从网上收集来的一些面试题和解题思路，加以整理，供参考。
Question 1:
Can you use a batch SQL or store procedure to calculating the Number of Days in a Month?
Answer 1：本月的最后一天(日)就等于该月的总天数
dateadd (dd, - 1 , -- 上个月的最后一天(日期)
dateadd (mm, 1 , cast (
cast ( year ( getdate ()) as varchar ) + ' - ' + -- 年份
cast ( month ( getdate ()) as varchar ) + ' -01 ' -- 月、日
as datetime )))) -- 下个月的第一天(日期)
Answer 2：下个月1号与本月1号之间的天数就等于该月的总天数
(
@Year int ,
@Month int
)
as
begin
if @Month > 12 or @Month < 1
begin
print ' Month值应该在1～12之间！ ' ;
return
end
declare @NextMonth int ;
declare @NextYear int ;
if @Month = 12
begin
set @NextMonth = 1 ;
set @NextYear = @Year + 1 ;
end
else
begin
set @NextMonth = @Month + 1 ;
set @NextYear = @Year
end
declare @StartDate nchar ( 9 );
declare @EndDate nchar ( 9 );
set @StartDate = Cast ( @Month as nchar ( 2 )) + ' /1/ ' + Cast ( @Year as nchar ( 4 ));
set @EndDate = Cast ( @NextMonth as nchar ( 2 )) + ' /1/ ' + Cast ( @NextYear as nchar ( 4 ));
SELECT DATEDIFF ( day , Cast ( @StartDate as datetime ), Cast ( @EndDate as datetime )) AS NumberOfDays
end
Question 2:
Can you use a SQL statement to calculating it! (SQL Server 2K中的pubs数据库，下同)
How can I print "10 to 20" for books that sell for between $10 and $20，"unknown" for books whose price is null, and "other" for all other prices?
Answer：
case
when price is null then ' unknown '
when price > 10 and price < 20 then ' 10 to 20 '
else ' other '
end
from titles
Question 3：
How can I find authors with the same last name?(SQL Server 2K中的pubs数据库，下同)
You can use the table authors in datatabase pubs. I want to get the result as below:
Output:
au_lname number_dups
---------------------------------------- -----------
Ringer 2
(1 row(s) affected)
Answer：
Question4：
Can you create a cross-tab report in my SQL Server!
How can I get the report about sale quality for each store and each quarterand the total sale quality for each quarter at year 1993?
You can use the table sales and stores in datatabase pubs.
Table Sales record all sale detail item for each store. Column store_id is the id of each store, ord_date is the order date of each sale item, and column qty is the sale qulity. Table stores record all store information.
I want to get the result look like as below:
Output:
stor_name Total Qtr1 Qtr2 Qtr3 Qtr4
------------------------------------------ ----------- ----------- ----------- ----------- -----------
Barnum's 50 0 50 0 0
Bookbeat 55 25 30 0 0
Doc-U-Mat: Quality Laundry and Books 85 0 85 0 0
Fricative Bookshop 60 35 0 0 25
Total 250 60 165 0 25
Answer：(可以动态拼接SQL来实现，也可以使用SQL Server 2K5中的PIVOT)
isnull ( [ 1 ] , 0 ) as ' Qtr1 ' , isnull ( [ 2 ] , 0 ) as ' Qtr2 ' , isnull ( [ 3 ] , 0 ) as ' Qtr3 ' , isnull ( [ 4 ] , 0 ) as ' Qtr4 '
from
( select stor_name,sales.stor_id, DATEPART (qq, ord_date) as Quarty,qty from sales,stores
where sales.stor_id = stores.stor_id and year (sales.ord_date) = 1993 ) as tmp
pivot
(
sum (qty) for Quarty in ( [ 0 ] , [ 1 ] , [ 2 ] , [ 3 ] , [ 4 ] )
) as pvt
union
select ' total ' , sum ( isnull ( [ 1 ] , 0 )) + sum ( isnull ( [ 2 ] , 0 )) + sum ( isnull ( [ 3 ] , 0 )) + sum ( isnull ( [ 4 ] , 0 )) as ' Total ' ,
sum ( isnull ( [ 1 ] , 0 )) as ' Qtr1 ' , sum ( isnull ( [ 2 ] , 0 )) as ' Qtr2 ' , sum ( isnull ( [ 3 ] , 0 )) as ' Qtr3 ' , sum ( isnull ( [ 4 ] , 0 )) as ' Qtr4 '
from
( select sales.stor_id, DATEPART (qq, ord_date) as Quarty,qty from sales,stores
where sales.stor_id = stores.stor_id and year (sales.ord_date) = 1993 ) as tmp
pivot
(
sum (qty) for Quarty in ( [ 0 ] , [ 1 ] , [ 2 ] , [ 3 ] , [ 4 ] )
) as pvt
Question 5:
How can I add row numbers to my result set?
In database pubs, have a table titles , now I want the result shown as below,each row have a row number, how can you do that?
Result:
line-no title_id
----------- --------
1 BU1032
2 BU1111
3 BU2075
4 BU7832
5 MC2222
6 MC3021
7 MC3026
8 PC1035
9 PC8888
10 PC9999
11 PS1372
12 PS2091
13 PS2106
14 PS3333
15 PS7777
16 TC3218
17 TC4203
18 TC7777
Answer：
select row_number() as line_no ,title_id from titles
-- SQL Server 2000中可以借助于临时表
select line_no identity ( int , 1 , 1 ),title_id into #t from titles
select * from #t
drop table #t
Question 6:
Can you tell me what the difference of two SQL statements at performance of execution?
Statement 1:
if NOT EXISTS ( select * from publishers where state = 'NY')
begin
SELECT 'Sales force needs to penetrate New York market'
end
else
begin
SELECT 'We have publishers in New York'
end
Statement 2:
if EXISTS ( select * from publishers where state = 'NY')
begin
SELECT 'We have publishers in New York'
end
else
begin
SELECT 'Sales force needs to penetrate New York market'
end
Answer：(我也不知道答案。)
一个说法是NOT EXISTS内部是由EXISTS来实现的，但我在publishers表中构造了4+百万条记录，上面两条SQL的执行计划完全相同(In Server Server 2005)，各执行计划的占50%，并没有发现有差异。
Question 7:
How can I list all California authors regardless of whether they have written a book?
Indatabase pubs, have a table authors and titleauthor , table authors has a column state, and titleauthor have books each author written.
CA behalf of california in table authors.
Answer ：
“标准”答案是：SELECT * FROM authors WHERE state='CA' AND EXISTS
(SELECT * FROM titleauthor WHERE authors.au_id=titleauthor.au_id)
但我觉得题目有问题，按照题目的意思，只要找出State='CA'的Authors就行了(regardless of whether they have written a book)，select * from authors where state='CA'，没有必要去Join其他的表。
Question 8：
How can I get a list of the stores that have bought both 'bussiness' and 'mod_cook' type books?
In database pubs, use three table stores,sales and titles to implement this requestment.
Now I want to get the result as below:
stor_id stor_name
------- ----------------------------------------
...
7896 Fricative Bookshop
...
...
...
Answer：(也可以不用临时表，而改用SQL Server 2005中的CTE)
where sales.title_id = titles.title_id;
select distinct stores.stor_id, stores.stor_name from stores,#SaleTitle
where
' business ' in ( select #SaleTitle. [ type ] from #SaleTitle where stores.stor_id = #SaleTitle.stor_id)
and
' mod_cook ' in ( select #SaleTitle. [ type ] from #SaleTitle where stores.stor_id = #SaleTitle.stor_id);
drop table #SaleTitle;
Question 9:
How can I list non-contignous data?
In database pubs, I create a table test using statement as below, and I insert several row as below
create table test
( id int primary key )
go
insert into test values (1 )
insert into test values (2 )
insert into test values (3 )
insert into test values (4 )
insert into test values (5 )
insert into test values (6 )
insert into test values (8 )
insert into test values (9 )
insert into test values (11)
insert into test values (12)
insert into test values (13)
insert into test values (14)
insert into test values (18)
insert into test values (19)
go
Now I want to list the result of the non-contignous row as below,how can I do it?
Missing after Missing before
------------- --------------
6 8
9 11
...
Answer：
from test as Test1,test as Test2 where Test1.id + 1 not in ( select ID from Test)
and Test2.id = ( select min (id) from test where id > Test1.id)
Question10:
How can I list all book with prices greather than the average price of books of the same type?
Indatabase pubs, have a table named titles , its column named price meanthe price of the book, and another named type mean the type of books.
Now I want to get the result as below:
type title price
------------ -------------------------------------------------------------------------------- ---------------------
business The Busy Executive's Database Guide 19.9900
...
...
...
...
Answer：
AS
(
SELECT [ type ] , AVG (price) AS ' price ' FROM titles GROUP BY [ type ]
)
SELECT titles.type,title,titles.price FROM titles
JOIN AvgPrice ON titles.price > AvgPrice.price AND titles. [ type ] = AvgPrice. [ type ] ;
或：
select a.type,a.title,a.price from titles a,
( select type,price = avg (price) from titles group by type)b
where a.type = b.type and a.price > b.price;