网上有很多SQL笔试经典50题及答案解析，并且答案和详解也有很多了，这里只是从巩固知识的角度从多方位想出解法，也算是将SQL知识进行系统的巩固，在不考虑运行效率的基础上，对于大部分题目本文将提供不同的解法。本文尽量基于标准SQL语句进行编写。
首先建立四张表：学生信息表 student、课程信息表 course、学生成绩信息表sc、教师信息表 teacher。

create table Student(sid varchar(10),sname varchar(10),sage datetime,ssex nvarchar(10));
insert into Student values('01' , '赵雷' , '1990-01-01' , '男');
insert into Student values('02' , '钱电' , '1990-12-21' , '男');
insert into Student values('03' , '孙风' , '1990-05-20' , '男');
insert into Student values('04' , '李云' , '1990-08-06' , '男');
insert into Student values('05' , '周梅' , '1991-12-01' , '女');
insert into Student values('06' , '吴兰' , '1992-03-01' , '女');
insert into Student values('07' , '郑竹' , '1989-07-01' , '女');
insert into Student values('08' , '王菊' , '1990-01-20' , '女');create table Course(cid varchar(10),cname varchar(10),tid varchar(10));
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');create table Teacher(tid varchar(10),tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');create table SC(sid varchar(10),cid varchar(10),score decimal(18,1));
insert into SC values('01' , '01' , 80);
insert into SC values('01' , '02' , 90);
insert into SC values('01' , '03' , 99);
insert into SC values('02' , '01' , 70);
insert into SC values('02' , '02' , 60);
insert into SC values('02' , '03' , 80);
insert into SC values('03' , '01' , 80);
insert into SC values('03' , '02' , 80);
insert into SC values('03' , '03' , 80);
insert into SC values('04' , '01' , 50);
insert into SC values('04' , '02' , 30);
insert into SC values('04' , '03' , 20);
insert into SC values('05' , '01' , 76);
insert into SC values('05' , '02' , 87);
insert into SC values('06' , '01' , 31);
insert into SC values('06' , '03' , 34);
insert into SC values('07' , '02' , 89);
insert into SC values('07' , '03' , 98);

1、查询“01”课程比“02”课程成绩高的所有学生的学号；

-- 关联子查询
SELECT sid FROM sc AWHERE cid='01'AND A.score>(SELECT scoreFROM sc BWHERE cid='02'AND A.sid=B.sid);

-- 联结
SELECTA.sid
FROM (SELECT * FROM sc WHERE cid = '01') A
JOIN (SELECT * FROM sc WHERE cid = '02') B ON A.sid = B.sidAND A.score > B.score;

SELECT A.sidFROM sc AWHERE EXISTS (SELECT *FROM sc BWHERE A.sid = B.sidAND A.cid = '01'AND B.cid = '02'AND A.score > B.score)

2、查询平均成绩大于60分的同学的学号和平均成绩；

SELECT sid,CAST(AVG(score) AS DECIMAL(15,2)) 均分 FROM scGROUP BY sid
HAVING AVG(score) > 60;

3、查询所有同学的学号、姓名、选课数、总成绩

 SELECT B.sid, B.sname,COUNT(*) 选课数, SUM(score) 总成绩 FROM sc A ,student BWHERE A.sid = B.sidGROUP BY B.sid , B.sname 

4、查询姓“李”的老师的个数；

SELECTCOUNT(distinct tid)  teacher_cntFROM  teacherWHERE tname like '李%'

5、查询没学过“张三”老师课的同学的学号、姓名；

--使用with表达式WITH new_sc AS(SELECT A.sid,A.cid,A.score,C.tnameFROM sc A,course B,teacher CWHERE A.cid=B.cidAND B.tid=C.tid)SELECT DISTINCT B.sid,B.sname FROM student BLEFT JOIN new_sc AON A.sid=B.sidWHERE NOT EXISTS(SELECT * FROM new_sc CWHERE A.sid=C.sidAND C.tname='张三');