--1.查找到同时存在" 01 "课程和" 02 "课程的情况
时间:2020-01-15 16:59:41
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-- 创建表及数据信息,在第一篇随笔中.
-- 1.通过两张子表得到课程01和课程02的成绩信息,


-- 2.笛卡儿积合并两张表,最后WHERE筛选
SELECT * FROM (SELECT * FROM sc WHERE Cid = 01) AS a,(SELECT * FROM sc WHERE Cid = 02) AS b WHERE a.Sid = b.SId;

-- 3.去除冗余;
SELECT a.Sid,a.score 01_score,b.score 02_score FROM (SELECT * FROM sc WHERE Cid = 01) AS a,(SELECT * FROM sc WHERE Cid = 02) AS b WHERE a.Sid = b.Sid ;

原文:https://www.cnblogs.com/zunfengliu/p/12197312.html
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