SQL 笔试题集锦

1.一道SQL语句面试题,关于group by表内容:

info 表

date result

2005-05-09 win

2005-05-09 lose

2005-05-09 lose

2005-05-09 lose

2005-05-10 win

2005-05-10 lose

2005-05-10 lose

 

如果要生成下列结果, 该如何写sql语句?

 

              win lose

2005-05-09 2 2

2005-05-10 1 2

答案:

(1) select date, sum(case when result = "win" then 1 else 0 end) as "win", sum(case when result = "lose" then 1 else 0 end) as "lose" from info group by date;

(2) select a.date, a.result as win, b.result as lose

  from

  (select date, count(result) as result from info where result = "win" group by date) as a

  join

  (select date, count(result) as result from info where result = "lose" group by date) as b

  on a.date = b.date;

 

2.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 else c end) from table;

 

3.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:chinese,math,english。其中有一条记录chinese 70分,math 80分,english 58分,请用一条sql语句查询出所有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excellent,大于或等于60表示pass,小于60分表示fail。 

       显示格式: 以上面的chinese 70分,math 80分,english 58分

       chinese              math                english 

       pass                  excellent           fail

 

select (case when chinese >= 80 then "excellent" when chinese >= 60 then "pass" else "fail" end) as chinese,

  (case when math >= 80 then "excellent" when math >= 60 then "pass" else "fail" end) as math,

  (case when english >= 80 then "excellent" when english >= 60 then "pass" else "fail" end) as english

  from grade;

 

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。

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

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

 

select t1.dep,

sum(case when mon = 1 then yj else 0 end) as jun,

sum(case when mon = 2 then yj else 0 end) as feb,

sum(case when mon = 3 then yj else 0 end) as mar

from

t1 right join t2 on t1.dep = t2.dep

group by t1.dep;

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