❶ sql笔试题寻求答案
select count(*) ,case Fwage > 5000 then 1 else 0 end as type1, case fage > 32 then 1 else 0 end as type2 from emploreeinfo group by type1,type2
有个问题啊 就是你的month是什么形式的 年月日 还是数字形式 我按数字形式给你写一个啊 如果是年-月的话 回头再给你写
我假设是同一年了 如果不是可以在left join 的on后面改条件
select s1.month,sum(s1.consume) as 当月消费, sum(s2.consume) as 上月消费, sum(s3.consume) as 下月消费, from student s1 left join student s2 on s1.month=s2.month+1 left join student s3 on s1.month=s3.month-1 group by s1.month
❷ sql语句 面试题
A.创建表格CODE省略
注明：学生表PK stu_id 课程表pk cos_id 分数表PK enrollment_id FK stu_id,cos_id
B.插入数据code省略
C.Query
select s.stu_id,stu_name,count(cos_id) from student s,enrollments e where s.stu_id = e.stu_id and e.grade>60 group by s.stu_id,stu_name;
select e.stu_id,s.stu_name,c.cos_name from student s,enrollments e,course c
where s.stu_id = e.stu_id
and e.cos_id = c.cos_id
and c.cos_name = 'CHINESE'
and s.stu_name like 'W%';
select stu_id,stu_name from (select e.stu_id,stu_name,cos_name from enrollments e,student s,course c
where s.stu_id = e.stu_id
and e.cos_id = c.cos_id
and c.cos_name IN ('CHINESE','MUSIC'))
group by stu_id,stu_name
having count(cos_name) = 2
select distinct e.cos_id,c.cos_name,count(e.stu_id) stu_count,count(e.stu_id)-NVL(A.FAIL,0) upscore,(count(e.stu_id)-NVL(A.FAIL,0))/count(e.stu_id) rate from
(select cos_id,count(stu_id) fail from enrollments where grade<60 group by cos_id) a,enrollments e,course c
where e.cos_id = a.cos_id(+)
and e.cos_id = c.cos_id
group by e.cos_id,NVL(a.fail,0),c.cos_name;
update student
set avg_grade =(select avg(grade) X from enrollments group by stu_id
having student.stu_id = enrollments.stu_id);
select stu_id,avg(grade) from
(select stu_id,cos_id,grade,row_number() over(partition by stu_id order by grade ) X from enrollments)
group by stu_id
having count(*)<=2
UNION
select A.stu_id,avg(A.grade)from
(select stu_id,cos_id,grade,row_number() over(partition by stu_id order by grade ) X from enrollments) A,
(select stu_id,count(*) c from
(select stu_id,cos_id,grade,row_number() over(partition by stu_id order by grade ) X from enrollments)
group by stu_id) B
where A.stu_id = B.stu_id
and A.x>1 and x<B.c
group by A.stu_id,b.c
_________________________________________________
环境：oracle 10g/TOAD 以上代码均通过测试，如有问题，请联系，谢谢
❸ 数据库SQL查询语句面试题
5.1
select a.username,b.deptname from users a,dept b where a.dept_id=b.id;
5.2
update users set dept_id='9' where dept_id='2';
5.3
select a.deptname,b.count_id from dept a,(select dept_id,count(id) as count_id from users group by dept_id having count(id)>1) b where a.id=b.dept_id;
5.4
select a.deptname,b.count_man,c.count_woman from dept a,(select dept_id,count(sex) as count_man from users where sex='男' group by dept_id) b,(select dept_id,count(sex) as count_woman from users where sex='女' group by dept_id) c where a.id=b.dept_id and a.id=c.dept_id;
5.5
添加历史记录表
create table history(
id number(8), -- 记录编号
dept_id varchar2(5), -- 部门ID
user_id varchar2(5), -- 用户ID
change_date date -- 变动日期
);
❹ 某公司SQL 笔试题寻找答案
第一道题是查看学生的姓名，选修的课程和对应的学分吧 select '姓名'=(select name from students_t where id=student_course_t.student_id ), '选修课程'=(select name from course_t where id=student_course_t.course_id ), '学分'=(select credit from course_t where id=student_course_t.course_id ) from student_course_t order by credit desc 只能给你第一题的了。后面的如果不自己打一次容易出现错误，用嵌套子查询加相关查询可以做到, oracle语法的话最后要加分号;
❺ SQL语句面试题
SELECT*,
(SELECTCOUNT(*)FROM(SELECTCOUNT(*),b.sidFROMscbLEFTJOINcoursecONb.Cid=c.CidGROUPBYb.sid,c.tid)owWHEREow.sid=a.sid)as'
选课数量',
(SELECTsum(Score)FROMScdWHEREa.sid=d.sid)as'总成绩'
FROMstudenta
;
SELECTd.sid,d.snamefromteachera
LEFTJOINcoursebona.tid=b.tid
LEFTJOINscconc.cid=b.cid
leftJOINstudentdONd.sid=c.sid
WHEREa.tname='叶萍';
SELECTsid,sname
FROMstudent
WHEREsidin(selecta.sid
FROMscAleftjoinscbona.sid=b.sid
WHEREa.cid=1andb.cid=2anda.score>b.score)
SELECTsc.Sid,sum(CASEWHENc.Cname='
数学'thensc.Scoreelse0end)数学,SUM(casewhenc.Cname='物理
'THENsc.ScoreELSE0END)物理,AVG(sc.Score)平均分
FROMsc
INNERJOINCourseconsc.Cid=c.Cid
WHEREc.Cnamein('数学','物理')
GROUPBYsc.Sid
ORDERBYAVG(sc.Score)DESC
insertintoSc(Sid,Cid,Score)values(003,3,85);
insertintoSc(Sid,Cid,Score)values(003,3,30);
首先这个不知道是你i写错了还是怎么一个人化学成绩有2个所以我在这里按照你这个上做的查询所以有一个人是选了单个课程
❻ 求救一条笔试SQL查询语句
DECLARE @Course varchar(20), --课程
@ETime datetime(20), --考试日期
@PResult decimal(5,2) --及格分数线
SET @Course = '英语'
SET @ETime = '2008-08-08'
SET @PResult = 60
SELECT C.Cid AS 班级编号,COUNT(S.Sid) AS 总人数,
COUNT(R.Sid) AS 实考人数,MAX(R.Results) AS 最高分,
MIN(R.Results) AS 最低分,
SUM(CASE WHEN R.Results<@PResult THEN 1 ELSE 0 END) AS 不及格人数,
SUM(CASE WHEN R.Results>=@PResult THEN 1 ELSE 0 END)/COUNT(R.SId) AS 及格率
FROM T_Class C INNER JOIN T_Student S ON C.CIid = S.Cid
INNER JOIN T_Results R ON S.Sid = R.Sid
WHERE R.Course = @Course
AND R.ETime = @ETime
GROUP BY C.Cid
❼ SQL语句分组查询笔试题
select match_date,
sum(case when match_result ='win' then 1 else 0 end) as amount_win,
sum(case when match_result='lose' then 1 else 0 end) as amount_lose
from t_match
group by match_date