User-Defined Variables

 1 mysql> SET @w := SELECT COUNT(*) FROM course WHERE cteacher=‘程军‘;
 2 ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version
 3  the right syntax to use near ‘SELECT COUNT(*) FROM course WHERE cteacher=‘程军‘‘ at line 1
 4 mysql> SELECT @w;
 5 +------+
 6 | @w   |
 7 +------+
 8 | NULL |
 9 +------+
10 1 row in set (0.00 sec)
11
12 mysql> SET @w := (SELECT COUNT(*) FROM course WHERE cteacher=‘程军‘);
13 Query OK, 0 rows affected (0.00 sec)
14
15 mysql> SELECT @w;
16 +------+
17 | @w   |
18 +------+
19 |    2 |
20 +------+
21 1 row in set (0.00 sec)
22
23 mysql>

 1 SHOW TABLES;
 2 DROP TABLE IF EXISTS student;
 3 SHOW TABLES;
 4 CREATE TABLE IF NOT EXISTS student
 5 (
 6     sno int primary key,
 7     sname VARCHAR(20) CHARACTER SET gbk,
 8     sage TINYINT,
 9     shair ENUM(‘黑‘,‘白‘) CHARACTER SET gbk
10 );
11 INSERT INTO student VALUES(‘1‘,‘李强‘,23,‘黑‘);
12 INSERT INTO student VALUES(‘2‘,‘刘丽‘,22,‘白‘);
13 INSERT INTO student VALUES(‘5‘,‘张友‘,22,‘黑‘);
14 SELECT * FROM student;
15
16 SHOW TABLES;
17 DROP TABLE IF EXISTS course;
18 SHOW TABLES;
19 CREATE TABLE IF NOT EXISTS course
20 (
21     cno VARCHAR(10) primary key,
22     cname VARCHAR(20) CHARACTER SET gbk,
23     cteacher  VARCHAR(20) CHARACTER SET gbk
24 );
25 INSERT INTO course VALUES(‘k1‘,‘c语言‘,‘王华‘);
26 INSERT INTO course VALUES(‘k5‘,‘数据库原理‘,‘程军‘);
27 INSERT INTO course VALUES(‘k8‘,‘编译原理‘,‘程军‘);
28 SELECT * FROM course;
29
30 SHOW TABLES;
31 DROP TABLE IF EXISTS sc;
32 SHOW TABLES;
33 CREATE TABLE IF NOT EXISTS sc
34 (
35     fk_sno INT NOT NULL,
36     fk_cno VARCHAR(10) NOT NULL,
37     score INT NOT NULL,
38     PRIMARY KEY(fk_sno,fk_cno),
39     FOREIGN KEY(fk_sno) REFERENCES student(sno),
40     FOREIGN KEY(fk_cno) REFERENCES course(cno)
41 );
42 INSERT INTO sc VALUES(‘1‘,‘K1‘,83);
43 INSERT INTO sc VALUES(‘2‘,‘K1‘,85);
44 INSERT INTO sc VALUES(‘5‘,‘K1‘,92);
45 INSERT INTO sc VALUES(‘2‘,‘K5‘,90);
46 INSERT INTO sc VALUES(‘5‘,‘K5‘,84);
47 INSERT INTO sc VALUES(‘5‘,‘K8‘,80);
48 SELECT * FROM sc;

1-检索至少选修＂程军＂老师所授全部课程的学生姓名（SNAME）
检索选修了“程军”老师所授全部课程的学生姓名

0-检索选修了“程军”老师所授课程的学生

1 SELECT * FROM sc RIGHT JOIN course ON  sc.fk_cno=course.cno;
2 SELECT * FROM (SELECT * FROM sc RIGHT JOIN course ON  sc.fk_cno=course.cno) w RIGHT JOIN student ON w.fk_sno=student.sno;
3 SELECT * FROM (SELECT * FROM sc RIGHT JOIN course ON  sc.fk_cno=course.cno) w RIGHT JOIN student ON w.fk_sno=student.sno WHERE cteacher=‘程军‘;

集合子集，逐步精确，缩小范围 1-是学生-2-选程老师的课-3-选了程老师所有的课

/*

假定

“ 某学生选程老师的课程门数等于程老师所开门数”等价于“该学生至少选了程老师的所有的课程”。

*/

 1             SELECT sname FROM student
 2             WHERE
 3             (
 4                 SELECT COUNT(*) FROM
 5                 (
 6                     SELECT  * FROM (SELECT * FROM sc RIGHT JOIN course ON  sc.fk_cno=course.cno) w RIGHT JOIN student ON w.fk_sno=student.sno
 7                     WHERE cteacher = ‘程军‘
 8                 ) wb
 9                 WHERE wb.sname = student.sname
10             )
11             =
12             (SELECT COUNT(*) FROM course WHERE cteacher=‘程军‘);

/* 0-澄清问题：找出至少选了程老师全部课程的学生：学生-没有选程老师的课+学生选了程老师的部分的课+学生选了程老师全部的课

等价转化：找出选了程老师部分的课或者没有选程老师的课的学生

数据关系：学生和老师没有数据关系，是两个完全独立的数据对象，分数是桥梁

等价转化：

　　选了部分程老师的课 ，NOT EXISTS (SELECT * FROM sc WHERE sc.sno = student AND sc.cno=course.cno)->FALSE (内层SELECT不空，TRUE,退出)

　　没有选程老师的课 teacher=‘程军‘->FALSE ,后边的不用看了(中层SELECT空，FALSE，退出)

　　外中内-三层SELECT，外层返回的是最终的结果

　　最外层选得的最终量result，学生姓名，它的措施是将各个学生名称带入，逐个检测

1-程序表达： */

 1 SELECT sname FROM student
 2 WHERE NOT EXISTS
 3 (
 4    SELECT * FROM course
 5    WHERE cteacher=‘程军‘ AND NOT EXISTS
 6    (
 7        SELECT * FROM sc
 8        WHERE sc.fk_sno = student.sno AND sc.fk_cno = course.cno
 9     )
10 ) ;

http://blog.csdn.net/wolfofsiberian/article/details/39346781

发问:

0-语言只是工具而已，不是解决问题方法的源泉，怎么减少对接语法API的时间？