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mysql oracle

abin — Sun, 11 Oct 2015 14:05:00 GMT

mysql�?nbsp;myisam 引擎不支持事务的概念�Q�多用于数据仓库�q�样查询多而事务少的情况，速度较快�?br />mysql�?nbsp;innoDB 引擎支持事务的概念，多用于web�|�站后台�{�实时的中小型事务处理后台�?br />
而oracle没有引擎的概念，oracle有OLTP和OLAP模式的区分，两者的差别不大�Q�只有参数设�|�上的不同�?br />oracle无论哪种模式都是支持事务概念的，oracle是一个不允许读脏的数据库�pȝ��?br />

当今的数据处理大致可以分成两大类�Q�联��Z��务处理OLTP�Q�on-line transaction processing�Q�、联机分析处理OLAP�Q�On-Line Analytical Processing�Q�。OLTP是传�l�的关系型数据库的主要应用，主要是基本的、日常的事务处理�Q�例如银行交易。OLAP是数据仓库系�l�的主要应用�Q�支持复杂的分析操作�Q�侧重决�{�支持，�q�且提供直观易懂的查询结�?
OLTP:
也称为面向交易的处理�pȝ��Q�其基本特征是顾客的原始数据可以立即传送到计算中心�q�行处理�Q��ƈ在很短的旉��内给出处理结果�?/span>
�q�样做的最大优�Ҏ��可以��x��地处理输入的数据�Q�及时地回答。也�U�Cؓ实时�pȝ��(Real time System)。衡量联��Z��务处理系�l�的一个重要性能指标是系�l�性能�Q�具体体��Cؓ实时响应旉��(Response Time)�Q�即用户在终端上送入数据之后�Q�到计算机对�q�个��h��l�出�{�复所需要的旉��。OLTP是由数据库引擎负责完成的�?/span>
OLTP 数据库旨在��事务应用�E�序仅写入所需的数据，以便��快处理单个事务�?/span>
OLAP:
��写�ؓOLAP,随着数据库技术的发展和应用，数据库存储的数据量从20世纪80�q�代的兆�Q�M�Q�字节及千兆�Q�G�Q�字节过渡到现在的兆兆（T�Q�字节和千兆兆（P�Q�字节，同时�Q�用��L��查询需求也��来��复杂，涉及的已不仅是查询或操纵一张关�p�表中的一条或几条记录�Q�而且要对多张表中千万条记录的数据�q�行数据分析和信息综合，关系数据库系�l�已不能全部满��q�一要求。在国外�Q�不��Y件厂商采取了发展其前端��品来弥补关系数据库管理系�l�支持的不��Q�力囄��一分散的公共应用逻辑�Q�在短时间内响应非数据处理专业�h员的复杂查询要求�?/span>
联机分析处理�Q�OLAP�Q�系�l�是数据仓库�pȝ��最主要的应用，专门设计用于支持复杂的分析操作，侧重对决�{��h员和高层��理人员的决�{�支持，可以�Ҏ��分析人员的要求快速、灵�z�d��q�行大数据量的复杂查询处理，�q�且以一�U�直观而易懂的形式��查询结果提供给决策人员�Q�以便他们准��掌握企业（公司�Q�的�l�营状况�Q�了解对象的需求，制定正确的方案�?/span>

abin 2015-10-11 22:05 发表评论

事务�q�发、事务隔��ȝ��?

abin — Sat, 31 Jan 2015 10:20:00 GMT

�q�发问题可归�U��ؓ以下几类:

A.丢失更新�Q�撤销一个事务时�Q�把其他事务已提交的更新数据覆盖�Q?/span>A�?/span>B事务�q�发执行�Q?/span>A事务执行更新后，提交�Q?/span>B事务�?/span>A事务更新后，B事务�l�束前也做了对该行数据的更新操作�Q�然后回滚，则两�ơ更新操作都丢失了）�?/span>

B.脏读�Q�一个事务读到另一个事务未提交的更新数据（A�?/span>B事务�q�发执行�Q?/span>B事务执行更新后，A事务查询B事务没有提交的数据，B事务回滚�Q�则A事务得到的数据不是数据库中的真实数据。也��是脏数据，卛_��数据库中不一致的数据�Q��?/span>

C.不可重复读：一个事务读到另一个事务已提交的更新数据（A�?/span>B事务�q�发执行�Q?/span>A事务查询数据�Q�然�?/span>B事务更新该数据，A再次查询该数据时�Q�发现该数据变化了）�?/span>

D. 覆盖更新�Q�这是不可重复读中的特例�Q�一个事务覆盖另一个事务已提交的更新数据（�?/span>A事务更新数据�Q�然�?/span>B事务更新该数据，A事务查询发现自己更新的数据变了）�?/span>

E.虚读�Q?/span>�q�读�Q�：一个事务读到另一个事务已提交的新插入的数据（A�?/span>B事务�q�发执行�Q?/span>A事务查询数据�Q?/span>B事务插入或者删除数据，A事务再次查询发现�l�果集中有以前没有的数据或者以前有的数据消�׃��Q��?/span>

数据库系�l�提供了四种事务隔离�U�别供用户选择�Q?/span>

A.Serializable�Q�串行化�Q�：一个事务在执行�q�程中完全看不到其他事务�Ҏ��据库所做的更新�Q�事务执行的时候不允许别的事务�q�发执行。事务串行化执行�Q�事务只能一个接着一个地执行�Q�而不能�ƈ发执行。）�?/span>

B.Repeatable Read�Q�可重复读）�Q�一个事务在执行�q�程中可以看到其他事务已�l�提交的新插入的记录�Q�但是不能看到其他其他事务对已有记录的更新�?/span>

C.Read Commited�Q�读已提交数据）�Q�一个事务在执行�q�程中可以看到其他事务已�l�提交的新插入的记录�Q�而且能看到其他事务已�l�提交的对已有记录的更新�?/span>

D.Read Uncommitted�Q�读未提交数据）�Q�一个事务在执行�q�程中可以看到其他事务没有提交的新插入的记录�Q�而且能看到其他事务没有提交的对已有记录的更新�?/span>

	丢失更新	脏读	非重复读	覆盖更新	�q�d��?/span>
未提交读	Y	Y	Y	Y	Y
已提交读	N	N	Y	Y	Y
可重复读	N	N	N	N	Y
串行�?/span>	N	N	N	N	N

隔离�U�别

数据库系�l�有四个隔离�U�别�Q�大多数数据库默认��别�ؓread commited�Q�。对数据库��用何�U�隔��ȝ�别要审慎分析�Q�因�?/span>

1. �l�护一个最高的隔离�U�别虽然会防止数据的出错�Q�但是却��D��了�ƈ行度的损失，以及��D��死锁出现的可能性增加�?/span>

2. 然而，降低隔离�U�别�Q�却会引起一些难以发现的bug�?/span>

SERIALIZABLE�Q�序列化�Q?/span>

��d��范围锁（比如表锁�Q�页锁等�Q�关于range lock�Q�我也没有很深入的研�IӞ��Q�直到transaction A�l�束。以此阻止其它transaction B�Ҏ��范围内的insert�Q�update�{�操作�?/span>

�q�读�Q�脏读，不可重复�ȝ��问题都不会发生�?/span>

REPEATABLE READ�Q�可重复读）

对于��d��的记录，��d��׃�n锁直到transaction A�l�束。其它transaction B对这个记录的试图修改会一直等待直到transaction A�l�束�?/span>

可能发生的问题：当执行一个范围查询时�Q�可能会发生�q�读�?/span>

READ COMMITTED�Q�提交读�Q?/span>

在transaction A中读取数据时对记录添加共享锁�Q�但��d��l�束立即释放。其它transaction B对这个记录的试图修改会一直等待直到A中的��d��q�程�l�束�Q�而不需要整个transaction A的结束。所以，在transaction A的不同阶�D�对同一记录的读取结果可能是不同的�?/span>

可能发生的问题：不可重复诅R�?/span>

READ UNCOMMITTED(未提交读)

不添加共享锁。所以其它transaction B可以在transaction A对记录的��d��q�程中修改同一记录�Q�可能会��D��A��d��的数据是一个被破坏的或者说不完整不正确的数据�?/span>

另外�Q�在transaction A中可以读取到transaction B�Q�未提交�Q�中修改的数据。比如transaction B对R记录修改了，但未提交。此�Ӟ��在transaction A中读取R记录�Q�读出的是被B修改�q�的数据�?/span>

可能发生的问题：脏读�?/span>

问题

我们看到�Q�当执行不同的隔��ȝ�别时�Q�可能会发生各种各样不同的问题。下面对它们�q�行�ȝ��q��D例说明�?/span>

�q�读

�q�读发生在当两个完全相同的查询执行时�Q�第二次查询所�q�回的结果集跟第一个查询不相同�?/span>

发生的情况：没有范围锁�?/span>

例子�Q?/span>

事务1	事务2
SELECT * FROM users WHERE age BETWEEN 10 AND 30
	INSERT INTO users VALUES ( 3 , 'Bob' , 27 ); COMMIT;
SELECT * FROM users WHERE age BETWEEN 10 AND 30;

如何避免�Q�实行序列化隔离模式�Q�在��M��一个低�U�别的隔��M��都可能会发生�?/span>

不可重复�?/span>

在基于锁的�ƈ行控制方法中�Q�如果在执行select时不��d��读锁�Q�就会发生不可重复读问题�?/span>

在多版本�q�行控制机制中，当一个遇到提交冲�H�的事务需要回退但却被释放时�Q�会发生不可重复读问题�?/span>

事务1	事务2
SELECT * FROM users WHERE id = 1;
	UPDATE users SET age = 21 WHERE id = 1 ; COMMIT; /* in multiversion concurrency/ control, or lock-based READ COMMITTED
SELECT * FROM users WHERE id = 1;
	COMMIT; /* lock-based REPEATABLE READ */

在上面这个例子中�Q�事�?提交成功�Q�它所做的修改已经可见。然而，事务1已经��d��了一个其它的倹{��在序列化和可重复读的隔��ȝ�别中�Q�数据库��理�pȝ��会返回旧��|��卛_��被事�?修改之前的倹{��在提交��d��未提交读隔离�U�别下，可能会返回被更新的��|��q�就�?#8220;不可重复�?#8221;�?/span>

有两个策略可以防止这个问题的发生�Q?/span>

1. 推迟事务2的执行，直至事务1提交或者回退。这�U�策略在使用锁时应用�?悲观锁机�Ӟ��比如用select for update为数据行加上一个排他锁)

2. 而在多版本�ƈ行控制中�Q�事�?可以被先提交。而事�?�Q��l�执行在旧版本的数据上。当事务1�l�于��试提交�Ӟ��数据库会��验它的结果是否和事务1、事�?��序执行时一栗��如果是�Q�则事务1提交成功。如果不是，事务1会被回退。（乐观锁机�Ӟ��

脏读

脏读发生在一个事务A��d��了被另一个事务B修改�Q�但是还未提交的数据。假如B回退�Q�则事务A��d��的是无效的数据。这跟不可重复读�c�M��Q�但是第二个事务不需要执行提交�?nbsp;

事务1	事务2
SELECT * FROM users WHERE id = 1�Q?/span>
	UPDATE users SET age = 21 WHERE id = 1
SELECT FROM users WHERE id = 1�Q?/span>
	COMMIT; /* lock-based DIRTY READ */

abin 2015-01-31 18:20 发表评论

isolation level

abin — Sat, 31 Jan 2015 10:15:00 GMT

ANSI/ISO SQL的四个isolation level

SERIALIZABLE

�q�是最高层�ơisolation level�Q�这个层�ơ的isolation实现了串行化的效果，卻I��几个Transcation在执行时�Q�其执行效果和某个串行序执行�q�几个Transaction的效果是一��L��。��用Serializable层次的事务，其中的查询语句在每次都必��d��同样的数据上执行�Q�从而防止了phantom read�Q�其实现机制是range lock。与下一个层�ơ的不同之处在于range lock�?/p>在基于锁的DBMS中，Serializable直到事务�l�束才释放select数据集上的读、写锁。而且select查询语句必须首先获得range-locks才能执行。在非锁�q�行的DBMS中，�pȝ��每当��到write collision�Ӟ��一�ơ只有一个write能被写入�?/span>

range lock

Range lock记录Serializable Transaction所��d��的key��D��_��L��其他在这个key��D��围内的记录被插入到表中。Range lock位于Index上，记录一个�v始Index和一个终止Index。在Range lock锁定的Index范围内，��M��Update、Insert和Delete操作都是被禁止的�Q�从而保证了Serializable中的操作每次都在同样的数据集上进行�?/p>

REPEATABLE READS

在基于锁的DBMS中，Repeatable Reads直到事务�l�束才释放select数据集上的读、写锁。但不会使用range lock�Q�因此会产生Phantom Read�?/span>

READ COMMITTED

在基于锁的DBMS中，Read Committed直到事务�l�束才释放select数据集上的写锁，但读锁会在一个select完成后才被释放。和Repeatable Read一��P��Read Committed不��用range lock�Q�会产生Phantom read和Unrepeatable read。（注：�q�段主要参考wikipedia�Q�我也搜索过Read Committed�Q�基本都采用了这�U�说法。但我有一个疑问，既然write锁直��C��务提交才释放�Q�那么在�q�个阶段是不会发生update操作的，在一个事务中的多个读应该也不会��生不同的�l�果才对。望知之者指点一二，��弟不胜感激。）

READ UNCOMMITTED

�q�是Isolation Level的最底层�Q�不仅会产生Phantom Read、Unrepeatable Read�Q�还会有Dirty Read的风险�?/span>

Read phenomena

SQL 92标准��事务T1��d��T2的操作分��Z��U�，Phantom Read、Unrepeatable Read和Dirty Read Users
id name age
1 Joe 20
2 Jill 25

id	name	age
1	Joe	20
2	Jill	25

Dirty reads (Uncommitted Dependency)

事务T1在读取事务T2已经修改、但T2�q�未提交的数据时会发生Dirty Read。这个Isolation Level唯一能保证的是Update操作按顺序执行，即事务中前面的update一定比后面的update先执行�?/p>

下面的这个例子是一个典型的Dirty Read

Transaction 1	Transaction 2
/* Query 1 / SELECT age FROM users WHERE id = 1; / will read 20 */
	/* Query 2 / UPDATE users SET age = 21 WHERE id = 1; / No commit here */
/* Query 1 / SELECT age FROM users WHERE id = 1; / will read 21 */
	ROLLBACK; /* lock-based DIRTY READ */

Non-repeatable reads

在一个事务T1中，如果对同一条记录读取两�ơ而��g��一��P��那么��发生了Non-repeatable read�?/p>

在基于锁的DBMS中，Non-repeatable read可能在未获得read lock时进行select操作�Q�或在select操作刚结束就释放read lock时发生。在多版本�ƈ行控制的DBMS中，non-repeatable read可能在违�?#8220;受commit conflict影响的事务必��d��?#8220;的原则时发生�?br />

Transaction 1	Transaction 2
/* Query 1 / SELECT FROM users WHERE id = 1;
	/* Query 2 / UPDATE users SET age = 21 WHERE id = 1; COMMIT; / in multiversion concurrency control, or lock-based READ COMMITTED */
/* Query 1 / SELECT FROM users WHERE id = 1; COMMIT; /* lock-based REPEATABLE READ */

Transaction 1

Transaction 2

/* Query 1 */ SELECT * FROM users WHERE id = 1;

/* Query 2 */ UPDATE users SET age = 21 WHERE id = 1; COMMIT; /* in multiversion concurrency    control, or lock-based READ COMMITTED */

/* Query 1 */ SELECT * FROM users WHERE id = 1; COMMIT; /* lock-based REPEATABLE READ */

有两�U�方式来防止non-repeatable read。第一�U�方式用锁��T1和T2串行。另外一�U�方式是在多版本�q�行控制的DBMS中��用的。在�q�里�Q�T2可以提交�Q�而先于T2启动的T1则在自己的Snapshot�Q�快照）里��l�执行，在T1提交�Ӟ��pȝ��会检查其�l�果是否与T1、T2序执行的�l�果一��P��如果一��P��则T1提交�Q�否则T1必须回滚�q�生成一个serialization failure�?br />

Phantom reads

Phantom read是指在一个事务中�Q�执行两个或多个同样的查询返回的�l�果集却不相同的现象。这�U�现象发生在没获得range lock卌��行select... where....操作�Ӟ��解决�Ҏ��是��用range lock�?br />

Transaction 1	Transaction 2
/* Query 1 / SELECT FROM users WHERE age BETWEEN 10 AND 30;
	/* Query 2 */ INSERT INTO users VALUES ( 3, 'Bob', 27 ); COMMIT;
/* Query 1 / SELECT FROM users WHERE age BETWEEN 10 AND 30;

参考：

http://msdn.microsoft.com/en-us/library/ms191272.aspx
http://en.wikipedia.org/wiki/Isolation_%28database_systems%29

abin 2015-01-31 18:15 发表评论

sql 解释执行计划

abin — Fri, 27 Dec 2013 17:40:00 GMT

//oracle解释执行计划
SQL> explain plan for select t.* from abin1 t where t.id<200;

Explained

SQL> select * from table(dbms_xplan.display);

PLAN_TABLE_OUTPUT

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

Plan hash value: 1270356885

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

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

| 0 | SELECT STATEMENT | | 98 | 1862 | 2 (0)| 00:00:01 |

|* 1 | TABLE ACCESS FULL| ABIN1 | 98 | 1862 | 2 (0)| 00:00:01 |

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

Predicate Information (identified by operation id):

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

1 - filter("T"."ID"<200)

13 rows selected

SQL>

//mysql解释执行计划

explain select t.* from user_rest t where id<>8

//oracle攉��表的�l�计信息

analyze table abin1 compute statistics;

abin 2013-12-28 01:40 发表评论

abin — Thu, 25 Apr 2013 06:29:00 GMT

查看oracle表结构：
desc ABIN_LEE;
select * from user_tab_columns where table_name='ABIN_LEE';
select dbms_metadata.get_ddl('TABLE','ABIN_LEE') from dual;
select * from cols where table_name= 'ABIN_LEE';

abin 2013-04-25 14:29 发表评论

数据库连接池参数

abin — Wed, 27 Mar 2013 02:32:00 GMT

　数据库连接池在初始化的时候会创徏initialSize个连接，当有数据库操作时�Q�会从池中取��Z��个连接。如果当前池中正在��用的�q�接数等于maxActive�Q�则会等待一�D�|��_��{�待其他操作释放掉某一个连接，如果�q�个�{�待旉��过了maxWait�Q�则会报错；如果当前正在使用的连接数没有辑ֈ�maxActive�Q�则判断当前是否�I�闲�q�接�Q�如果有则直接��用空闲连接，如果没有则新建立一个连接。在�q�接使用完毕后，不是��其物理�q�接关闭�Q�而是��其攑օ�池中�{�待其他操作复用�?br />　　对于一个数据库�q�接池，一般包括一下属性�?br />　　
　　
　　名称　　说明　　备注
minIdle 最��活跃数可以允许的空闲连接数�?br />maxActive 最大连接数相当于池的大��?br />initialSize 初始化连接大��?
maxWait 获取�q�接的最大等待时�?
timeBetweenEvictionRunsMillis �I�闲�q�接的最大生存时�?指定�I�闲�q�接在空闲多长时间后�Q�关闭其物理�q�接
testWhileIdle 是否在空闲时��连接可用�?�׃��|�络或者数据库配置的原因（比如mysql�q�接�?��时限制�Q�，开启这个开兛_��以定期检��连接的可用�?br />

abin 2013-03-27 10:32 发表评论

sql去除重复语句

abin — Wed, 21 Nov 2012 15:20:00 GMT

sql 单表/多表查询去除重复记录

单表distinct

多表group by

group by 必须攑֜� order by �?limit之前�Q�不然会报错

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

1、查找表中多余的重复记录�Q�重复记录是�Ҏ��单个字段�Q�peopleId�Q�来判断

select * from people
where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)

2、删除表中多余的重复记录�Q�重复记录是�Ҏ��单个字段�Q�peopleId�Q�来判断�Q�只留有rowid最��的记录
delete from people
where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)
and rowid not in (select min(rowid) from people group by peopleId having count(peopleId )>1)

3、查找表中多余的重复记录�Q�多个字�D�）
select * from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)

4、删除表中多余的重复记录�Q�多个字�D�）�Q�只留有rowid最��的记录
delete from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)
and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(*)>1)

5、查找表中多余的重复记录�Q�多个字�D�）�Q�不包含rowid最��的记录
select * from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)
and rowid not in (select min(rowid) from vitae group by peopleId,seq having count(*)>1)

(�?
比方�?br />在A表中存在一个字�D?#8220;name”�Q?br />而且不同记录之间�?#8220;name”值有可能会相同，
现在��是需要查询出在该表中的各记录之间�Q?#8220;name”值存在重复的��；
Select Name,Count(*) From A Group By Name Having Count(*) > 1

如果�q�查性别也相同大则如�?
Select Name,sex,Count(*) From A Group By Name,sex Having Count(*) > 1

(�?
�Ҏ��一

declare @max integer,@id integer

declare cur_rows cursor local for select ��d��D?count(*) from 表名 group by ��d��D?having count(*) >�Q?1

open cur_rows

fetch cur_rows into @id,@max

while @@fetch_status=0

begin

select @max = @max -1

set rowcount @max

delete from 表名 where ��d��D?= @id

fetch cur_rows into @id,@max
end

close cur_rows

set rowcount 0

�Ҏ��?/p>

�Q�重复记录＂有两个意义上的重复记录，一是完全重复的记录�Q�也��x��有字�D�均重复的记录，二是部分关键字段重复的记录，比如Name字段重复�Q�而其他字�D�不一定重复或都重复可以忽略�?/p>

1、对于第一�U�重复，比较�Ҏ��解决�Q��?/p>

select distinct * from tableName

��可以得到无重复记录的结果集�?/p>

如果该表需要删除重复的记录�Q�重复记录保�?条）�Q�可以按以下�Ҏ��删除

select distinct * into #Tmp from tableName

drop table tableName

select * into tableName from #Tmp
drop table #Tmp

发生�q�种重复的原因是表设计不周��生的�Q�增加唯一索引列即可解冟�?/p>

2、这�c�重复问题通常要求保留重复记录中的�W�一条记录，操作�Ҏ��如下

假设有重复的字段为Name,Address�Q�要求得到这两个字段唯一的结果集

select identity(int,1,1) as autoID, * into #Tmp from tableName

select min(autoID) as autoID into #Tmp2 from #Tmp group by Name,autoID

select * from #Tmp where autoID in(select autoID from #tmp2)

最后一个select卛_��C��Name�Q�Address不重复的�l�果集（但多了一个autoID字段�Q�实际写时可以写在select子句中省��L��列）

(�?
查询重复

select * from tablename where id in (select id from tablename

group by id

having count(id) > 1

)

3、查找表中多余的重复记录�Q�多个字�D�）
select * from vitae a
where (a.peopleId,a.seq) in (select peopleId,seq from vitae group by peopleId,seq having count(*) > 1)

�q�行会��生问题，where(a.peopleId,a.seq)�q�样的写发是通不�q�的�Q�！�Q?/p>

abin 2012-11-21 23:20 发表评论

一道关于员工与部门查询的SQL�W�试�?

abin — Wed, 18 Jul 2012 06:05:00 GMT

create table dept
(
deptno varchar(10) primary key,
dname varchar(10)
);
create table emp
(
empno varchar(10) primary key,
ename varchar(10),
job varchar(10),
mgr varchar(10),
sal varchar(10),
deptno varchar(10) references dept(deptno)
);
drop table dept;
drop table emp;
insert into dept values ('1','事业�?);
insert into dept values ('2','销售部');
insert into dept values ('3','技术部');

insert into emp values ('01','jacky','clerk','tom','1000','1');
insert into emp values ('02','tom','clerk','','2000','1');
insert into emp values ('07','biddy','clerk','','2000','1');
insert into emp values ('03','jenny','sales','pretty','600','2');
insert into emp values ('04','pretty','sales','','800','2');
insert into emp values ('05','buddy','jishu','canndy','1000','3');
insert into emp values ('06','canndy','jishu','','1500','3');
select * from dept;
select * from emp;

1.列出emp表中各部门的部门��P��最高工资，最低工�?/p>

select e.deptno ,max(sal) ,min(sal) from emp as e group by e.deptno;

解析�Q?span style="color: rgb(255,0,0)">各部�?/span> 提示�Q�分�l�查询，所以用group by

2 列出emp表中各部门job�?CLERK'的员工的最低工资，最高工�?br /> select e.deptno ,max(sal) ,min(sal) from emp as e where e.job="CLERK" group by e.deptno;

解析�Q?span style="color: rgb(255,0,0)">各部�?/span> 提示�Q�分�l�查询， job提示�Q�条件查询，所以用�?group by �?where

3 对于emp中最低工资小�?000的部门，列出job�?CLERK'的员工的部门��P��最低工资，最高工�?/p>

select b.deptno ,max(sal) ,min(sal) from emp as b where b.job="CLERK"
and (select min(sal) from emp as a where a.deptno=b.deptno)<2000 group by b.deptno;

解析�Q�以上面两个题�ؓ基础�Q?#8220;emp中最低工资小�?000的部�?/span>”只是一个条件而已�?/p>

4 据部门号由高而低�Q�工资有低而高列出每个员工的姓名，部门��P��工资

select * from emp as b order by b.deptno desc ,b.sal asc;

解析�Q?#8220;部门��L��高而低�Q�工资有�?/span>”提示�Q�需要排序，所以用�?order by �?desc asc

5 列出'buddy'所在部门中每个员工的姓名与部门�?/p>

select * from emp as b where b.deptno=(select a.deptno from emp as a where a.ename="buddy");

解析�Q?#8220;'buddy'所在部�?#8221;提示扑և�他所在的部门��P��所以可以理�?#8220;扑և�所有和buddy在同一部门的所有�h”

6 列出每个员工的姓名，工作�Q�部门号�Q�部门名

select ename,job,dept.deptno,dept.dname from emp,dept where emp.deptno=dept.deptno;

7 列出emp中工作�ؓ'CLERK'的员工的姓名�Q�工作，部门��P��部门�?/p>

select ename,job,dept.deptno,dept.dname from emp,dept where emp.deptno=dept.deptno where job="CLERK";

8 对于emp中有��理者的员工�Q�列出姓名，��理者姓名（��理者外键�ؓmgr�Q?/p>

select a.deptno,a.ename ,b.ename from emp as a,emp as b where a.mgr is not null and a.mgr=b.ename;

解析�Q�本句可以理�?#8220;一个员工他也是领导�Q�所以emp.ename,emp.mgr 都有值�ƈ且有相等的员�?#8221;

9 对于dept表中�Q�列出所有部门名�Q�部门号�Q�同时列出各部门工作�?CLERK'的员工名与工�?/p>

select a.deptno ,a.dname, b.ename ,b.job from dept as a,emp as b where a.deptno=b.deptno and b.job="clerk";

解析�Q�连表即�?br />

http://blog.csdn.net/shiyuntian_wang/article/details/4636799

��Q?/p>

DEPARTMENTS: DEPARTMENT_ID(primary key), DEPARTMENT_NAME, LOCATION

EMPLOYEES: EMPLOYEE_ID(primary key), EMPLOYEE_NAME, EMPLOYEE_JOB, MANAGER, SALARY, DEPARTMENT_ID

列出EMPLOYEES表中各部门的部门��P��最高工资，最低工�?/li>

select max(SALARY) as 最高工�?min(SALARY) as 最低工�?DEPARTMENT_ID from EMPLOYEES group by DEPARTMENT_ID;

列出EMPLOYEES表中各部门EMPLOYEE_JOB�?CLERK'的员工的最低工资，最高工�?/li>

select max(SALARY) as 最高工�?min(SALARY) as 最低工�?DEPARTMENT_ID as 部门�?from EMPLOYEES where EMPLOYEE_JOB = 'CLERK' group by DEPARTMENT_ID;

对于EMPLOYEES中最低工资小�?000的部门，列出EMPLOYEE_JOB�?CLERK'的员工的部门��P��最低工资，最高工�?/li>

select max(SALARY) as 最高工�?min(SALARY) as 最低工�?DEPARTMENT_ID as 部门�?from EMPLOYEES as b where EMPLOYEE_JOB ='CLERK' and 1000>(select min(SALARY) from EMPLOYEES as a where a.DEPARTMENT_ID = b.DEPARTMENT_ID) group by b.DEPARTMENT_ID

�Ҏ��部门��L��高而低�Q�工资有低而高列出每个员工的姓名，部门��P��工资

select DEPARTMENT_ID as 部门�?EMPLOYEE_NAME as 姓名,SALARY as 工资 from EMPLOYEES order by DEPARTMENT_ID desc,SALARY asc

列出'张三'所在部门中每个员工的姓名与部门�?/li>

select EMPLOYEE_NAME,DEPARTMENT_ID from EMPLOYEES where DEPARTMENT_ID = (select DEPARTMENT_ID from EMPLOYEES where EMPLOYEE_NAME = '张三')

列出每个员工的姓名，工作�Q�部门号�Q�部门名

select EMPLOYEE_NAME,EMPLOYEE_JOB,EMPLOYEES.DEPARTMENT_ID,DEPARTMENTS.DEPARTMENT_NAME from EMPLOYEES,DEPARTMENTS where EMPLOYEES.DEPARTMENT_ID = DEPARTMENTS.DEPARTMENT_ID

列出EMPLOYEES中工作�ؓ'CLERK'的员工的姓名�Q�工作，部门��P��部门�?/li>

select EMPLOYEE_NAME,EMPLOYEE_JOB,DEPARTMENTS.DEPARTMENT_ID,DEPARTMENT_NAME from EMPLOYEES,DEPARTMENTS where DEPARTMENTS.DEPARTMENT_ID = EMPLOYEES.DEPARTMENT_ID and DEPARTMENT_JOB = 'CLERK'

对于EMPLOYEES中有��理者的员工�Q�列出姓名，��理者姓名（��理者外键�ؓMANAGER�Q?/li>

select a.EMPLOYEE_NAME as 姓名,b.EMPLOYEE_NAME as ���理�?from EMPLOYEES as a,EMPLOYEES as b where a.MANAGER is not null and a.MANAGER = b.EMPLOYEE_ID

对于DEPARTMENTS表中�Q�列出所有部门名�Q�部门号�Q�同时列出各部门工作�?CLERK'的员工名与工�?/li>

select DEPARTMENT_NAME as 部门�?DEPARTMENTS.DEPARTMENT_ID as 部门�?EMPLOYEE_NAME as 员工�?EMPLOYEE_JOB as 工作 from DEPARTMENTS,EMPLOYEES where DEPARTMENTS.DEPARTMENT_ID *= EMPLOYEES.DEPARTMENT_ID and EMPLOYEE_JOB = 'CLERK'

对于工资高于本部门��^均水�q�的员工�Q�列出部门号�Q�姓名，工资�Q�按部门��h��?/li>

select a.DEPARTMENT_ID as 部门�?a.EMPLOYEE_NAME as 姓名,a.SALARY as 工资 from EMPLOYEES as a where a.SALARY>(select avg(SALARY) from EMPLOYEES as b where a.DEPARTMENT_ID = b.DEPARTMENT_ID) order by a.DEPARTMENT_ID

对于EMPLOYEES�Q�列出各个部门中�q�_��工资高于本部门��^均水�q�的员工数和部门��P��按部门号排序

select count(a.SALARY) as 员工�?a.DEPARTMENT_ID as 部门�?from EMPLOYEES as a where a.SALARY>(select avg(SALARY) from EMPLOYEES as b where a.DEPARTMENT_ID = b.DEPARTMENT_ID) group by a.DEPARTMENT_ID order by a.DEPARTMENT_ID

对于EMPLOYEES中工资高于本部门�q�_��水��^�Q��h数多�?人的�Q�列出部门号�Q��h敎ͼ�按部门号排序

select count(a.EMPLOYEE_ID) as 员工�?a.DEPARTMENT_ID as 部门�?avg(SALARY) as �q�_��工资 from EMPLOYEES as a where (select count(c.EMPLOYEE_ID) from EMPLOYEES as c where c.DEPARTMENT_ID = a.DEPARTMENT_ID and c.SALARY>(select avg(SALARY) from EMPLOYEES as b where c.DEPARTMENT_ID = b.DEPARTMENT_ID))>1 group by a.DEPARTMENT_ID order by a.DEPARTMENT_ID

对于EMPLOYEES中低于自己工资至��?人的员工�Q�列出其部门��P��姓名�Q�工资，以及工资��于自己的�h�?/li>

select a.DEPARTMENT_ID,a.EMPLOYEE_NAME,a.SALARY,(select count(b.EMPLOYEE_NAME) from EMPLOYEES as b where b.SALARY < a.SALARY) as 人数 from EMPLOYEES as a where (select count(b.EMPLOYEE_NAME) from EMPLOYEES as b where b.SALARY5

http://blog.sina.com.cn/s/blog_4979ec3e0100r574.html

abin 2012-07-18 14:05 发表评论

sql 常见面试题一

abin — Sun, 08 Jul 2012 02:39:00 GMT

　�l�常会看到这��L��SQL面试题，��L��一条SQL语句�l�计出学生的��L��L��l�，��L��一条sql语句删除表中重复的内容，但第一条保留。最�q�得�Ԍ��p��着写了�q�么个demo�Q�今天来�q�和大家分��n下，如果大家有其他的sql题也可以拿出来大家一赯��论，一起分享�?

先创��Z��个表�Q?/p>

CREATE TABLE [dbo].[Score](
    [ID] [int] IDENTITY(1,1) PRIMARY KEY NOT NULL,
    [Name] [nvarchar](50) NULL,
    [CID] [int] NULL,
    [Score] [int] NULL
�Q?br />

INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('张三',1,60)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('张三',2,70)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('张三',3,80)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('张三',4,90)

INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('李四',1,60)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('李四',2,70)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('李四',3,80)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('李四',4,90)

INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('王五',1,60)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('王五',2,70)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('王五',3,80)
INSERT INTO [Test].[dbo].[Score]([Name],[CID],[Score])VALUES('王五',4,90)

好了�Q�准备工作做完了�Q�下面我们来写两条Sql语句�Q�解军_��始提出的那两个问�?/p>

A、统计学生的成�W

select name,SUM(Score)Score from Score group by Name

B、删除表中重复的记录�Q�因��表中的name是有重复的，所以我们就直接用这表来test

delete from Score where Name in 
(select Name from Score group by Name having COUNT(name)>0)and ID 
 not in (select MIN(id) from Score group by Name having COUNT(Name)>0)

好了�Q�两条语句解决了两个问题.

abin 2012-07-08 10:39 发表评论

abin — Sun, 08 Jul 2012 02:26:00 GMT

Sql常见面试题（�ȝ��Q?br />1.用一条SQL语句查询出每门课都大�?0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    ��p��    90

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

2.学生�?如下:
自动�~�号学号姓名评��~�号评��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�~�号不同,其他都相同的学生冗余信息

A: delete tablename where 自动�~�号 not in(select min(自动�~�号) from tablename group by 学号,姓名,评��~�号,评��名称,分数)
一个叫department的表�Q�里面只有一个字�D�name,一共有4条纪录，分别是a,b,c,d,对应四个球对�Q�现在四个球对进行比赛，用一条sql语句昄��所有可能的比赛�l�合.
你先按你自己的想法做一下，看结果有我的�q�个��单吗�Q?br />�{�：select a.name, b.name
from team a, team b
where a.name < b.name

��L��SQL语句实现�Q�从TestDB数据表中查询出所有月份的发生额都�?01�U�目相应月䆾的发生额高的�U�目。请注意�Q�TestDB中有很多�U�目�Q�都�?�Q?2月䆾的发生额�?br />AccID�Q�科目代码，Occmonth�Q�发生额月䆾�Q�DebitOccur�Q�发生额�?br />数据�?/span>名：JcyAudit�Q�数据集�Q�Select * from TestDB
�{�：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�?br />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�q�样一个结�?br />year m1  m2  m3  m4
1991 1.1 1.2 1.3 1.4
1992 2.1 2.2 2.3 2.4

�{�案一�?br />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

�q�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;
************************************************************************************
�_�֦�的SQL语句�Q?br />�_�֦�SQL语句
作者：不详发文旉��Q?003.05.29 10:55:05

说明�Q�复制表(只复制结�?源表名：a 新表名：b)

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

说明�Q�拷贝表(拯��数据,源表名：a 目标表名�Q�b)

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

说明�Q�显�C�文章、提交�h和最后回复时�?

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�Q�外�q�接查询(表名1�Q�a 表名2�Q�b)

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�Q�日�E�安排提前五分钟提醒

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

说明�Q�两张关联表�Q�删除主表中已经在副表中没有的信�?

SQL:

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

说明�Q?-

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 �Q?�Q?

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

WHERE A.NUM = B.NUM

说明�Q?-

SQL:

select * from studentinfo where not exists(select * from student where studentinfo.id=student.id) and �p�d��U?'"&strdepartmentname&"' and 专业名称='"&strprofessionname&"' order by 性别,生源�?高考��L��l?

说明�Q?

从数据库中去一�q�的各单位电话费�l�计(电话费定额贺电化肥清单两个表来源�Q?

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�Q�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�Q�得到表中最��的未��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A和B�Q�均有key和value两个字段�Q�如果B的key在A中也有，��把B的value换�ؓA中对应的value
�q�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��sql面试�?br />原表:
courseid coursename score
-------------------------------------
1 java 70
2 oracle 90
3 xml 40
4 jsp 30
5 servlet 80
-------------------------------------
��Z��便于阅读,查询此表后的�l�果昑ּ�如下(及格分数�?0):
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�?br />没有装Ｏ�QԌ��Q�Ｌ�Q�，没试�q?
select courseid, coursename ,score ,decode�Q�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�?

id pro1 pro2
1 M F
2 N G
3 B A
写出查询语句
解决�Ҏ��

sql求解
表a
�?a1 a2
记录 1 a
1 b
2 x
2 y
2 z
用select能选成以下�l�果吗？
1 ab
2 xyz
使用pl/sql代码实现�Q�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�?
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;

用sql�Q?

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
先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
*******************************************************************************
一个SQL 面试�?
��d��应聘一个职位未�?光��被考了一个看似简单的�?但我没有扑ֈ�好的大案.
不知各位大虾有无好的解法?

题�ؓ:
有两个表, t1, t2,
Table t1:

SELLER | NON_SELLER
----- -----

A B
A C
A D
B A
B C
B D
C A
C B
C D
D A
D B
D C

Table t2:

SELLER | COUPON | BAL
----- --------- ---------
A 9 100
B 9 200
C 9 300
D 9 400
A 9.5 100
B 9.5 20
A 10 80

要求用SELECT 语句列出如下�l�果:------如A的SUM(BAL)为B,C,D的和,B的SUM(BAL)为A,C,D的和.......
且用的方法不要增加数据库负担,如用临时表等.

NON-SELLER| COUPON | SUM(BAL) ------- --------
A 9 900
B 9 800
C 9 700
D 9 600
A 9.5 20
B 9.5 100
C 9.5 120
D 9.5 120
A 10 0
B 10 80
C 10 80
D 10 80
关于论坛上那个SQL微��Y面试�?br />问题�Q?br />
一百个账户各有100$�Q�某个�̎��h��天如有支出则��d��一条新记录�Q�记录其余额。一癑֤�后，误��出每天所有�̎��L��余额信息

�q�个问题的难点在于每个用户在某天可能有多条纪录，也可能一条纪录也没有�Q�不包括�W�一天）

�q�回的记录集是一�?00�?100个用��L��U�录�?br />
下面是我的思�\�Q?br />
1.创徏表�ƈ插入��试数据�Q�我们要求username�?-100
CREATE TABLE [dbo].[TABLE2] (
[username] [varchar] (50) NOT NULL , --用户�?br />[outdate] [datetime] NOT NULL , --日期
[cash] [float] NOT NULL --余额
) ON [PRIMARY

declare @i int
set @i=1
while @i<=100
  begin
insert table2 values(convert(varchar(50),@i),'2001-10-1',100)
insert table2 values(convert(varchar(50),@i),'2001-11-1',50)
set @i=@i+1
  end
insert table2 values(convert(varchar(50),@i),'2001-10-1',90)

select * from table2 order by outdate,convert(int,username)

2.�l�合查询语句�Q?br />a.我们必须�q�回一个从�W�一天开始到100天的�U�录集：
如：2001-10-1�Q�这个日期是��L��的） �?2002-1-8
�׃��W�一天是��L��一天，所以我们需要下面的SQL语句�Q?br />select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
�q�里的奥妙在于：
convert(int,username)-1�Q�记得我们指定用户名�?-100 :-))
group by username,min(outdate):�W�一天就可能每个用户有多个纪录�?br />�q�回的结果：
outdate
------------------------------------------------------
2001-10-01 00:00:00.000
.........
2002-01-08 00:00:00.000

b.�q�回一个所有用户名的纪录集�Q?br />select distinct username from table2
�q�回�l�果�Q?br />username
--------------------------------------------------
1
10
100
......
99

c.�q�回一�?00天记录集�?00个用戯��录集的笛卡尔集合�Q?br />select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as A
CROSS join
(
select distinct username from table2
) as B
order by outdate,convert(int,username)
�q�回�l�果100*100条纪录：
outdate                         username
2001-10-01 00:00:00.000          1
......
2002-01-08 00:00:00.000          100

d.�q�回当前所有用户在数据库的有的�U�录�Q?br />select outdate,username,min(cash) as cash from table2
group by outdate,username

order by outdate,convert(int,username)
�q�回�U�录�Q?br />outdate                         username cash
2001-10-01 00:00:00.000          1       90
......
2002-01-08 00:00:00.000          100       50

e.��c中返回的�W�卡��集和d中返回的�U�录做left join:
select C.outdate,C.username,
D.cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as A
CROSS join
(
select distinct username from table2
) as B
) as C
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as D
on(C.username=D.username and datediff(d,C.outdate,D.outdate)=0)

order by C.outdate,convert(int,C.username)
注意�Q�用户在当天如果没有�U�录�Q�cash字段�q�回NULL�Q�否则cash�q�回每个用户当天的余�?br />outdate                         username cash
2001-10-01 00:00:00.000          1       90
2001-10-01 00:00:00.000          2       100
......
2001-10-02 00:00:00.000          1       90
2001-10-02 00:00:00.000          2       NULL  <--注意�q�里
......

2002-01-08 00:00:00.000          100       50

f.好了�Q�现在我们最后要做的��是�Q�如果cash为NULL�Q�我们要�q�回��于当前�U�录日期的第一个用户余�?�׃��我们使用order by cash,所以返回top 1�U�录卛_��Q��用min应该也可�?�Q�这个余额即为当前的余额�Q?br />case isnull(D.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=C.username
and datediff(d,C.outdate,table2.outdate)<0
order by table2.cash
)
else D.cash
end as cash

g.最后组合的完整语句��是
select C.outdate,C.username,
case isnull(D.cash,0)
when 0 then
(
select top 1 cash from table2 where table2.username=C.username
and datediff(d,C.outdate,table2.outdate)<0
order by table2.cash
)
else D.cash
end as cash
from
(
select * from
(
select top 100 dateadd(d,convert(int,username)-1,min(outdate)) as outdate
from table2
group by username
order by convert(int,username)
) as A
CROSS join
(
select distinct username from table2
) as B
) as C
left join
(
select outdate,username,min(cash) as cash from table2
group by outdate,username
) as D
on(C.username=D.username and datediff(d,C.outdate,D.outdate)=0)

order by C.outdate,convert(int,C.username)

�q�回�l�果�Q?br />outdate                               username       cash
2001-10-01 00:00:00.000 1                   90
2001-10-01 00:00:00.000 2                100
......
2002-01-08 00:00:00.000 100             50
***********************************************************************************
取出sql表中�W?1�?0的记录（以自动增长ID��Z��键）
*从数据表中取出第n条到�W�m条的记录*/

declare @m int
declare @n int
declare @sql varchar(800)
set @m=40
set @n=31
set @sql='select top '+str(@m-@n+1) + '* from idetail where autoid not in(
select top '+ str(@n-1) + 'autoid from idetail)'
exec(@sql)

select top 10 * from t where id not in (select top 30 id from t order by id ) orde by id
--------------------------------------------------------------------------------
select top 10 * from t where id in (select top 40 id from t order by id) order by id desc

*******************************************************************************
一道面试题,写sql语句

有表a存储二叉树的节点,要用一条sql语句查出所有节点及节点所在的�?
表a
c1 c2 A ----------1
---- ---- / \
A B B C --------2
A C / / \
B D D N E ------3
C E / \ \
D F F K I ---4
E I
D K
C N

所要得到的�l�果如下

jd cs
----- ----
A 1
B 2
C 2
D 3
N 3
E 3
F 4
K 4
I 4
有高手指��g��?我只能用pl/sql写出�?��h��用一条sql语句的写�?br />SQL> select c2, level + 1 lv
2 from test start
3 with c1 = 'A'
4 connect by c1 = prior c2
5 union
6 select 'A', 1 from dual
7 order by lv;

C2 LV
-- ----------
A 1
B 2
C 2
D 3
E 3
N 3
F 4
I 4
K 4

已选择9行�?/div>

abin 2012-07-08 10:26 发表评论

abin — Sun, 08 Jul 2012 02:26:00 GMT

Student(S#,Sname,Sage,Ssex) 学生�?br /> Course(C#,Cname,T#) 评��?br /> SC(S#,C#,score) 成�W�?br /> Teacher(T#,Tname) 教师�?

问题�Q?br /> 1、查�?#8220;001”评��?#8220;002”评��成�W高的所有学生的学号�Q?br /> select a.S#
from (select s#,score from SC where C#=’001′) a,
(select s#,score from SC where C#=’002′) b
where a.score>b.score and a.s#=b.s#;

2、查询��^均成�l�大�?0分的同学的学号和�q�_��成�W�Q?br /> select S#,avg(score)
from sc
group by S# having avg(score) >60;

3、查询所有同学的学号、姓名、选课数、��L��l�；
select Student.S#,Student.Sname,count(SC.C#),sum(score)
from Student left Outer join SC on Student.S#=SC.S#
group by Student.S#,Sname

4、查询姓“�?#8221;的老师的个敎ͼ�
select count(distinct(Tname))
from Teacher
where Tname like ‘�?’;

5、查询没学过“叶��^”老师评��同学的学受��姓名；
select Student.S#,Student.Sname
from Student
where S# not in (select distinct( SC.S#) from SC,Course,Teacher where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname=’叶��^’);

6、查询学�q?#8220;001”�q�且也学�q�编�?#8220;002”评��的同学的学号、姓名；
select Student.S#,Student.Sname
from Student,SC
where Student.S#=SC.S# and SC.C#=’001′and exists( Select * from SC as SC_2 where SC_2.S#=SC.S# and SC_2.C#=’002′);
7、查询学�q?#8220;叶��^”老师所教的所有课的同学的学号、姓名；
select S#,Sname
from Student
where S# in
(select S#
from SC ,Course ,Teacher
where SC.C#=Course.C# and Teacher.T#=Course.T# and Teacher.Tname=’叶��^’ group by S# having count(SC.C#)=(select count(C#) from Course,Teacher where Teacher.T#=Course.T# and Tname=’叶��^’));

8、查询所有课�E�成�l�小�?0分的同学的学受��姓名；
select S#,Sname
from Student
where S# not in (select Student.S# from Student,SC where S.S#=SC.S# and score>60);

9、查询没有学全所有课的同学的学号、姓名；
select Student.S#,Student.Sname
from Student,SC
where Student.S#=SC.S#
group by Student.S#,Student.Sname having count(C#) <(select count(C#) from Course);

10、查询至��有一门课与学号�ؓ“1001”的同学所学相同的同学的学号和姓名�Q?br /> select S#,Sname
from Student,SC
where Student.S#=SC.S# and C# in �Q�select C# from SC where S#='1001'�Q?

11、删除学�?#8220;叶��^”老师评��SC表记录；
Delect SC
from course ,Teacher
where Course.C#=SC.C# and Course.T#= Teacher.T# and Tname='叶��^';

12、查询各�U�成�l�最高和最低的分：以如下�Ş式显�C�：评��ID�Q�最高分�Q�最低分
SELECT L.C# 评��ID,L.score 最高分,R.score 最低分
FROM SC L ,SC R
WHERE L.C# = R.C#
and
L.score = (SELECT MAX(IL.score)
FROM SC IL,Student IM
WHERE IL.C# = L.C# and IM.S#=IL.S#
GROUP BY IL.C#)
and
R.Score = (SELECT MIN(IR.score)
FROM SC IR
WHERE IR.C# = R.C#
GROUP BY IR.C# );

13、查询学生��^均成�l�及其名��?br /> SELECT 1+(SELECT COUNT( distinct �q�_��成�W)
FROM (SELECT S#,AVG(score) �q�_��成�W
FROM SC
GROUP BY S# ) T1
WHERE �q�_��成�W > T2.�q�_��成�W) 名次, S# 学生学号,�q�_��成�W
FROM (SELECT S#,AVG(score) �q�_��成�W FROM SC GROUP BY S# ) T2
ORDER BY �q�_��成�W desc;

14、查询各�U�成�l�前三名的记�?(不考虑成�W�q�列情况)
SELECT t1.S# as 学生ID,t1.C# as 评��ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 3 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC)
ORDER BY t1.C#;

15、查询每门功成�W最好的前两�?br /> SELECT t1.S# as 学生ID,t1.C# as 评��ID,Score as 分数
FROM SC t1
WHERE score IN (SELECT TOP 2 score
FROM SC
WHERE t1.C#= C#
ORDER BY score DESC )
ORDER BY t1.C#;

补充�Q?br /> 已经知道原表
year salary
——————
2000 1000
2001 2000
2002 3000
2003 4000

解：
select b.year,sum(a.salary)
from salary a,salary b
where a.year<=b.year
group by b.year
order by b.year;

在面试过�E�中多次��到一道SQL查询的题目，查询A(ID,Name)表中�W?1�?0条记录，ID作�ؓ主键可能是不是连�l�增长的列，完整的查询语句如下：
�Ҏ��一�Q?br /> select top 10 *
from A
where ID >(select max(ID) from (select top 30 ID from A order by ID ) T) order by ID
�Ҏ��二：
select top 10 *
from A
where ID not In (select top 30 ID from A order by ID)
order by ID

abin 2012-07-08 10:26 发表评论

国产高清精品在线,久久精品亚洲热,99热99re6国产在线播放

mysql oracle

事务�q�发、事务隔��ȝ��?

isolation level

ANSI/ISO SQL的四个isolation level

SERIALIZABLE

range lock

REPEATABLE READS

在基于锁的DBMS中，Repeatable Reads直到事务�l�束才释放select数据集上的读、写锁。但不会使用range lock�Q�因此会产生Phantom Read�?/span>

READ COMMITTED

READ UNCOMMITTED

�q�是Isolation Level的最底层�Q�不仅会产生Phantom Read、Unrepeatable Read�Q�还会有Dirty Read的风险�?/span>

Read phenomena

SQL 92标准���事务T1��d��T2的操作分��Z���U�，Phantom Read、Unrepeatable Read和Dirty Read Usersidnameage1Joe202Jill25

Dirty reads (Uncommitted Dependency)

Non-repeatable reads

Phantom reads

sql 解释执行计划

数据库连接池参数

sql去除重复语句

一道关于员工与部门查询的SQL�W�试�?

sql 常见面试题一

SQL 92标准��事务T1��d��T2的操作分��Z��U�，Phantom Read、Unrepeatable Read和Dirty Read Users
id name age
1 Joe 20
2 Jill 25