经典面试题

金蝶面试,出了这样一道SQL题,共享之

给下面这样的一个表记录:

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

购物人      商品名称     数量
A            甲          2
B            乙          4
C            丙          1
A            丁          2
B            丙          5

给出所有购入商品为两种或两种以上的购物人记录


select 购物人 as 顾客,count(商品名称) as 商品种类数 from 表 group by 购物人 having count(商品名称)>=2

1.一道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 varchar(10),shengfu nchar(1))

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')

1)select rq, sum(case when shengfu='胜' then 1 else 0 end)'胜',sum(case when shengfu='负' then 1 else 0 end)'负' from #tmp group by rq
2) select N.rq,N.勝,M.負 from (
select rq,勝=count(*) from #tmp where shengfu='胜'group by rq)N inner join
(select rq,負=count(*) from #tmp where shengfu='负'group by rq)M on N.rq=M.rq
3)select a.col001,a.a1 胜,b.b1 负 from
(select col001,count(col001) a1 from temp1 where col002='胜' group by col001) a,
(select col001,count(col001) b1 from temp1 where col002='负' group by col001) b
where a.col001=b.col001

2.请教一个面试中遇到的SQL语句的查询问题
表中有A B C三列,用SQL语句实现:当A列大于B列时选择A列否则选择B列,当B列大于C列时选择B列否则选择C列。
------------------------------------------
select (case when a>b then a else b end ),
(case when b>c then b esle c end)
from table_name

3.面试题:一个日期判断的sql语句?
请取出tb_send表中日期(SendTime字段)为当天的所有记录?(SendTime字段为datetime型,包含日期与时间)
------------------------------------------
select * from tb where datediff(dd,SendTime,getdate())=0

4.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    
------------------------------------------
select
(case when 语文>=80 then '优秀'
        when 语文>=60 then '及格'
else '不及格') as 语文,
(case when 数学>=80 then '优秀'
        when 数学>=60 then '及格'
else '不及格') as 数学,
(case when 英语>=80 then '优秀'
        when 英语>=60 then '及格'
else '不及格') as 英语,
from table

5.在sqlserver2000中请用sql创建一张用户临时表和系统临时表,里面包含两个字段ID和IDValues,类型都是int型,并解释下两者的区别?
------------------------------------------
用户临时表:create table #xx(ID int, IDValues int)
系统临时表:create table ##xx(ID int, IDValues int)

区别:
用户临时表只对创建这个表的用户的Session可见,对其他进程是不可见的.
当创建它的进程消失时这个临时表就自动删除.

全局临时表对整个SQL Server实例都可见,但是所有访问它的Session都消失的时候,它也自动删除.

6.sqlserver2000是一种大型数据库,他的存储容量只受存储介质的限制,请问它是通过什么方式实现这种无限容量机制的。
------------------------------------------
它的所有数据都存储在数据文件中(*.dbf),所以只要文件够大,SQL    Server的存储容量是可以扩大的.

SQL Server 2000 数据库有三种类型的文件:

主要数据文件
主要数据文件是数据库的起点,指向数据库中文件的其它部分。每个数据库都有一个主要数据文件。主要数据文件的推荐文件扩展名是 .mdf。

次要数据文件
次要数据文件包含除主要数据文件外的所有数据文件。有些数据库可能没有次要数据文件,而有些数据库则有多个次要数据文件。次要数据文件的推荐文件扩展名是 .ndf。

日志文件
日志文件包含恢复数据库所需的所有日志信息。每个数据库必须至少有一个日志文件,但可以不止一个。日志文件的推荐文件扩展名是 .ldf。

7.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。
如使用存储过程也可以。

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        null      null
      02      10         8        null
      03      null       5        8
      04      null      null      9

------------------------------------------
1)
select a.部门名称dname,b.业绩yj as '一月份',c.业绩yj as '二月份',d.业绩yj as '三月份'
from table1 a,table2 b,table2 c,table2 d
where a.部门dep = b.部门dep and b.月份mon = '一月份' and
a.部门dep = c.部门dep and c.月份mon = '二月份' and
a.部门dep = d.部门dep and d.月份mon = '三月份' and
2)
select a.dep,
sum(case when b.mon=1 then b.yj else 0 end) as '一月份',
sum(case when b.mon=2 then b.yj else 0 end) as '二月份',
sum(case when b.mon=3 then b.yj else 0 end) as '三月份',
sum(case when b.mon=4 then b.yj else 0 end) as '四月份',
sum(case when b.mon=5 then b.yj else 0 end) as '五月份',
sum(case when b.mon=6 then b.yj else 0 end) as '六月份',
sum(case when b.mon=7 then b.yj else 0 end) as '七月份',
sum(case when b.mon=8 then b.yj else 0 end) as '八月份',
sum(case when b.mon=9 then b.yj else 0 end) as '九月份',
sum(case when b.mon=10 then b.yj else 0 end) as '十月份',
sum(case when b.mon=11 then b.yj else 0 end) as '十一月份',
sum(case when b.mon=12 then b.yj else 0 end) as '十二月份',
from table2 a left join table1 b on a.dep=b.dep

8.华为一道面试题
一个表中的Id有多个记录,把所有这个id的记录查出来,并显示共有多少条记录数。
------------------------------------------
select id, Count(*) from tb group by id having count(*)>1
select * from(select count(ID) as count from table group by ID)T where T.count>1

1.用一条SQL语句 查询出每门课都大于80分的学生姓名

name   kecheng   fenshu
张三     语文       81
张三     数学
       75
李四     语文
       76
李四     数学
       90
王五     语文
       81
王五     数学
       100
王五     英语
       90

A: select distinct name from table where name not in (select distinct name from table where fenshu<=80)

2.
学生表 如下:
自动编号   学号   姓名 课程编号 课程名称 分数

1        2005001
张三 0001      数学    69
2        2005002
李四 0001      数学
    89
3        2005001
张三 0001      数学
    69
删除除了自动编号不同,其他都相同的学生冗余信息


A: delete tablename where
自动编号 not in(select min(自动编号) from tablename group by 学号,姓名,课程编号,课程名称,分数)

一个叫department的表,里面只有一个字段name,一共有4条纪录,分别是a,b,c,d,对应四个球对,现在四个球对进行比赛,用一条sql语句显示所有可能的比赛组合.
你先按你自己的想法做一下,看结果有我的这个简单吗?

答:select a.name, b.name
from team a, team b
where a.name < b.name

 

请用SQL语句实现:从TestDB数据表中查询出所有月份的发生额都比101科目相应月份的发生额高的科目。请注意:TestDB中有很多科目,都有112月份的发生额。
AccID
:科目代码,Occmonth:发生额月份,DebitOccur:发生额。
数据库名:JcyAudit,数据集:Select * from TestDB

答:select a.*
from TestDB a
,(select Occmonth,max(DebitOccur) Debit101ccur from TestDB where AccID='101' group by Occmonth) b
where a.Occmonth=b.Occmonth and a.DebitOccur>b.Debit101ccur

************************************************************************************

面试题:怎么把这样一个表儿
year   month amount
1991   1     1.1
1991   2     1.2
1991   3     1.3
1991   4     1.4
1992   1     2.1
1992   2     2.2
1992   3     2.3
1992   4     2.4
查成这样一个结果
year m1   m2   m3   m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4

答案一、
select year,
(select amount from   aaa m where month=1   and m.year=aaa.year) as m1,
(select amount from   aaa m where month=2   and m.year=aaa.year) as m2,
(select amount from   aaa m where month=3   and m.year=aaa.year) as m3,
(select amount from   aaa m where month=4   and m.year=aaa.year) as m4
from aaa   group by year

 

这个是ORACLE  中做的:
select * from (select name, year b1, lead(year) over
(partition by name order by year) b2, lead(m,2) over(partition by name order by year) b3,rank()over(
partition by name order by year) rk from t) where rk=1;

************************************************************************************

精妙的SQL语句!
精妙SQL语句  
作者:不详 发文时间:2003.05.29 10:55:05

说明:复制表(只复制结构,源表名:a 新表名:b)

SQL: select * into b from a where 1<>1

说明:拷贝表(拷贝数据,源表名:a 目标表名:b)

SQL: insert into b(a, b, c) select d,e,f from b;

说明:显示文章、提交人和最后回复时间

SQL: select a.title,a.username,b.adddate from table a,(select max(adddate) adddate from table where table.title=a.title) b

说明:外连接查询(表名1a 表名2b)

SQL: select a.a, a.b, a.c, b.c, b.d, b.f from a LEFT OUT JOIN b ON a.a = b.c

说明:日程安排提前五分钟提醒

SQL: select * from
日程安排 where datediff('minute',f开始时间,getdate())>5

说明:两张关联表,删除主表中已经在副表中没有的信息

SQL:

delete from info where not exists ( select * from infobz where info.infid=infobz.infid )

说明:--

SQL:

SELECT A.NUM, A.NAME, B.UPD_DATE, B.PREV_UPD_DATE

FROM TABLE1,

(SELECT X.NUM, X.UPD_DATE, Y.UPD_DATE PREV_UPD_DATE

FROM (SELECT NUM, UPD_DATE, INBOUND_QTY, STOCK_ONHAND

FROM TABLE2

WHERE TO_CHAR(UPD_DATE,'YYYY/MM') = TO_CHAR(SYSDATE, 'YYYY/MM')) X,

(SELECT NUM, UPD_DATE, STOCK_ONHAND

FROM TABLE2

WHERE TO_CHAR(UPD_DATE,'YYYY/MM') =

TO_CHAR(TO_DATE(TO_CHAR(SYSDATE, 'YYYY/MM') ¦¦ '/01','YYYY/MM/DD') - 1, 'YYYY/MM') ) Y,

WHERE X.NUM = Y.NUM
+

AND X.INBOUND_QTY + NVL(Y.STOCK_ONHAND,0) <> X.STOCK_ONHAND ) B

WHERE A.NUM = B.NUM

说明:--

SQL:

select * from studentinfo where not exists(select * from student where studentinfo.id=student.id) and
系名称='"&strdepartmentname&"' and 专业名称='"&strprofessionname&"' order by 性别,生源地,高考总成绩

说明:

从数据库中去一年的各单位电话费统计(电话费定额贺电化肥清单两个表来源)

SQL:

SELECT a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy') AS telyear,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '01', a.factration)) AS JAN,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '02', a.factration)) AS FRI,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '03', a.factration)) AS MAR,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '04', a.factration)) AS APR,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '05', a.factration)) AS MAY,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '06', a.factration)) AS JUE,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '07', a.factration)) AS JUL,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '08', a.factration)) AS AGU,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '09', a.factration)) AS SEP,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '10', a.factration)) AS OCT,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '11', a.factration)) AS NOV,

SUM(decode(TO_CHAR(a.telfeedate, 'mm'), '12', a.factration)) AS DEC

FROM (SELECT a.userper, a.tel, a.standfee, b.telfeedate, b.factration

FROM TELFEESTAND a, TELFEE b

WHERE a.tel = b.telfax) a

GROUP BY a.userper, a.tel, a.standfee, TO_CHAR(a.telfeedate, 'yyyy')

说明:四表联查问题:

SQL: select * from a left inner join b on a.a=b.b right inner join c on a.a=c.c inner join d on a.a=d.d where .....

说明:得到表中最小的未使用的ID

SQL:

SELECT (CASE WHEN EXISTS(SELECT * FROM Handle b WHERE b.HandleID = 1) THEN MIN(HandleID) + 1 ELSE 1 END) as HandleID

FROM Handle

WHERE NOT HandleID IN (SELECT a.HandleID - 1 FROM Handle a)

 

*******************************************************************************

有两个表AB,均有keyvalue两个字段,如果BkeyA中也有,就把Bvalue换为A中对应的value
这道题的SQL语句怎么写?

update b set b.value=(select a.value from a where a.key=b.key) where b.id in(select b.id from b,a where b.key=a.key);

***************************************************************************

高级sql面试题

原表:
courseid coursename score
-------------------------------------
1 java 70
2 oracle 90
3 xml 40
4 jsp 30
5 servlet 80
-------------------------------------
为了便于阅读,查询此表后的结果显式如下(及格分数为60):
courseid coursename score mark
---------------------------------------------------
1 java 70 pass
2 oracle 90 pass
3 xml 40 fail
4 jsp 30 fail
5 servlet 80 pass
---------------------------------------------------
写出此查询语句

没有装ORACLE,没试过
select courseid, coursename ,score ,decode
sign(score-60),-1,'fail','pass') as mark from course

完全正确

SQL> desc course_v
Name Null? Type
----------------------------------------- -------- ----------------------------
COURSEID NUMBER
COURSENAME VARCHAR2(10)
SCORE NUMBER

SQL> select * from course_v;

COURSEID COURSENAME SCORE
---------- ---------- ----------
1 java 70
2 oracle 90
3 xml 40
4 jsp 30
5 servlet 80

SQL> select courseid, coursename ,score ,decode(sign(score-60),-1,'fail','pass') as mark from course_v;

COURSEID COURSENAME SCORE MARK
---------- ---------- ---------- ----
1 java 70 pass
2 oracle 90 pass
3 xml 40 fail
4 jsp 30 fail
5 servlet 80 pass

*******************************************************************************

原表:

id proid proname
1 1 M
1 2 F
2 1 N
2 2 G
3 1 B
3 2 A
查询后的表:

id pro1 pro2
1 M F
2 N G
3 B A
写出查询语句

解决方案

sql
求解
a
a1 a2
记录 1 a
1 b
2 x
2 y
2 z
select能选成以下结果吗?
1 ab
2 xyz
使用pl/sql代码实现,但要求你组合后的长度不能超出oracle varchar2长度的限制。
下面是一个例子
create or replace type strings_table is table of varchar2(20);
/
create or replace function merge (pv in strings_table) return varchar2
is
ls varchar2(4000);
begin
for i in 1..pv.count loop
ls := ls || pv(i);
end loop;
return ls;
end;
/
create table t (id number,name varchar2(10));
insert into t values(1,'Joan');
insert into t values(1,'Jack');
insert into t values(1,'Tom');
insert into t values(2,'Rose');
insert into t values(2,'Jenny');

column names format a80;
select t0.id,merge(cast(multiset(select name from t where t.id = t0.id) as strings_table)) names
from (select distinct id from t) t0;

drop type strings_table;
drop function merge;
drop table t;




sql

Well if you have a thoretical maximum, which I would assume you would given the legibility of listing hundreds of employees in the way you describe then yes. But the SQL needs to use the LAG function for each employee, hence a hundred emps a hundred LAGs, so kind of bulky.

This example uses a max of 6, and would need more cut n pasting to do more than that.

SQL> select deptno, dname, emps
2 from (
3 select d.deptno, d.dname, rtrim(e.ename ||', '||
4 lead(e.ename,1) over (partition by d.deptno
5 order by e.ename) ||', '||
6 lead(e.ename,2) over (partition by d.deptno
7 order by e.ename) ||', '||
8 lead(e.ename,3) over (partition by d.deptno
9 order by e.ename) ||', '||
10 lead(e.ename,4) over (partition by d.deptno
11 order by e.ename) ||', '||
12 lead(e.ename,5) over (partition by d.deptno
13 order by e.ename),', ') emps,
14 row_number () over (partition by d.deptno
15 order by e.ename) x
16 from emp e, dept d
17 where d.deptno = e.deptno
18 )
19 where x = 1
20 /

DEPTNO DNAME EMPS
------- ----------- ------------------------------------------
10 ACCOUNTING CLARK, KING, MILLER
20 RESEARCH ADAMS, FORD, JONES, ROONEY, SCOTT, SMITH
30 SALES ALLEN, BLAKE, JAMES, MARTIN, TURNER, WARD

also
create function get_a2;
create or replace function get_a2( tmp_a1 number)
return varchar2
is
Col_a2 varchar2(4000);
begin
Col_a2:='';
for cur in (select a2 from unite_a where a1=tmp_a1)
loop
Col_a2=Col_a2||cur.a2;
end loop;
return Col_a2;
end get_a2;

select distinct a1 ,get_a2(a1) from unite_a
1 ABC
2 EFG
3 KMN

 

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