15道常用sql面试题
- 二、题目：
-- 01）查询 " 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
SELECT
st.*,
s1.score '01课程',
s2.score '02课程'
FROM
( SELECT * FROM t_mysql_score WHERE cid = "01" ) s1,
( SELECT * FROM t_mysql_score WHERE cid = "02" ) s2,
t_mysql_student st
WHERE
s1.sid = s2.sid
AND s1.sid = st.sid
AND s1.score > s2.score
-- 02）查询同时存在 " 01 "课程和" 02 "课程的情况
SELECT
s.*,
( CASE WHEN s1.cid = "01" THEN s1.score END ) '01课程',
( CASE WHEN s2.cid = "02" THEN s1.score END ) '02课程'
FROM
( SELECT * FROM t_mysql_score WHERE cid = "01" ) s1,
( SELECT * FROM t_mysql_score WHERE cid = "02" ) s2,
t_mysql_student s
WHERE
s1.sid = s2.sid
AND s1.sid = s.sid
-- 03）查询存在 " 01 "课程但可能不存在" 02 "课程的情况 ( 不存在时显示为 NULL )
SELECT
s1.sid,
s1.cid,
s1.score '01课程',
s2.score '02课程'
FROM
( SELECT * FROM t_mysql_score sc WHERE sc.cid = "01" ) s1
LEFT JOIN ( SELECT * FROM t_mysql_score sc WHERE sc.cid = "02" ) s2 ON s1.sid = s2.sid
-- 04）查询不存在 " 01 "课程但存在" 02 "课程的情况
SELECT
*
FROM
t_mysql_score
WHERE
sid IN ( SELECT sid FROM t_mysql_student WHERE sid NOT IN ( SELECT sid FROM t_mysql_score WHERE cid = '01' ) )
AND cid = '02'
-- 05）查询平均成绩大于等于 60 分的同学的学生编号和学生姓名和平均成绩
SELECT
s.sid,
s.sname,
ROUND( AVG( c.score ), 2 ) '平均成绩'
FROM
t_mysql_student s,
t_mysql_score c
WHERE
s.sid = c.sid
GROUP BY
s.sid,
s.sname
HAVING
avg( c.score ) >= 60
-- 06）查询在 t_mysql_score表存在成绩的学生信息
SELECT
s.*
FROM
t_mysql_score sc,
t_mysql_student s
WHERE
sc.sid = s.sid
GROUP BY
s.sid
-- 07）查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩 ( 没成绩的显示为 NULL )
SELECT
s.sid,
s.sname,
count( sc.cid ) '选课总数',
sum( sc.score ) '总成绩'
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
s.sid = sc.sid
GROUP BY
s.sid,
s.sname
-- 08）查询「李」姓老师的数量
SELECT
*
FROM
t_mysql_teacher t
WHERE
tname LIKE '李%'
-- 09）查询学过「张三」老师授课的同学的信息
SELECT
*
FROM
t_mysql_student
WHERE
sid IN (
SELECT
sc.sid
FROM
t_mysql_teacher t,
t_mysql_score sc,
t_mysql_course c
WHERE
t.tid = c.tid
AND sc.cid = c.cid
AND t.tname = "张三"
GROUP BY
sc.sid
)
-- 10）查询没有学全所有课程的同学的信息
SELECT
s.*,
count( sc.score ) '课程数量'
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
s.sid = sc.sid
GROUP BY
s.sid,
s.sname
HAVING
count( sc.score ) < ( SELECT count( 1 ) FROM t_mysql_course )
11）查询没学过 "张三"老师讲授的任一门课程的学生姓名
SELECT
sname
FROM
t_mysql_student
WHERE
sid NOT IN (
SELECT
sc.sid
FROM
t_mysql_score sc,
t_mysql_course c,
t_mysql_teacher t
WHERE
t.tid = c.tid
AND sc.cid = c.cid
AND t.tname = "张三"
)
-- 12）查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩
SELECT
s.sid,
s.sname,
ROUND( AVG( sc.score ), 2 ) 平均成绩
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
sc.sid = s.sid
AND sc.score < 60 GROUP BY s.sid, s.sname HAVING count( sc.score ) >= 2
-- 13）检索 " 01 "课程分数小于 60，按分数降序排列的学生信息
SELECT
s.*,
sc.score
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
s.sid = sc.sid
AND sc.cid = "01"
AND sc.score < 60
ORDER BY
sc.score DESC
-- 14）按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
SELECT
s.sid,
s.sname,
IF
(
max( CASE WHEN sc.cid = '01' THEN sc.score END ) > 0,
max( CASE WHEN sc.cid = '01' THEN sc.score END ),
0
) 语文,
IF
(
max( CASE WHEN sc.cid = '02' THEN sc.score END ) > 0,
max( CASE WHEN sc.cid = '02' THEN sc.score END ),
0
) 数学,
IF
(
max( CASE WHEN sc.cid = '03' THEN sc.score END ) > 0,
max( CASE WHEN sc.cid = '03' THEN sc.score END ),
0
) 英语,
ROUND( AVG( sc.score ), 2 ) 平均成绩
FROM
t_mysql_student s,
t_mysql_score sc
WHERE
s.sid = sc.sid
GROUP BY
s.sid,
s.sname
-- 15）查询各科成绩最高分、最低分和平均分： 以如下形式显示：课程 ID，课程 NAME，最高分，最低分，平均分，及格率，中等率，优良率，优秀率及格为 >= 60，中等为：70-80，优良为：80-90，优秀为： >= 90 要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
SELECT
c.cid,
c.cname,
max( sc.score ) '最高分',
min( sc.score ) '最低分',
ROUND( AVG( sc.score ), 2 ) '平均分',
CONCAT(ROUND(sum( IF ( sc.score >= 60, 1, 0 ) ) / COUNT( sc.score ) * 100,2 ),'%' ) 及格率,
CONCAT(ROUND(sum( IF ( sc.score >= 70 AND sc.score < 80, 1, 0 ) ) / COUNT( sc.score ) * 100,2 ),'%' ) 中等率,
CONCAT(ROUND(sum( IF ( sc.score >= 80 AND sc.score < 90, 1, 0 ) ) / COUNT( sc.score ) * 100,2 ),'%' ) 优良率,
CONCAT(ROUND(sum( IF ( sc.score >= 90, 1, 0 ) ) / COUNT( sc.score ) * 100,2 ),'%' ) 优秀率,
COUNT( sc.score ) 选修人数
FROM
t_mysql_score sc,
t_mysql_course c
WHERE
sc.cid = c.cid
GROUP BY
sc.cid
ORDER BY
选修人数 DESC,
c.cid ASC;
总结：实现SQL代码的思路可以分为以下几个步骤：
