数据库SQL学习的经典案例：学生专业老师分数表的 *** 练

概述数据库SQL学习的经典案例：学生专业老师分数表的 *** 练

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MysqL> desc SC;+-------+---------------+------+-----+---------+-------+| FIEld | Type          | Null | Key | Default | Extra |+-------+---------------+------+-----+---------+-------+| S_ID  | varchar(10)   | YES  |     | NulL    |       || C_ID  | varchar(10)   | YES  |     | NulL    |       || score | decimal(18,1) | YES  |     | NulL    |       |+-------+---------------+------+-----+---------+-------+3 rows in set (0.00 sec)MysqL> select * from SC;+------+------+-------+| S_ID | C_ID | score |+------+------+-------+| 01   | 01   |  80.0 || 01   | 02   |  90.0 || 01   | 03   |  99.0 || 02   | 01   |  70.0 || 02   | 02   |  60.0 || 02   | 03   |  80.0 || 03   | 01   |  80.0 || 03   | 02   |  80.0 || 03   | 03   |  80.0 || 04   | 01   |  50.0 || 04   | 02   |  30.0 || 04   | 03   |  20.0 || 05   | 01   |  76.0 || 05   | 02   |  87.0 || 06   | 01   |  31.0 || 06   | 03   |  34.0 || 07   | 02   |  89.0 || 07   | 03   |  98.0 || 08   | 04   |  79.0 || 11   | 03   |  77.9 || 12   | 02   |  47.9 || 12   | 04   |  47.9 || 11   | 01   |  77.9 || 01   | 04   |  73.9 || 01   | 05   |  83.9 || 06   | 04   |  75.0 || 06   | 05   |  85.0 || 11   | 05   |  81.0 || 11   | 04   |  91.0 |+------+------+-------+29 rows in set (0.00 sec)--1select B.* from SC B where B.C_ID = '01' and not exists(select * from SC B2 where B.S_ID = B2.S_ID and B2.C_ID = '02');--2select * from (select B.S_ID,B.C_ID aC_ID,B.score ascore,B2.C_ID bC_ID,B2.score bscore from SC B inner join SC B2 on B.S_ID = B2.S_ID and B.C_ID = '01' and B2.C_ID != '01') BBB where not exists (select * from (select B.S_ID,B2.score bscore from SC B inner join SC B2 on B.S_ID = B2.S_ID and B.C_ID = '01' and B2.C_ID != '01') CCC where BBB.S_ID = CCC.S_ID and CCC.bC_ID = '02' );--3select * from SC where C_ID = '01' and S_ID not in (select distinct S_ID from SC where C_ID = '02');--4select B.S_ID,B2.score bscore from SC B inner join SC B2 on B.S_ID = B2.S_ID and B.C_ID = '01' and B2.C_ID != '01' group by S_ID having (aC_ID,bC_ID) != ('01','02');--5select B.*,B2.* from SC B left join SC B2 on B.S_ID = B2.S_ID and B2.C_ID = '02' where B.C_ID = '01' and B2.C_ID is null;--6select B.*,B2.* from SC B right join SC B2 on B.S_ID = B2.S_ID and B.C_ID = '02' where B2.C_ID = '01' and B.C_ID is null;--7select B.S_ID,B2.score bscore from SC B inner join SC B2 on B.S_ID = B2.S_ID and B.C_ID = '01' and B2.C_ID != '01' group by S_ID having nullif(bC_ID,'02') >> 1;

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