sql题目练习 数据库脚本和答案
文章目录
-
- 一、班级、学生、成绩表查询
- 二、学生、课程、成绩、教师表查询
- 三、学生成绩表查询
- 四、成绩表（4_score）查询
- 五、 学生表（student）查询
- 六、 部门员工表（emp、dept）查询
- 七、SQL数据库表名为guest，请简答
- 数据库脚本
不带答案版本，访问这个
https://blog.csdn.net/shuai8624/article/details/116836863
一、班级、学生、成绩表查询
已知以下表结构
班级表（1_class）
| 编号 | 班级名称 |
|---|---|
| id | class_name |
| 1 | 一班 |
| 2 | 二班 |
| 3 | 三班 |
| 4 | 四班 |
| 5 | 五班 |
学生表（1_student）
| 编号 | 学号 | 姓名 | 性别 | 所属班级 |
|---|---|---|---|---|
| id | stu_no | stu_name | stu_gender | class_id |
| 1 | 2020001 | 张三 | 男 | 1 |
| 2 | 2020002 | 李四 | 男 | 1 |
| 3 | 2020003 | 李丽 | 女 | 2 |
| 4 | 2020004 | 赵婷 | 女 | 3 |
| 5 | 2020005 | 王五 | 男 | 3 |
成绩表（1_score）
| 学生 | 语文 | 数学 |
|---|---|---|
| stu_id | chinese | math |
| 1 | 70 | 47 |
| 2 | 80 | 60 |
| 3 | 50 | 82 |
| 4 | 80 | 90 |
业务场景限制：
- 一个班级有多名学生，一名学生只属于一个班级
- 学生有可能没有成绩
题目：
- 查询所有学生的信息（学号，姓名，性别，班级名称）
SELECT
stu_no,
stu_name,
stu_gender,
class_name
FROM
1_student s
JOIN 1_class c ON s.class_id = c.id
-
查询所有人(包括没有成绩的学生)的课程分数（学号，姓名，性别，班级名称，语文分数，数学分数）
SELECT stu_no, stu_name, stu_gender, class_name, se.chiness, se.math FROM 1_student s JOIN 1_class c ON s.class_id = c.id LEFT JOIN 1_score se on se.stu_id = s.id -
查询语文分数比“张三”高的学生（学号，姓名，性别，班级名称，语文分 数）
# 先查询出张三的语文分数，再查比他分数高的 SELECT stu_no, stu_name, stu_gender, class_name, se.chiness FROM 1_student s JOIN 1_class c ON s.class_id = c.id LEFT JOIN 1_score se ON se.stu_id = s.id WHERE se.chiness > ( SELECT se.chiness FROM 1_student s JOIN 1_class c ON s.class_id = c.id LEFT JOIN 1_score se ON se.stu_id = s.id WHERE s.stu_name = '张三' ) -
查询各科都合格（分数>=60）的学生（学号，姓名，语文分数，数学分数）
SELECT stu_no, stu_name, se.chiness, se.math FROM 1_student s LEFT JOIN 1_score se ON se.stu_id = s.id WHERE se.chiness >= 60 AND se.math >= 60 -
查询出所有班级的人数（若没有人，人数显示为0）
SELECT c.class_name, count( s.id ) FROM 1_class c LEFT JOIN 1_student s ON c.id = s.class_id GROUP BY c.id -
查询班级人数>=2的班级（班级编号，班级名称，人数）
SELECT c.class_name, count( s.id ) FROM 1_class c LEFT JOIN 1_student s ON c.id = s.class_id GROUP BY c.id HAVING count( s.id ) >= 2
二、学生、课程、成绩、教师表查询
有以下四个表，表名称及表结构如下：
student(sno,sname,sage,ssex) 学生表
course(cno,cname,tno) 课程表
sc(sno,cno,score) 成绩表
teacher(tno,tname) 教师表
-
查询课程1的成绩比课程2的成绩高的所有学生的信息
# 本题使用表的自连接查询，将sc1表看成课程1的成绩表，将sc2表看成课程2的成绩表，表连接后，再条件查询部分实现这一步，具体为：where sc1.cno=1 and sc2.cno=2 SELECT s.sno, sname, sage, ssex, sc1.score '课程1', sc2.score '课程2' FROM student s JOIN sc sc1 JOIN sc sc2 ON sc1.sno = sc2.sno AND sc1.sno = s.sno WHERE sc1.cno = 1 AND sc2.cno = 2 AND sc1.score > sc2.score -
查询平均成绩大于60分的同学的学号和平均成绩
# 对分组后的数据筛选，使用having SELECT sno, AVG( DISTINCT score ) AS avg_score FROM sc GROUP BY sno HAVING avg_score >= 60 -
查询学过‘李四’老师所教课程的所有同学的学号，姓名
SELECT DISTINCT s.sno, sname FROM teacher t JOIN course c JOIN sc JOIN student s ON t.tno = c.tno AND c.cno = sc.cno AND sc.sno = s.sno WHERE tname = '李四' -
查询姓“李”得老师的个数
SELECT count( tno ) FROM teacher WHERE tname LIKE '李%' -
查询每门课程的选修人数(课程名称，学生数量)–存在没有人选的课程
SELECT c.cname, count( sc.cno ) FROM sc JOIN course c ON c.cno = sc.cno GROUP BY sc.cno -
删除“1002”同学的“1”课程的成绩
delete from sc where sno='1002' and cno=1 -
查询选修人数最多的课程（课程id，课程名称，学生数量）–考虑有多门课程都是选修最多的情况
1. 查询出每个课程的选修人数，找到最大的选修人数 select count(sno) cou from sc group by cno order by cou desc limit 1; 2. 写主sql -- 查询选修人数值=max的课程信息 SELECT c.cno, cname, count( sno ) cou FROM course c JOIN sc ON c.cno = sc.cno GROUP BY sc.cno HAVING cou = ( SELECT count( sno ) cou FROM sc GROUP BY cno ORDER BY cou DESC LIMIT 1 )
三、学生成绩表查询
下面是学生成绩表(3_student_score)结构说明
| 字段名称 | 字段解释 | 字段类型 | 字段长度 |
|---|---|---|---|
| student_id | 学号 | 字符 | 8 |
| student_name | 姓名 | 字符 | 50 |
| student_gender | 性别 | 字符(男/女) | 4 |
| course_id | 课程号 | 字符 | 5 |
| score | 分数 | 数值 | 3 |
| ismakeup | 当前考试是否为补考 | 字符(补考:1;非补考:0) | 2 |
下面是课程表(3_course)说明
| 字段名称 | 字段解释 | 字段类型 | 字段长度 | 约束 |
|---|---|---|---|---|
| course_id | 课程号 | 字符 | 5 | PK |
| course_name | 课程名 | 字符 | 30 | Not null |
| course_desc | 课程介绍 | 字符 | 60 |
1、查找第一次考试后所有需要补考(小于60分)的学生姓名和这门课程的名称和成绩；
SELECT
s.student_name,
c.course_name,
s.score
FROM
3_student_score s
JOIN 3_course c ON c.course_id = s.course_id
WHERE
s.ismakeup = 0
AND s.score < 60
2、查询每个学生第一次考试后需要补考(小于60分)的课程平均分和科目数
SELECT
student_name,
avg( score ),
count( course_id )
FROM
3_student_score
WHERE
ismakeup = 0
AND score < 60
GROUP BY
student_id
3、查询所有参加了补考的学生的学生姓名，课程名称，补考成绩和非补考成绩；
SELECT
s.student_name,
c.course_name,
s.score AS '非补考成绩',
s2.score '补考成绩'
FROM
3_student_score s
JOIN 3_student_score s2 ON s.course_id = s2.course_id
AND s.student_id = s2.student_id
JOIN 3_course c ON c.course_id = s.course_id
WHERE
s.ismakeup = 0
AND s2.ismakeup = 1
四、成绩表（4_score）查询
有一张表4_score，三个字段名，姓名，课程，分数，数据如下，请写一条sql语句，查询出每门课程都大于等于80分的学生信息
| name | course | score |
|---|---|---|
| 张三 | 语文 | 81 |
| 张三 | 数学 | 75 |
| 李四 | 语文 | 76 |
| 王五 | 语文 | 81 |
| 王五 | 数学 | 100 |
| 王五 | 英语 | 90 |
SELECT NAME
FROM
`score`
GROUP BY
NAME
HAVING
min( score ) >= 80
五、 学生表（student）查询
有一张表5_student，包括字段id和name，请写一条sql语句，将表中name字段中重复的记录删除，只保留重复数据中的id最大的那一条数据。
| id | name |
|---|---|
| 1 | 张三 |
| 2 | 张三 |
| 3 | 李四 |
| 4 | 王五 |
| 5 | 王五 |
| 6 | 王五 |
要求只留下：2 张三， 3 李四， 6 王五 这三条记录
SELECT max(id), name FROM `5_student` GROUP BY name
六、 部门员工表（emp、dept）查询
emp:
empno int(员工编号) ,ename varchar(50)(员工姓名) ，job varchar(100) （工作岗位），mgr int (上级领导编号)，hiredate date（雇佣日期），sal int（薪金），comm int（佣金） deptno int (部门编号)
提示：工资=薪金+佣金
dept表：
deptno int (部门编号) ， dname 部门名称 loc 地点
-
列出在每个部门工作的员工数量，平均工资
SELECT d.dname, count(e.empno), avg( e.sal + e.comm ) FROM emp e right JOIN dept d ON d.deptno = e.deptno GROUP BY d.deptno -
列出所有员工的姓名，部门名称和工资
SELECT e.ename, d.dname, e.sal + e.comm FROM emp e INNER JOIN dept d ON d.deptno = e.deptno -
列出所有部门的详细信息和部门人数
SELECT d.*, count( e.empno ) '部门人数' FROM emp e INNER JOIN dept d ON d.deptno = e.deptno GROUP BY d.deptno -
列出各种工作的最低工资
SELECT e.job, min( e.sal + e.comm ) '工资' FROM emp e GROUP BY e.job -
列出各个部门的manager的最低薪金(若是manager,其job的值为manageer)
SELECT e.ename, e.job, min( e.sal + e.comm ) '工资' FROM emp e WHERE e.job = 'manageer' GROUP BY deptno -
列出受雇日期早于其直接上级的所有员工
# join自己 行转列 SELECT e1.* FROM emp e1 JOIN emp e2 ON e1.mgr = e2.empno WHERE e1.hiredate < e2.hiredate -
列出部门名称和这些部门的员工信息，同时列出那些没有员工的部门
# 因为要列出那些没有员工的部门，所以部门为主表 左连接 SELECT d.dname, e.* FROM dept d LEFT JOIN emp e ON d.deptno = e.deptno ORDER BY d.deptno -
列出所有‘clerk’（办事员）岗位的姓名以及部门名称
SELECT e.ename, d.dname FROM emp e JOIN dept d ON d.deptno = e.deptno WHERE e.job = 'clerk' -
列出最低薪金大于6500的各种工作
SELECT job FROM emp GROUP BY job HAVING min( sal + comm ) > 6500 -
列出在研发部工作的员工的姓名，假定不知道研发部的部门编号
SELECT e.*, d.dname FROM emp e JOIN dept d ON d.deptno = e.deptno WHERE d.dname = '研发部'
七、SQL数据库表名为guest，请简答
| 账号 | 消费 | 时间 | 金额 | 班次 |
|---|---|---|---|---|
| accounts | details | date | money | class |
| s0001 | 房费 | 2020-01-01 | 280 | 001 |
| s0001 | 酒水 | 2020-01-02 | 120 | 001 |
| s0001 | 房费 | 2020-01-08 | 300 | 003 |
| s0002 | 酒水 | 2020-01-29 | 50 | |
| s0003 | 房费 | 2020-01-31 | 180 | 002 |
| s0004 | 房费 | 2020-02-01 | 230 | 001 |
| s0005 | 酒水 | 2020-02-01 | 100 | |
| s0005 | 房费 | 2020-02-02 | 128 | 001 |
-
查询出房费都大于200的账号
# 先过滤出所有的房费记录，然后按账号分组，取 最小的金额 >200 的账号 SELECT accounts FROM guest WHERE details = '房费' GROUP BY accounts HAVING min( money ) > 200 -
查询出1月份每个账号酒水和房费的总金额
查找某段时间内的数据，可以使用between ..and..,也可以使用> < 数据库中between..and..会涉及到2个边界值，2个边界值都包括 select accounts,sum(money) from guest where date between '2020-01-01' and '2020-01-31' group by accounts -
将不是房费的班次都更改为‘001’
update guest set class='001' where details!='房费' -
查询出消费都大于100的账号
# 思路：查询出每个账户的最低消费金额，判断最低消费金额是否大于100 # 步骤：1. 求每个账户的最低消费金额 select accounts from guest group by accounts having min(money)>100
数据库脚本
-- ----------------------------
-- Table structure for 1_class
-- ----------------------------
DROP TABLE IF EXISTS `1_class`;
CREATE TABLE `1_class` (
`id` int(11) NOT NULL,
`class_name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of 1_class
-- ----------------------------
INSERT INTO `1_class` VALUES (1, '一班');
INSERT INTO `1_class` VALUES (2, '二班');
INSERT INTO `1_class` VALUES (3, '三班');
INSERT INTO `1_class` VALUES (4, '四班');
INSERT INTO `1_class` VALUES (5, '五班');
-- ----------------------------
-- Table structure for 1_score
-- ----------------------------
DROP TABLE IF EXISTS `1_score`;
CREATE TABLE `1_score` (
`stu_id` int(11) NULL DEFAULT NULL,
`chiness` int(3) NULL DEFAULT NULL,
`math` int(3) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of 1_score
-- ----------------------------
INSERT INTO `1_score` VALUES (1, 70, 47);
INSERT INTO `1_score` VALUES (2, 80, 60);
INSERT INTO `1_score` VALUES (3, 50, 82);
INSERT INTO `1_score` VALUES (4, 80, 90);
-- ----------------------------
-- Table structure for 1_student
-- ----------------------------
DROP TABLE IF EXISTS `1_student`;
CREATE TABLE `1_student` (
`id` int(11) NOT NULL,
`stu_no` int(11) NULL DEFAULT NULL,
`stu_name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`stu_gender` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`class_id` int(11) NULL DEFAULT NULL,
PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of 1_student
-- ----------------------------
INSERT INTO `1_student` VALUES (1, 2020001, '张三', '男', 1);
INSERT INTO `1_student` VALUES (2, 2020002, '李四', '男', 1);
INSERT INTO `1_student` VALUES (3, 2020003, '李丽', '女', 2);
INSERT INTO `1_student` VALUES (4, 2020004, '赵婷', '女', 3);
INSERT INTO `1_student` VALUES (5, 2020005, '王五', '男', 3);
-- ----------------------------
-- Table structure for 3_course
-- ----------------------------
DROP TABLE IF EXISTS `3_course`;
CREATE TABLE `3_course` (
`course_id` int(11) NOT NULL COMMENT '课程号',
`course_name` varchar(30) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL COMMENT '课程名',
`course_desc` varchar(60) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '课程介绍',
PRIMARY KEY (`course_id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of 3_course
-- ----------------------------
INSERT INTO `3_course` VALUES (1, 'Java基础', NULL);
INSERT INTO `3_course` VALUES (2, 'Java高级', NULL);
INSERT INTO `3_course` VALUES (3, 'C#', NULL);
INSERT INTO `3_course` VALUES (4, '计算机网络', NULL);
INSERT INTO `3_course` VALUES (5, '算法与结构', NULL);
INSERT INTO `3_course` VALUES (6, '数字模拟电子', NULL);
-- ----------------------------
-- Table structure for 3_student_score
-- ----------------------------
DROP TABLE IF EXISTS `3_student_score`;
CREATE TABLE `3_student_score` (
`student_id` varchar(8) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '学号',
`student_name` varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '姓名',
`student_gender` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '性别',
`course_id` int(11) NULL DEFAULT NULL COMMENT '课程号',
`score` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '分数',
`ismakeup` int(11) NULL DEFAULT NULL COMMENT '是否为补考(补考:1;非补考:0)'
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of 3_student_score
-- ----------------------------
INSERT INTO `3_student_score` VALUES ('1', '张三', '男', 1, '80', 1);
INSERT INTO `3_student_score` VALUES ('1', '张三', '男', 1, '50', 0);
INSERT INTO `3_student_score` VALUES ('1', '张三', '男', 2, '50', 0);
INSERT INTO `3_student_score` VALUES ('1', '张三', '男', 3, '80', 0);
INSERT INTO `3_student_score` VALUES ('3', '王二', '女', 1, '20', 0);
INSERT INTO `3_student_score` VALUES ('1', '张三', '男', 2, '90', 1);
INSERT INTO `3_student_score` VALUES ('3', '王二', '女', 1, '50', 1);
INSERT INTO `3_student_score` VALUES ('4', '李四', '男', 4, '50', 0);
-- ----------------------------
-- Table structure for 5_student
-- ----------------------------
DROP TABLE IF EXISTS `5_student`;
CREATE TABLE `5_student` (
`id` int(11) NOT NULL,
`name` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
PRIMARY KEY (`id`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of 5_student
-- ----------------------------
INSERT INTO `5_student` VALUES (1, '张三');
INSERT INTO `5_student` VALUES (2, '张三');
INSERT INTO `5_student` VALUES (3, '李四');
INSERT INTO `5_student` VALUES (4, '王五');
INSERT INTO `5_student` VALUES (5, '王五');
INSERT INTO `5_student` VALUES (6, '王五');
-- ----------------------------
-- Table structure for course
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
`cno` int(11) NOT NULL,
`cname` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`tno` int(11) NULL DEFAULT NULL,
PRIMARY KEY (`cno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of course
-- ----------------------------
INSERT INTO `course` VALUES (1, 'Java基础', 1);
INSERT INTO `course` VALUES (2, 'Java高级', 1);
INSERT INTO `course` VALUES (3, 'C#', 2);
INSERT INTO `course` VALUES (4, '计算机网络', 2);
INSERT INTO `course` VALUES (5, '算法与结构', 4);
INSERT INTO `course` VALUES (6, '数字模拟电子', 3);
-- ----------------------------
-- Table structure for dept
-- ----------------------------
DROP TABLE IF EXISTS `dept`;
CREATE TABLE `dept` (
`deptno` int(11) NOT NULL COMMENT '部门编号',
`dname` varchar(100) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`loc` varchar(100) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '地点',
PRIMARY KEY (`deptno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of dept
-- ----------------------------
INSERT INTO `dept` VALUES (1, '研发部', NULL);
INSERT INTO `dept` VALUES (2, '部门2', NULL);
INSERT INTO `dept` VALUES (3, '部门3', NULL);
INSERT INTO `dept` VALUES (4, '部门4', NULL);
-- ----------------------------
-- Table structure for emp
-- ----------------------------
DROP TABLE IF EXISTS `emp`;
CREATE TABLE `emp` (
`empno` int(11) NOT NULL COMMENT '员工编号',
`ename` varchar(50) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '员工姓名',
`job` varchar(100) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '工作岗位',
`mgr` int(11) NULL DEFAULT NULL COMMENT '上级领导编号',
`hiredate` date NULL DEFAULT NULL COMMENT '雇佣日期',
`sal` int(11) NULL DEFAULT NULL COMMENT '薪金',
`comm` int(11) NULL DEFAULT NULL COMMENT '佣金',
`deptno` int(11) NULL DEFAULT NULL COMMENT '部门编号',
PRIMARY KEY (`empno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of emp
-- ----------------------------
INSERT INTO `emp` VALUES (1, '员工1', 'clerk', 3, '2021-05-11', 1100, 100, 1);
INSERT INTO `emp` VALUES (2, '员工2', '2', 3, '2021-05-12', 1200, 200, 1);
INSERT INTO `emp` VALUES (3, '领导3', 'manageer', NULL, '2021-05-12', 6000, 600, 1);
INSERT INTO `emp` VALUES (4, '员工4', 'clerk', 5, '2021-05-14', 1400, 400, 2);
INSERT INTO `emp` VALUES (5, '领导5', 'manageer', 6, '2021-05-13', 6300, 600, 2);
INSERT INTO `emp` VALUES (6, '领导6', 'manageer', NULL, '2021-05-14', 7300, 700, 2);
-- ----------------------------
-- Table structure for guest
-- ----------------------------
DROP TABLE IF EXISTS `guest`;
CREATE TABLE `guest` (
`accounts` varchar(10) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '账号',
`details` varchar(30) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL COMMENT '消费',
`date` date NULL DEFAULT NULL COMMENT '时间',
`money` int(11) NULL DEFAULT NULL COMMENT '金额',
`class` int(11) NULL DEFAULT NULL COMMENT '班次'
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of guest
-- ----------------------------
INSERT INTO `guest` VALUES ('s0001', '房费', '2020-01-01', 280, 1);
INSERT INTO `guest` VALUES ('s0001', '酒水', '2020-01-02', 120, 1);
INSERT INTO `guest` VALUES ('s0001', '房费', '2020-01-08', 300, 3);
INSERT INTO `guest` VALUES ('s0002', '酒水', '2020-01-29', 50, NULL);
INSERT INTO `guest` VALUES ('s0003', '房费', '2020-01-31', 180, 2);
INSERT INTO `guest` VALUES ('s0004', '房费', '2020-02-01', 230, 1);
INSERT INTO `guest` VALUES ('s0005', '酒水', '2020-02-01', 100, NULL);
INSERT INTO `guest` VALUES ('s0005', '房费', '2020-02-02', 128, 1);
-- ----------------------------
-- Table structure for sc
-- ----------------------------
DROP TABLE IF EXISTS `sc`;
CREATE TABLE `sc` (
`sno` int(11) NOT NULL,
`cno` int(11) NULL DEFAULT NULL,
`score` int(255) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of sc
-- ----------------------------
INSERT INTO `sc` VALUES (1001, 1, 90);
INSERT INTO `sc` VALUES (1001, 2, 80);
INSERT INTO `sc` VALUES (1001, 3, 60);
INSERT INTO `sc` VALUES (1002, 1, 56);
INSERT INTO `sc` VALUES (1002, 2, 90);
INSERT INTO `sc` VALUES (1002, 4, 66);
INSERT INTO `sc` VALUES (1003, 1, 90);
INSERT INTO `sc` VALUES (1003, 2, 70);
INSERT INTO `sc` VALUES (1003, 4, 60);
INSERT INTO `sc` VALUES (1003, 3, 80);
-- ----------------------------
-- Table structure for score
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
`name` varchar(200) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`course` varchar(200) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`score` int(200) NULL DEFAULT NULL
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Dynamic;
-- ----------------------------
-- Records of score
-- ----------------------------
INSERT INTO `score` VALUES ('张三', '语文', 81);
INSERT INTO `score` VALUES ('张三', '数学', 75);
INSERT INTO `score` VALUES ('李四', '语文', 76);
INSERT INTO `score` VALUES ('王五', '语文', 81);
INSERT INTO `score` VALUES ('王五', '数学', 100);
INSERT INTO `score` VALUES ('王五', '英语', 90);
-- ----------------------------
-- Table structure for student
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
`sno` int(11) NOT NULL,
`sname` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
`sage` int(255) NULL DEFAULT NULL,
`ssex` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
PRIMARY KEY (`sno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of student
-- ----------------------------
INSERT INTO `student` VALUES (1001, 'tom', 22, '男');
INSERT INTO `student` VALUES (1002, 'jack', 23, '男');
INSERT INTO `student` VALUES (1003, 'rose', 24, '女');
-- ----------------------------
-- Table structure for teacher
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
`tno` int(11) NOT NULL,
`tname` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NULL DEFAULT NULL,
PRIMARY KEY (`tno`) USING BTREE
) ENGINE = InnoDB CHARACTER SET = utf8 COLLATE = utf8_general_ci ROW_FORMAT = Compact;
-- ----------------------------
-- Records of teacher
-- ----------------------------
INSERT INTO `teacher` VALUES (1, '张力');
INSERT INTO `teacher` VALUES (2, '李四');
INSERT INTO `teacher` VALUES (3, '王强');
INSERT INTO `teacher` VALUES (4, '李晓燕');