一、造数据
DROP TABLE IF EXISTS sc;
CREATE TABLE sc (
stu_no VARCHAR(4) COMMENT '学号',
class_no VARCHAR(4) COMMENT '课程号',
grade INT(2) COMMENT '成绩'
);
DROP TABLE IF EXISTS class;
CREATE TABLE class (
class_no VARCHAR(4) COMMENT '课程号',
class_name VARCHAR(8) COMMENT '课程名称'
);
INSERT INTO sc VALUES
('0001','C001',89),
('0001','C002',85),
('0001','C003',99),
('0002','C001',78),
('0002','C002',83),
('0002','C003',86),
('0002','C004',77),
('0003','C001',88),
('0003','C003',68),
('0003','C004',55);
INSERT INTO class VALUES
('C001','语文'),
('C002','数学'),
('C003','英语'),
('C004','科学');
二、查看数据结构
SELECT * FROM sc;
SELECT * FROM class;
三、需求 （查询出每个学生每一们课程的分数，如果该学生在数据库中没有那门课的分数，则那门课分数置为0）
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1.使用 cross join 对两张表做笛卡尔积。
SELECT * FROM sc CROSS JOIN class+--------+----------+-------+----------+------------+ | stu_no | class_no | grade | class_no | class_name | +--------+----------+-------+----------+------------+ | 0001 | C001 | 89 | C001 | 语文 | | 0001 | C001 | 89 | C002 | 数学 | | 0001 | C001 | 89 | C003 | 英语 | | 0001 | C001 | 89 | C004 | 科学 | | 0001 | C002 | 85 | C001 | 语文 | | 0001 | C002 | 85 | C002 | 数学 | | 0001 | C002 | 85 | C003 | 英语 | | 0001 | C002 | 85 | C004 | 科学 | | 0001 | C003 | 99 | C001 | 语文 | | 0001 | C003 | 99 | C002 | 数学 | | 0001 | C003 | 99 | C003 | 英语 | | 0001 | C003 | 99 | C004 | 科学 | | 0002 | C001 | 78 | C001 | 语文 | | 0002 | C001 | 78 | C002 | 数学 | | 0002 | C001 | 78 | C003 | 英语 | | 0002 | C001 | 78 | C004 | 科学 | | 0002 | C002 | 83 | C001 | 语文 | | 0002 | C002 | 83 | C002 | 数学 | | 0002 | C002 | 83 | C003 | 英语 | | 0002 | C002 | 83 | C004 | 科学 | | 0002 | C003 | 86 | C001 | 语文 | | 0002 | C003 | 86 | C002 | 数学 | | 0002 | C003 | 86 | C003 | 英语 | | 0002 | C003 | 86 | C004 | 科学 | | 0002 | C004 | 77 | C001 | 语文 | | 0002 | C004 | 77 | C002 | 数学 | | 0002 | C004 | 77 | C003 | 英语 | | 0002 | C004 | 77 | C004 | 科学 | | 0003 | C001 | 88 | C001 | 语文 | | 0003 | C001 | 88 | C002 | 数学 | | 0003 | C001 | 88 | C003 | 英语 | | 0003 | C001 | 88 | C004 | 科学 | | 0003 | C003 | 68 | C001 | 语文 | | 0003 | C003 | 68 | C002 | 数学 | | 0003 | C003 | 68 | C003 | 英语 | | 0003 | C003 | 68 | C004 | 科学 | | 0003 | C004 | 55 | C001 | 语文 | | 0003 | C004 | 55 | C002 | 数学 | | 0003 | C004 | 55 | C003 | 英语 | | 0003 | C004 | 55 | C004 | 科学 | +--------+----------+-------+----------+------------+ -
2. 将上表中两个class_no 不相等的数据行，grade 置为0
SELECT stu_no, t1.class_no, t2.class_no, CASE WHEN t1.class_no=t2.class_no THEN grade ELSE 0 END AS grade, class_name FROM sc AS t1 CROSS JOIN class AS t2+--------+----------+----------+-------+------------+ | stu_no | class_no | class_no | grade | class_name | +--------+----------+----------+-------+------------+ | 0001 | C001 | C001 | 89 | 语文 | | 0001 | C001 | C002 | 0 | 数学 | | 0001 | C001 | C003 | 0 | 英语 | | 0001 | C001 | C004 | 0 | 科学 | | 0001 | C002 | C001 | 0 | 语文 | | 0001 | C002 | C002 | 85 | 数学 | | 0001 | C002 | C003 | 0 | 英语 | | 0001 | C002 | C004 | 0 | 科学 | | 0001 | C003 | C001 | 0 | 语文 | | 0001 | C003 | C002 | 0 | 数学 | | 0001 | C003 | C003 | 99 | 英语 | | 0001 | C003 | C004 | 0 | 科学 | | 0002 | C001 | C001 | 78 | 语文 | | 0002 | C001 | C002 | 0 | 数学 | | 0002 | C001 | C003 | 0 | 英语 | | 0002 | C001 | C004 | 0 | 科学 | | 0002 | C002 | C001 | 0 | 语文 | | 0002 | C002 | C002 | 83 | 数学 | | 0002 | C002 | C003 | 0 | 英语 | | 0002 | C002 | C004 | 0 | 科学 | | 0002 | C003 | C001 | 0 | 语文 | | 0002 | C003 | C002 | 0 | 数学 | | 0002 | C003 | C003 | 86 | 英语 | | 0002 | C003 | C004 | 0 | 科学 | | 0002 | C004 | C001 | 0 | 语文 | | 0002 | C004 | C002 | 0 | 数学 | | 0002 | C004 | C003 | 0 | 英语 | | 0002 | C004 | C004 | 77 | 科学 | | 0003 | C001 | C001 | 88 | 语文 | | 0003 | C001 | C002 | 0 | 数学 | | 0003 | C001 | C003 | 0 | 英语 | | 0003 | C001 | C004 | 0 | 科学 | | 0003 | C003 | C001 | 0 | 语文 | | 0003 | C003 | C002 | 0 | 数学 | | 0003 | C003 | C003 | 68 | 英语 | | 0003 | C003 | C004 | 0 | 科学 | | 0003 | C004 | C001 | 0 | 语文 | | 0003 | C004 | C002 | 0 | 数学 | | 0003 | C004 | C003 | 0 | 英语 | | 0003 | C004 | C004 | 55 | 科学 | +--------+----------+----------+-------+------------+ -
3.观察上一个步骤中的数据不难发现，我们已经将每个学生缺失的课程分数用0补全，但是原本就正常的数据又重复了一些课程为0的数据，所以需要去重，去重无非就是用group by 或者 distinct ,但是distinct 不可控制，所以我们这里选用 group by 来去重得出我们想要的数据。
SELECT stu_no, t2.class_no, SUM(CASE WHEN t1.class_no=t2.class_no THEN grade ELSE 0 END ) AS grade, class_name FROM sc AS t1 CROSS JOIN class AS t2 GROUP BY stu_no,t2.class_no,class_name;
