sql行转列面试题

2020-06-10

字数统计: 5.9k字 | 阅读时长≈ 30分

前言

我今天面试遇到sql的面试题，基本都要行转列，行转列子查询也可以实现，但是写起来实在太麻烦，我因为没有掌握case when语法，今天的大部分行转列都没有做出来，于是我决定学习一下这个语法

其实mysql也有一个decode函数，也可以实现行转列，但是有平台限制，case when 主流的关系型数据库都进行了实现，所以我选择学习case when

case when

Case具有两种格式。简单Case函数和Case搜索函数。

简单case 函数
CASE expression
    WHEN condition1 THEN result1
    WHEN condition2 THEN result2
   ...
    WHEN conditionN THEN resultN
    ELSE result
END
-- expression为需要匹配的字段，当与when后面的某一个条件值相等时，返回then后面的结果  

Case搜索函数
CASE 
    WHEN expression1=condition1 THEN result1
    WHEN expression2=condition2 THEN result2
   ...
    WHEN expressionN=conditionN THEN resultN
    ELSE result
END
-- 这种就是把匹配的字段和值都放在when后面进行判断，这种灵活性强，匹配的字段可以换,就是expression1可以与expression2一样，也可以不一样

-- CASE 表示函数开始，END 表示函数结束。如果 condition1 成立，则返回 result1, 如果 condition2 成立，则返回 result2，当全部不成立则返回 result，而当有一个成立之后，后面的就不执行了。

练习题

1、表结构

–1.学生表 
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 
–2.课程表 
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
–3.教师表 
Teacher(t_id,t_name) –教师编号,教师姓名 
–4.成绩表 
Score(s_id,c_id,s_score) –学生编号,课程编号,分数

2、测试数据

--建表
--学生表
CREATE TABLE `Student`(
    `s_id` VARCHAR(20),
    `s_name` VARCHAR(20) NOT NULL DEFAULT '',
    `s_birth` VARCHAR(20) NOT NULL DEFAULT '',
    `s_sex` VARCHAR(10) NOT NULL DEFAULT '',
    PRIMARY KEY(`s_id`)
);
--课程表
CREATE TABLE `Course`(
    `c_id`  VARCHAR(20),
    `c_name` VARCHAR(20) NOT NULL DEFAULT '',
    `t_id` VARCHAR(20) NOT NULL,
    PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
    `t_id` VARCHAR(20),
    `t_name` VARCHAR(20) NOT NULL DEFAULT '',
    PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
    `s_id` VARCHAR(20),
    `c_id`  VARCHAR(20),
    `s_score` INT(3),
    PRIMARY KEY(`s_id`,`c_id`)
);
--插入学生表测试数据
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');
--课程表测试数据
insert into Course values('01' , '语文' , '02');
insert into Course values('02' , '数学' , '01');
insert into Course values('03' , '英语' , '03');
--教师表测试数据
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');
--成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);

3、测试题

总之行转列  聚合函数加case when

-- 5、查询所有同学的姓名,语文成绩，数学成绩，英语成绩
select  st.s_name
			 ,sum(case when s.c_id='01' then s.s_score end ) '语文'
			 ,sum(case when s.c_id='02' then s.s_score end ) '数学'
			 ,sum(case when s.c_id='03' then s.s_score end ) '英语'

from student st join score s on st.s_id=s.s_id group by st.s_name


-- 2、查询所有学生的课程及分数情况； 
        select a.s_id,a.s_name,
                    SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',
                    SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',
                    SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',
                    SUM(b.s_score) as  '总分'
        from student a left join score b on a.s_id = b.s_id 
        left join course c on b.c_id = c.c_id 
        GROUP BY a.s_id,a.s_name

扩展题

-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数  
select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
    student a 
    join score b on a.s_id=b.s_id and b.c_id='01'
    left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score
-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数 
select a.* ,b.s_score as 01_score,c.s_score as 02_score from 
    student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL 
     join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score
-- 3、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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    student b 
    join score a on b.s_id = a.s_id
    GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60; 

-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
        -- (包括有成绩的和无成绩的) 
select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from 
    student b 
    left join score a on b.s_id = a.s_id
    GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60
    union
select a.s_id,a.s_name,0 as avg_score from 
    student a 
    where a.s_id not in (
                select distinct s_id from score);  
                
-- 5、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
select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from 
    student a 
    left join score b on a.s_id=b.s_id
    GROUP BY a.s_id,a.s_name;
-- 6、查询"李"姓老师的数量 
select count(t_id) from teacher where t_name like '李%';
-- 7、查询学过"张三"老师授课的同学的信息 
select a.* from 
    student a 
    join score b on a.s_id=b.s_id where b.c_id in(
        select c_id from course where t_id =(
            select t_id from teacher where t_name = '张三'));
-- 8、查询没学过"张三"老师授课的同学的信息 
select * from 
    student c 
    where c.s_id not in(
        select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(
            select c_id from course where t_id =(
                select t_id from teacher where t_name = '张三')));
-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息 
select a.* from 
    student a,score b,score c 
    where a.s_id = b.s_id  and a.s_id = c.s_id and b.c_id='01' and c.c_id='02'; 
-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
select a.* from 
    student a 
    where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')
-- 11、查询没有学全所有课程的同学的信息 
select s.* from 
    student s where s.s_id in(
        select s_id from score where s_id not in(
            select a.s_id from score a 
                join score b on a.s_id = b.s_id and b.c_id='02'
                join score c on a.s_id = c.s_id and c.c_id='03'
            where a.c_id='01'))
-- 12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息 
select * from student where s_id in(
    select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')
    );
-- 13、查询和"01"号的同学学习的课程完全相同的其他同学的信息  
select a.* from student a where a.s_id in(
    select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01')
    group by s_id 
    having count(1)=(select count(1) from score where s_id='01'));
-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 
select a.s_name from student a where a.s_id not in (
    select s_id from score where c_id = 
                (select c_id from course where t_id =(
                    select t_id from teacher where t_name = '张三')) 
                group by s_id);
-- 15、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 a.s_id,a.s_name,ROUND(AVG(b.s_score)) from 
    student a 
    left join score b on a.s_id = b.s_id
    where a.s_id in(
            select s_id from score where s_score<60 GROUP BY  s_id having count(1)>=2)
    GROUP BY a.s_id,a.s_name 
-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select a.*,b.c_id,b.s_score from 
    student a,score b 
    where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;
-- 17、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 a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,
                (select s_score from score where s_id=a.s_id and c_id='02') as 数学,
                (select s_score from score where s_id=a.s_id and c_id='03') as 英语,
            round(avg(s_score),2) as 平均分 from score a  GROUP BY a.s_id ORDER BY 平均分 DESC;
-- 18.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
select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
    ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
    ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
    ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
    ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
    from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name
-- 19、按各科成绩进行排序，并显示排名(实现不完全)
-- mysql没有rank函数
    select a.s_id,a.c_id,
        @i:=@i +1 as i保留排名,
        @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
    union
    select a.s_id,a.c_id,
        @i:=@i +1 as i,
        @k:=(case when @score=a.s_score then @k else @i end) as rank,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
    union
    select a.s_id,a.c_id,
        @i:=@i +1 as i,
        @k:=(case when @score=a.s_score then @k else @i end) as rank,
        @score:=a.s_score as score
    from (
        select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s
-- 20、查询学生的总成绩并进行排名
select a.s_id,
    @i:=@i+1 as i,
    @k:=(case when @score=a.sum_score then @k else @i end) as rank,
    @score:=a.sum_score as score
from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,
    (select @k:=0,@i:=0,@score:=0)s
-- 21、查询不同老师所教不同课程平均分从高到低显示 
    select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
        left join score b on a.c_id=b.c_id 
        left join teacher c on a.t_id=c.t_id
        GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
-- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩 
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3
            UNION
            select d.*,c.排名,c.s_score,c.c_id from (
                select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'    
            )c
            left join student d on c.s_id=d.s_id
            where 排名 BETWEEN 2 AND 3;
-- 23、统计各科成绩各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比
        select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a
                left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,
                                            ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比
                                from score GROUP BY c_id)b on a.c_id=b.c_id
                left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,
                                            ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比
                                from score GROUP BY c_id)c on a.c_id=c.c_id
                left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,
                                            ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比
                                from score GROUP BY c_id)d on a.c_id=d.c_id
                left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,
                                            ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
                                from score GROUP BY c_id)e on a.c_id=e.c_id
                left join course f on a.c_id = f.c_id
-- 24、查询学生平均成绩及其名次 
        select a.s_id,
                @i:=@i+1 as '不保留空缺排名',
                @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',
                @avg_score:=avg_s as '平均分'
        from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;
-- 25、查询各科成绩前三名的记录
            -- 1.选出b表比a表成绩大的所有组
            -- 2.选出比当前id成绩大的 小于三个的
        select a.s_id,a.c_id,a.s_score from score a 
            left join score b on a.c_id = b.c_id and a.s_score<b.s_score
            group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3
            ORDER BY a.c_id,a.s_score DESC 
-- 26、查询每门课程被选修的学生数  
        select c_id,count(s_id) from score a GROUP BY c_id 
-- 27、查询出只有两门课程的全部学生的学号和姓名 
        select s_id,s_name from student where s_id in(
                select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2); 
-- 28、查询男生、女生人数 
        select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex
-- 29、查询名字中含有"风"字的学生信息
        select * from student where s_name like '%风%';
-- 30、查询同名同性学生名单，并统计同名人数 
        select a.s_name,a.s_sex,count(*) from student a  JOIN 
                    student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
        GROUP BY a.s_name,a.s_sex
-- 31、查询1990年出生的学生名单 
        select s_name from student where s_birth like '1990%' 
-- 32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列 
 
    select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC
-- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 
 
    select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
        left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85
-- 34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 
 
        select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(
                    select c_id from course where c_name ='数学') and b.s_score<60   
 -- 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数； 
            select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id
                left join student a on a.s_id=c.s_id where c.s_score>=70 
-- 37、查询不及格的课程
        select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id
            where a.s_score<60 
-- 38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名； 
        select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id
            where a.c_id = '01' and a.s_score>80
-- 39、求每门课程的学生人数 
        select count(*) from score GROUP BY c_id; 
-- 40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩
        -- 查询老师id   
        select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'
        -- 查询最高分（可能有相同分数）
        select MAX(s_score) from score where c_id='02'
        -- 查询信息
        select a.*,b.s_score,b.c_id,c.c_name from student a
            LEFT JOIN score b on a.s_id = b.s_id
            LEFT JOIN course c on b.c_id=c.c_id
            where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')
            and b.s_score in (select MAX(s_score) from score where c_id='02')
-- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 
    select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
-- 42、查询每门功成绩最好的前两名 
        -- 牛逼的写法
    select a.s_id,a.c_id,a.s_score from score a
        where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id
-- 43、统计每门课程的学生选修人数（超过5人的课程才统计）。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_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC
-- 44、检索至少选修两门课程的学生学号 
        select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2
-- 45、查询选修了全部课程的学生信息 
        select * from student where s_id in(        
            select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))
--46、查询各学生的年龄
    -- 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
    select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - 
                (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age
        from student;
-- 47、查询本周过生日的学生
    select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)
    select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))
    select WEEK(DATE_FORMAT(NOW(),'%Y%m%d')) 
-- 48、查询下周过生日的学生
    select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth) 
-- 49、查询本月过生日的学生
    select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)
-- 50、查询下月过生日的学生
    select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

中国移动sql面试题

没写答案，小伙伴们感兴趣自己完成哦


中国移动sql面试题：
create table test(
   id number(10) primary key,
   type number(10) ,
   t_id number(10),
   value varchar2(5)
);
insert into test values(100,1,1,'张三');
insert into test values(200,2,1,'男');
insert into test values(300,3,1,'50');

insert into test values(101,1,2,'刘二');
insert into test values(201,2,2,'男');
insert into test values(301,3,2,'30');

insert into test values(102,1,3,'刘三');
insert into test values(202,2,3,'女');
insert into test values(302,3,3,'10');
请写出一条查询语句结果如下：

姓名      性别     年龄
--------- -------- ----
张三       男        50

 

------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
 

2.一道SQL语句面试题，关于group by
表内容：
2005-05-09 胜
2005-05-09 胜
2005-05-09 负
2005-05-09 负
2005-05-10 胜
2005-05-10 负
2005-05-10 负

如果要生成下列结果, 该如何写sql语句?

          胜 负
2005-05-09 2 2
2005-05-10 1 2
------------------------------------------
create table tmp(rq varchar2(10),shengfu varchar2(5));

insert into tmp values('2005-05-09','胜');
insert into tmp values('2005-05-09','胜');
insert into tmp values('2005-05-09','负');
insert into tmp values('2005-05-09','负');
insert into tmp values('2005-05-10','胜');
insert into tmp values('2005-05-10','负');
insert into tmp values('2005-05-10','负');


 4.create table STUDENT_SCORE
(
  name    VARCHAR2(20),
  subject VARCHAR2(20),
  score   NUMBER(4,1)
);
insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '语文', 78.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '数学', 88.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('张三', '英语', 98.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '语文', 89.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '数学', 76.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('李四', '英语', 90.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '语文', 99.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '数学', 66.0);
insert into student_score (NAME, SUBJECT, SCORE) values ('王五', '英语', 91.0);

4.1得到类似下面的结果
姓名   语文  数学  英语

王五    89    56    89

4.2有一张表，里面有3个字段：语文，数学，英语。其中有3条记录分别表示语文70分，数学80分，英语58分，请用一条sql语句查询出这三条记录并按以下条件显示出来（并写出您的思路）：  
   大于或等于80表示优秀，大于或等于60表示及格，小于60分表示不及格。  
       显示格式：  
       语文              数学                英语  
       及格              优秀                不及格    
------------------------------------------



5.请用一个sql语句得出结果
从table1,table2中取出如table3所列格式数据，注意提供的数据及结果不准确，
只是作为一个格式向大家请教。
 

table1

月份mon 部门dep 业绩yj
-------------------------------
一月份      01      10
一月份      02      10
一月份      03      5
二月份      02      8
二月份      04      9
三月份      03      8

table2

部门dep      部门名称dname
--------------------------------
      01      国内业务一部
      02      国内业务二部
      03      国内业务三部
      04      国际业务部

table3 （result）

部门dep 一月份      二月份      三月份
--------------------------------------
      01      10                  
      02      10         8         
      03                 5        8
      04                          9

------------------------------------------

create table yj01(
       month varchar2(10),
       deptno number(10),
       yj number(10)
)

insert into yj01(month,deptno,yj) values('一月份',01,10);
insert into yj01(month,deptno,yj) values('二月份',02,10);
insert into yj01(month,deptno,yj) values('二月份',03,5);
insert into yj01(month,deptno,yj) values('三月份',02,8);
insert into yj01(month,deptno,yj) values('三月份',04,9);
insert into yj01(month,deptno,yj) values('三月份',03,8);

create table yjdept(
       deptno number(10),
       dname varchar2(20)
)

insert into yjdept(deptno,dname) values(01,'国内业务一部');
insert into yjdept(deptno,dname) values(02,'国内业务二部');
insert into yjdept(deptno,dname) values(03,'国内业务三部');
insert into yjdept(deptno,dname) values(04,'国际业务部');

本文作者： 忘忧症
本文链接： https://NepenthesZGW.github.io/2020/06/10/database/面试题/sql行转列/
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