stuInfo( CLASSNO VARCHAR(50), STUDENTNO INT(11), `NAME` VARCHAR(50), SEX VARCHAR(50), COURSE VARCHAR(50), GRADE INT(11) ); #向学生信息表插入数据 INSERT INTO stuInfo VALUES ('Class1',1001,'Jennifer','F','English',98), ('Class1',1002,'Tommy','M','English',89), ('Class1',1003,'Alex','M','English',91), ('Class2',2001,'Emar','F','English',91), ('Class2',2002,'Racher','F','English',80), ('Class2',2003,'Chandler','M','English',81), ('Class3',3001,'Roben','M','English',65), ('Class3',3002,'Hugo','M','English',66), ('Class3',3003,'Ellen','F','English',67), ('Class1',1004,'Jonny','F','French',88), ('Class1',1005,'Tedesco','M','French',88), ('Class1',1006,'Scofield','M','French',89), ('Class2',2001,'Emar','F','French',80), ('Class2',2002,'Racher','F','French',78), ('Class2',2003,'Chandler','M','French',84), ('Class3',3001,'Roben','M','French',67), ('Class3',3002,'Hugo','M','French',70), ('Class3',3003,'Ellen','F','French',71); #1)查询Class1班级英语平均分与Class2中英语平均分的差值 SELECT ABS( ( SELECT AVG(s.`GRADE`) FROM stuInfo s WHERE s.`COURSE`='English' AND s.`CLASSNO`='Class1') - ( SELECT AVG(s.`GRADE`) FROM stuInfo s WHERE s.`COURSE`='English' AND s.`CLASSNO`='Class2')) 1班和2班英语平均分差值; #2)查询各班级法语成绩中低于班级法语平均分人员数 SELECT SUM(CASE WHEN s.`GRADE`<a.avgg THEN 1 ELSE 0 END) 人数,s.`CLASSNO` 班级 FROM stuInfo s , (SELECT AVG(s.`GRADE`) avgg,s.`CLASSNO` CLASS FROM stuInfo s WHERE s.`COURSE`='French' GROUP BY s.`CLASSNO`) a WHERE s.`COURSE`='French' AND s.`CLASSNO`=a.CLASS GROUP BY s.`CLASSNO`; #3)查询每个班级英语成绩最好的同学输出:班级,姓名 SELECT a.n 姓名,a.c 班级 FROM ( SELECT MAX(s.`GRADE`),s.`NAME` n,s.`CLASSNO` c FROM stuInfo s WHERE s.`COURSE`='English' GROUP BY s.`CLASSNO`) a; #4)对所有班级同学英语成绩从高往低排序,结果输出:排名 班级 姓名 分数 SELECT @b:=@b+1 排名,s.`CLASSNO` 班级,s.`NAME` 姓名,s.`GRADE` 分数 FROM stuInfo s ,(SELECT @b:=0) a WHERE s.`COURSE`='English' ORDER BY s.`GRADE` DESC; #创建考试表 CREATE TABLE IF NOT EXISTS exam( examNo INT(10) PRIMARY KEY, `Name` VARCHAR(10), exam1 INT(10), exam2 INT(10) ); #向考试表插入数据 INSERT INTO exam VALUES (1,'A',35,43), (2,'B',78,65), (3,'C',43,48), (4,'A',59,67), (5,'A',84,81), (6,'C',58,59), (7,'B',89,60), (8,'D',59,94), (9,'D',65,75), (10,'D',84,63); /* 5) 考试通过的定义:exam1成绩>=60exam2>=60 每个人有多次考试记录,有一次通过则通过,若全部未通过,则为未通过 无论哪种情况都输出模块1成绩+模块2成绩和成绩最高的记录 查询所有成员最终的考核结果,结果如模板所示写出查询语句 */ #通过的人需要找最好成绩 SELECT d.examNo 考试编号,d.name 姓名,d.exam1 模块1考试成绩,d.exam2 模块2考试成绩, CASE WHEN TRUE THEN '考试通过' END 考试结果 FROM ( SELECT * FROM ( SELECT DISTINCT a.*,a.exam1+a.exam2 su FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过') b) d, ( SELECT MAX(c.su) su,c.name FROM ( SELECT * FROM ( SELECT DISTINCT a.*,a.exam1+a.exam2 su FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过') b) c GROUP BY c.name) e WHERE d.name=e.name AND e.su=d.su ORDER BY d.examNo; #没通过的人需要找最好成绩 SELECT d.examNo 考试编号,d.name 姓名,d.exam1 模块1考试成绩,d.exam2 模块2考试成绩, CASE WHEN TRUE THEN '考试未通过' END 考试结果 FROM ( SELECT *,a.exam1+a.exam2 FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='未通过' AND a.name NOT IN (SELECT DISTINCT a.name FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过' ) GROUP BY a.name HAVING MAX(a.exam1+a.exam2)) d; #联合表 SELECT * FROM (SELECT d.examNo 考试编号,d.name 姓名,d.exam1 模块1考试成绩,d.exam2 模块2考试成绩, CASE WHEN TRUE THEN '考试通过' END 考试结果 FROM ( SELECT * FROM ( SELECT DISTINCT a.*,a.exam1+a.exam2 su FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过') b) d, ( SELECT MAX(c.su) su,c.name FROM ( SELECT * FROM ( SELECT DISTINCT a.*,a.exam1+a.exam2 su FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过') b) c GROUP BY c.name) e WHERE d.name=e.name AND e.su=d.su ORDER BY 考试编号 ASC) a1 UNION SELECT * FROM ( SELECT d.examNo 考试编号,d.name 姓名,d.exam1 模块1考试成绩,d.exam2 模块2考试成绩, CASE WHEN TRUE THEN '考试未通过' END 考试结果 FROM ( SELECT *,a.exam1+a.exam2 FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='未通过' AND a.name NOT IN (SELECT DISTINCT a.name FROM (SELECT e.*,(CASE WHEN e.exam1>=60 AND e.exam2>=60 THEN '通过' ELSE '未通过' END) 考试结果 FROM exam e) a WHERE a.考试结果='通过' ) GROUP BY a.name HAVING MAX(a.exam1+a.exam2)) d ) a2; #创建订单表 CREATE TABLE IF NOT EXISTS orderList( orderId INT(10), orderDate DATETIME, orderPerson VARCHAR(11), orderAddress VARCHAR(11) ); #向订单表插入数据 INSERT INTO orderList VALUES (1001,'2020/4/1 16:58:00','A','中国'), (1002,'2020/5/14 16:58:00','A','美国'), (1003,'2020/5/20 16:58:00','B','英国'), (1004,'2020/6/10 16:58:00','C','中国'), (1005,'2020/4/23 16:58:00','D','法国'), (1006,'2020/6/9 16:58:00','C',NULL); #创建订单明细表 CREATE TABLE IF NOT EXISTS orderProduct( orderId INT(10), productId INT(10), productNum INT(10), productPrice DOUBLE(8,2) ); #插入数据 INSERT INTO orderProduct VALUES (1001,001,5,6), (1001,002,8,23), (1002,003,12,17), (1003,001,3,6), (1003,004,4,36), (1004,005,9,15), (1004,002,14,23), (1005,003,23,17), (1006,003,14,17); #6)查询订单申请时间在2020年4月和5月,各申请人的合计订单总额结果以订单总额降序排序 SELECT c.orderperson 姓名,SUM(c.s) 45俩月总金额 FROM (SELECT o.`orderPerson`,o.`orderId`,b.s FROM (SELECT SUM(a.产品总价) s, a.orderid FROM (SELECT *,p.`productNum`*p.`productPrice` 产品总价 FROM orderProduct p WHERE p.`orderId` IN (SELECT o.`orderId` FROM orderList o WHERE CONVERT(DATE_FORMAT(CONVERT(o.`orderDate`,DATE),'%m'),UNSIGNED INTEGER) IN (4,5))) a GROUP BY a.orderid ) b JOIN orderList o ON o.`orderId`=b.orderid) c GROUP BY c.orderperson ORDER BY 45俩月总金额 DESC; #7)去除英国法国的订单,查询各年月的订单申请总额 SELECT a.时间,a.orderid,SUM(a.productNum*a.productPrice) 订单申请总金额 FROM ( SELECT o.*,p.`productNum`,p.`productPrice`,CONCAT(DATE_FORMAT(o.`orderDate`,'%Y'),'年',DATE_FORMAT(o.`orderDate`,'%m'),'月') 时间 FROM orderList o JOIN orderProduct p ON p.`orderId`=o.`orderId` WHERE o.`orderAddress` NOT IN ('英国','法国') OR o.`orderAddress` IS NULL) a GROUP BY a.时间,a.orderid; #创建销售表 CREATE TABLE IF NOT EXISTS sale( proId VARCHAR(10), saleNum INT(10), saleTime DATETIME ); #插入数据 INSERT INTO sale VALUES ('C1001',15,'2020/6/1 0:00'), ('C1002',26,'2020/5/2 0:00'), ('C1003',21,'2020/4/3 0:00'), ('C1003',23,'2020/4/4 0:00'), ('C1003',0,'2020/3/5 0:00'), ('C1001',16,'2020/2/6 0:00'), ('C1002',32,'2020/1/7 0:00'), ('C1001',16,'2019/12/8 0:00'), ('C1001',32,'2019/6/9 0:00'), ('C1002',17,'2019/5/9 0:00'); #创建产品明细表 CREATE TABLE IF NOT EXISTS product( productId VARCHAR(10) PRIMARY KEY NOT NULL, productName VARCHAR(10), productPrice DOUBLE(8,2) ); #插入数据 INSERT INTO product VALUES ('C1001','产品A',45), ('C1002','产品B',52), ('C1003','产品C',39); #8)查询2020年,每月的销售额,以及同比环比,并按照年月进行升序排序-->同比环比啥玩意?先写个总销售额 #9)查询2020年起,每月的累计销售额-->第八题一样的啊 SELECT 2020,a.月份 月,a.总销售数*p.productprice 总销售额 FROM ( SELECT proid,SUM(salenum) 总销售数,2020,CONVERT(DATE_FORMAT(s.`saleTime`,'%m'),UNSIGNED INTEGER)月份 FROM sale s WHERE DATE_FORMAT(s.`saleTime`,'%Y')=2020 GROUP BY 月份,proid ORDER BY 月份 ) a JOIN product p ON p.`productId`=a.proid ORDER BY;
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