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공지사항

최근에 올라온 글


1. 두 테이블에 인덱스가 없으면, hash 조인으로만 푼다.
   아니다 오히려 hash 로 못 푸는 경우가 발생 할 수도 있다.

if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int
)
go

create table tbly
(idx int
,c1 int
)
go

insert into tblx values (1,1)
insert into tblx values (1,2)
insert into tbly values (1,1)
insert into tbly values (2,null)
go

set statistics profile on

if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int
)
go

create table tbly
(idx int
,c1 int
)
go

insert into tblx values (1,1)
insert into tblx values (1,2)
insert into tbly values (1,1)
insert into tbly values (2,null)
go

set statistics profile on

select *
  from tblx a
  join tbly b
    on a.idx = b.idx
 where a.idx > 1
option (hash join)

select *    from tblx a    join tbly b      on a.idx = b.idx    where a.idx > 1   option (hash join)
  |--Hash Match(Inner Join, HASH:([a].[idx])=([b].[idx]), RESIDUAL:([a].[dbo].[tbly].[idx] as [b].[idx]=[a].[dbo].[tblx].[idx] as [a].[idx]))
       |--Table Scan(OBJECT:([a].[dbo].[tblx] AS [a]), WHERE:([a].[dbo].[tblx].[idx] as [a].[idx]>(1)))
       |--Table Scan(OBJECT:([a].[dbo].[tbly] AS [b]), WHERE:([a].[dbo].[tbly].[idx] as [b].[idx]>(1)))
      
      
select *
  from tblx a
  join tbly b
    on a.idx = b.idx
 where a.idx = 1
option (hash join)

메시지 8622, 수준 16, 상태 1, 줄 1
이 쿼리에 정의된 힌트로 인해 쿼리 프로세서에서 쿼리 계획을 생성할 수 없습니다. 힌트를 지정하거나 SET FORCEPLAN을 사용하지 않고 쿼리를 다시 전송하십시오.


select *
  from tblx a
  join tbly b
    on a.idx = b.idx
 where a.idx = 1

select *    from tblx a    join tbly b      on a.idx = b.idx    where a.idx = 1
  |--Nested Loops(Inner Join) -- 뒤에 어떠한 join predicate 가 없다.
       |--Table Scan(OBJECT:([a].[dbo].[tbly] AS [b]), WHERE:([a].[dbo].[tbly].[idx] as [b].[idx]=(1)))
       |--Table Scan(OBJECT:([a].[dbo].[tblx] AS [a]), WHERE:([a].[dbo].[tblx].[idx] as [a].[idx]=(1)))

hash join 은 적어도 하나의 equijoin predicate 가 있어야만 풀 수 있기 때문이다. merge join 도 마찬가지 이다.

이것은 프로시저로 작성할 때 변수에 의해서 쿼리가 실행되고 안되고 하는 샘플을 만들 수 있고 변수에 따라서 쿼리가 동작 할 수 있고 없는 조건이 발생 하기도 한다.

a.idx between 1 and 100 이 입력되는 경우도 있고, 1 and 1 이라고 입력 된다고 생각해 보라. 이 또한 이행적 폐쇄(transitive closure) 법칙이 적용될 수 있고, 이것은 sql server 에서는 힌트로 막을 방법이 없다. 이러한 변경이 일어나는 경우 hash join, merge join 의 기본 원칙인 equijoin 이 predicate 로 반드시 들어가야 하는 기본 원칙을 위배 하므로 실행 될 수 없다.

(oracle 의 경우 query_rewrite_enabled 파라미터를 이용해 막을 수 있다.)


select *
  from tblx a
  join tbly b
    on a.idx = b.idx
 where a.idx = 1
option (merge join)

메시지 8622, 수준 16, 상태 1, 줄 1
이 쿼리에 정의된 힌트로 인해 쿼리 프로세서에서 쿼리 계획을 생성할 수 없습니다. 힌트를 지정하거나 SET FORCEPLAN을 사용하지 않고 쿼리를 다시 전송하십시오.


select *
  from tblx a
  join tbly b
    on a.idx = b.idx


select *    from tblx a    join tbly b      on a.idx = b.idx
  |--Hash Match(Inner Join, HASH:([b].[idx])=([a].[idx]), RESIDUAL:([pc].[dbo].[tbly].[idx] as [b].[idx]=[pc].[dbo].[tblx].[idx] as [a].[idx]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly] AS [b]))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx] AS [a]))


select *
  from tblx a
  inner loop join tbly b
    on a.idx = b.idx


select *    from tblx a    inner loop join tbly b      on a.idx = b.idx
  |--Nested Loops(Inner Join, WHERE:([pc].[dbo].[tbly].[idx] as [b].[idx]=[pc].[dbo].[tblx].[idx] as [a].[idx]))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx] AS [a]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly] AS [b]))


select *
  from tblx a
  inner merge join tbly b
    on a.idx = b.idx


select *    from tblx a    inner merge join tbly b      on a.idx = b.idx
  |--Merge Join(Inner Join, MANY-TO-MANY MERGE:([a].[idx])=([b].[idx]), RESIDUAL:([pc].[dbo].[tbly].[idx] as [b].[idx]=[pc].[dbo].[tblx].[idx] as [a].[idx]))
       |--Sort(ORDER BY:([a].[idx] ASC))
       |    |--Table Scan(OBJECT:([pc].[dbo].[tblx] AS [a]))
       |--Sort(ORDER BY:([b].[idx] ASC))
            |--Table Scan(OBJECT:([pc].[dbo].[tbly] AS [b]))

또하나의 간단한 이야기를 해보자.

if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int
)
go

create table tbly
(idx int
,c1 int
)
go

insert into tblx values (1,1)
insert into tblx values (2,2)
insert into tbly values (1,1)

select *
  from tblx a
  join tbly b
    on a.idx = 1
option (hash join)
go

select *    from tblx a    join tbly b      on a.idx = 1
  |--Nested Loops(Inner Join) -- 여기에 아무것도 없다.
       |--Table Scan(OBJECT:([a].[dbo].[tblx] AS [a]), WHERE:([a].[dbo].[tblx].[idx] as [a].[idx]=(1)))
       |--Table Scan(OBJECT:([a].[dbo].[tbly] AS [b]))

해당 쿼리는 a.idx = 1 인것을 찾고 그냥 cross join 을 수행한다.
그러므로 hash function 을 수행 할 equijoin predicate 가 역시 없다. 그러므로 hash join 으로 풀 수 없다.
아래 not exists 에서 hash join 으로 못 푸는 것 역시 이렇게 해석 할 수 있다.

그러나 full outer merge join 의 경우 non equijoin 도 가능하기 때문에 다음 쿼리가 가능하다.

select *
  from tblx a
  full outer join tbly b
    on a.idx = 1
option (merge join)


필터 조건이 없으면 hash 만 되는가?
아니다 다음 쿼리의 예제를 보자


select *
  from tblx a
  join tbly b
    on a.idx > b.idx


select *    from tblx a    join tbly b      on a.idx > b.idx
  |--Nested Loops(Inner Join, WHERE:([pc].[dbo].[tblx].[idx] as [a].[idx]>[pc].[dbo].[tbly].[idx] as [b].[idx]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly] AS [b]))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx] AS [a]))


힌트주면 되지요? 
아니다. 해쉬펑션을 통과한 값의 크다 작다를 비교 할 수 있으면 넌 신이다. 난 신 많이 봤다. ;)


select *
  from tblx a
  inner hash join tbly b
    on a.idx > b.idx


메시지 8622, 수준 16, 상태 1, 줄 1
이 쿼리에 정의된 힌트로 인해 쿼리 프로세서에서 쿼리 계획을 생성할 수 없습니다. 힌트를 지정하거나 SET FORCEPLAN을 사용하지 않고 쿼리를 다시 전송하십시오.


2. in 과 not in 은 반대 인가요?
   아니다


다음을 살펴보자.



if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int
)
go

create table tbly
(idx int
,c1 int
)
go

insert into tblx values (1,1)
insert into tblx values (2,2)
insert into tbly values (1,1)
insert into tbly values (2,null)
go



select *
  from tblx
 where c1 in (select c1 from tbly)


idx         c1
----------- -----------
1           1

(1개 행이 영향을 받음)



select *
  from tblx
 where c1 not in (select c1 from tbly)


idx         c1
----------- -----------

(0개 행이 영향을 받음)


2 는 어디로 갔나?
왜 그런가? null 이있다는 생각을 하자 null 의 오묘함에 빠져보면 재미있을 것이다.



3. table 의 check 조건 써야 하나요?
   그렇다. 꼭 써 주는게 좋습니다. 이러한 체크조건이 있냐 없냐에 따라 join 방법도 틀리게 할 수 있다.
   써주면 optimizer 가 매우 효율적으로 동작 할 수 있는 실마리를 준다.


if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int
)
go

create table tbly
(idx int
,c1 int
)
go

insert into tblx values (1,1)
insert into tblx values (2,2)
insert into tbly values (1,1)

go

select *
  from tblx
 where c1 not in (select c1 from tbly)


select *     from tblx    where c1 not in (select c1 from tbly)
  |--Nested Loops(Left Anti Semi Join, WHERE:([pc].[dbo].[tblx].[c1] IS NULL OR [pc].[dbo].[tbly].[c1] IS NULL OR [pc].[dbo].[tblx].[c1]=[pc].[dbo].[tbly].[c1]))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly]))


hash join 으로 풀 수 있는가?

select *
  from tblx
 where c1 not in (select c1 from tbly)
option (hash join)

메시지 8622, 수준 16, 상태 1, 줄 2
이 쿼리에 정의된 힌트로 인해 쿼리 프로세서에서 쿼리 계획을 생성할 수 없습니다. 힌트를 지정하거나 SET FORCEPLAN을 사용하지 않고 쿼리를 다시 전송하십시오.


테이블에 not null 체크 조건이 있으면 어떻게 되는가?
잘푼다.
if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 int not null
)
go

create table tbly
(idx int
,c1 int not null
)
go

insert into tblx values (1,1)
insert into tblx values (2,2)
insert into tbly values (1,1)
go

-- 아까 위헤서 아래 퀴리는 위에서 실행계획을 생성 할 수 없습니다. 그 쿼리이다.


select *
  from tblx
 where c1 not in (select c1 from tbly)
option (hash join)


select *     from tblx    where c1 not in (select c1 from tbly)  option (hash join)
  |--Hash Match(Right Anti Semi Join, HASH:([pc].[dbo].[tbly].[c1])=([pc].[dbo].[tblx].[c1]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly]))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx]))


테이블 제약 조건 주지말고 hash join 을 하고 싶은가?


select *
  from tblx
 where c1 not in (select c1 from tbly where c1 is not null)
   and c1 is not null
option (hash join)


select *     from tblx    where c1 not in (select c1 from tbly where c1 is not null)     and c1 is not null  option (hash join)
  |--Hash Match(Right Anti Semi Join, HASH:([pc].[dbo].[tbly].[c1])=([pc].[dbo].[tblx].[c1]), RESIDUAL:([pc].[dbo].[tblx].[c1]=[pc].[dbo].[tbly].[c1]))
       |--Table Scan(OBJECT:([pc].[dbo].[tbly]), WHERE:([pc].[dbo].[tbly].[c1] IS NOT NULL))
       |--Table Scan(OBJECT:([pc].[dbo].[tblx]), WHERE:([pc].[dbo].[tblx].[c1] IS NOT NULL))


풀리긴 풀리는데 residual 이나 where 필터가 형성된다. 


체크조건이 있으면 residual 등이 없고 깔끔하지 않은가? 어떤가? check 조건 꼭 기술하는게 좋을것 같지 않은가?
이 외 다양한 이슈가 있다. 재미 있는가? 나머지 이슈는 나중에 책을 통해서 만나보자! (sqltag 에서는 sql 책을 준비하고 있어요 ;) 교성아 join 다 적어가나?????


if object_id('tblx') is not null
drop table tblx
go

if object_id('tbly') is not null
drop table tbly
go


create table tblx
(idx int
,c1 char(8000) not null
)
go

create table tbly
(idx int
,c1 char(8000) not null
)
go

insert into tblx values (1,1)
insert into tblx values (2,2)
insert into tbly values (1,1)
go

select *
  from tblx a
  join tbly b
    on a.idx = b.idx
 order by a.idx    
go
 

select *
  from tblx a
 inner hash join tbly b
    on a.idx = b.idx
 order by a.idx    
option(hash join)        

메시지 8618, 수준 16, 상태 2, 줄 1
작업 테이블이 필요하며 최소 행 크기가 허용되는 최대 크기 8060바이트를 초과하므로 쿼리 프로세서에서 쿼리 계획을 생성할 수 없습니다. 작업 테이블이 필요한 일반적인 이유는 쿼리의 GROUP BY 또는 ORDER BY 절 때문입니다. 쿼리에 GROUP BY 또는 ORDER BY 절이 있을 경우 이 절의 필드 수 및/또는 크기를 줄이십시오. 또한 필드의 접두사(LEFT()) 또는 해시(CHECKSUM())를 사용하여 그룹화하거나 접두사를 사용하여 정렬하십시오. 그러나 이 작업을 수행하면 쿼리의 동작이 변경됩니다.

다음은 full outer join 의 실행 예이다.



재미 있지 않은가? join 된 결과를 넣어두고 sort 해야 하는데 두 컬럼의 합이 8060을 넘으니 hash join 한 후 work table 을 만들수 없어 일어나는 에러이다.  

교성이의 join advanced 챕터로 쓸 예정 입니다.


Posted by 보미아빠
, |
Error: 18456, Severity: 14, State: 8 에서 State 를 자세히 봅니다. bol 이나 다음 사이트에서 18456 의 설명을 참고한다.
http://blogs.msdn.com/b/sql_protocols/archive/2006/02/21/536201.aspx

runas 를 통해서 로그인을 가장하고, 11 번 에러를 고의로 발생시킨 경우이다.
그러므로 11번 에러가 나면 다음 그림을 참고해서 데이터베이스 엔진에 연결할 권한을 주도록 한다.

이 외, 윈도우 계정의 암호를 바꾸면 Service 시작 계정의 암호도 바꾸어 주어야 하며, 특별한 경우에는 sql server 에 등록된 윈도우 계정을 삭제 했다가 다시 넣어줘야 되는 경우도 있다.


Aaron is a senior consultant for SQL Sentry, Inc., makers of performance monitoring and event management software for SQL Server, Analysis Services, and Windows. He has been blogging here at sqlblog.com since 2006, focusing on manageability, performance, and new features; has been a Microsoft MVP since 1997; tweets as @AaronBertrand; and speaks frequently at user group meetings and SQL Saturday events.

SQL Server v.Next (Denali) : Troubleshooting Error 18456

I think we've all dealt with error 18456, whether it be an application unable to access SQL Server, credentials changing over time, or a user who can't type a password correctly.  The trick to troubleshooting this error number is that the error message returned to the client or application trying to connect is intentionally vague (the error message is similar for most errors, and the state is always 1).  In a few cases, some additional information is included, but for the most part several of these conditions appear the same to the end user.  In order to figure out what is really going wrong, you need to have alternative access to the SQL Server and inspect the log for the true state in the error message.  I helped our support team just today solve a client's 18456 issues - once we tracked down the error log and saw that it was state 16, it was easy to determine that their login had been setup with a default database that had been detached long ago.

In the next version of SQL Server, there is a new feature called "contained databases" - I've blogged about it here and here.  With this feature comes a new layer of security that may creep onto your radar if you use this functionality: contained user login failures.  There are a variety of things that can go wrong here.  If you connect with a contained user but forget to specify a database name, SQL Server will attempt to authorize you as a SQL login, and you will fail with state 5 (if there is no SQL login with that name) or state 8 (if there is also a SQL login with the same name and the password doesn't match).  There is also a new state 65 which occurs if you have specified the correct username and contained database, but entered an incorrect password.

When I see folks struggling with this problem, I almost always see the answer point to this blog post, which has a very brief partial list and a lot of unanswered questions:

http://blogs.msdn.com/b/sql_protocols/archive/2006/02/21/536201.aspx

So here is what I consider a more complete listing of all the various states for login failures, including changes for Denali (I tried to highlight them, but Community Server doesn't want to obey simple HTML).  I included an instance of 18470 under state 1 for completeness.

State Description Example (note: the verbose message always has [CLIENT: <IP>] suffix)
1 State 1 now occurs when a login is disabled - but actually, the error in the log is 18470, not 18456 - because the login is disabled, it doesn't get that far.  See state

Pre-SQL Server 2005, State 1 always appeared in the log for all login failures, making for fun troubleshooting.  :-)
Error: 18470, Severity: 14, State: 1.
Login failed for user '<x>'.
Reason: The account is disabled.
2 The login (whether using SQL or Windows Authentication) does not exist.  For Windows Auth, it likely means that the login hasn't explicitly been given access to SQL Server - which may mean it is not a member of an appropriate domain group.  State 2 indicates that the login attempt came from a remote machine.
Error: 18456, Severity: 14, State: 2.
Login failed for user '<x>'.
Reason: Could not find a login matching the name provided.
5 Like state 2, the login does not exist in SQL Server, but the login attempt came from the local machine.  For both state 2 and 5, prior to SQL Server 2008, the reason was not included in the error log - just the login failed message.  And starting in Denali, for both state 2 and 5, this error can happen if you specify the correct username and password for a contained database user, but the wrong (or no) database. Error: 18456, Severity: 14, State: 5.
Login failed for user '<x>'.
Reason: Could not find a login matching the name provided.
6 This means you tried to specify SQL authentication but entered a Windows-style login in the form of Domain\Username.  Make sure you choose Windows Authentication (and you shouldn't have to enter your domain / username when using Win Auth).
Error: 18456, Severity: 14, State: 6.
Login failed for user '<x\y>'.
Reason: Attempting to use an NT account name with SQL Server Authentication.
7 The login is disabled *and* the password is incorrect.  This shows that password validation occurs first, since if the password is correct and the login is disabled, you get error 18470 (see state 1 above).
Error: 18456, Severity: 14, State: 7.
Login failed for user '<x>'.
Reason: An error occurred while evaluating the password.
8 Probably the simplest of all: the password is incorrect (cASe sEnsiTiVitY catches a lot of folks here).  Note that it will say "login" even if you attempted to connect as a contained user but forgot to specify a database (or specified the wrong database).
Error: 18456, Severity: 14, State: 8.
Login failed for user '<x>'.
Reason: Password did not match that for the login provided.
9 Like state 2, I have not seen this in the wild. It allegedly means that the password violated a password policy check, but I tried creating a login conforming to a weak password policy, strengthened the policy, and I could still log in fine. And obviously you can't create a login with, or later set, a password that doesn't meet the policy. Let me know if you've seen it. Error: 18456, Severity: 14, State: 9.
???
10 This is a rather complicated variation on state 9; as KB #925744 states, this means that password checking could not be performed because the domain account being used for the SQL Server service is disabled or locked on the domain controller. No reason is given in the verbose message in the error log. Error: 18456, Severity: 14, State: 10.
Login failed for user '<x\y>'.
11 States 11 & 12 mean that SQL Server was able to authenticate you, but weren't able to validate with the underlying Windows permissions. It could be that the Windows login has no profile or that permissions could not be checked due to UAC. Try running SSMS as administrator and/or disabling UAC. Another reason could be that the domain controller could not be reached. You may need to resort to re-creating the login (see this post from Simon Sabin).
Error: 18456, Severity: 14, State: 11.
Login failed for user '<x>'.
Reason: Login-based server access validation failed with an infrastructure error. Check for previous errors.
12 See state 11 above.
Error: 18456, Severity: 14, State: 12.
Login failed for user '<x>'.
Reason: Token-based server access validation failed with an infrastructure error. Check for previous errors.
13 This state occurs when the SQL Server service has been paused (which you can do easily from the context menu in Object Explorer). Error: 18456, Severity: 14, State: 13.
Login failed for user '<x>'.
Reason: SQL Server service is paused. No new connections can be accepted at this time.
16 State 16, which only occurs prior to SQL Server 2008, means that the default database was inaccessible. This could be because the database has been removed, renamed, or is offline. This state does not indicate a reason in the error log.  In 2008 and beyond, this is reported as state 40 (see below), with a reason.  In SQL Server 2005, this state may also be reported if the user's default database is online but the database they explicitly requested is not available (see state 27).
Error: 18456, Severity: 14, State: 16.
Login failed for user '<x>'.
18 Supposedly this indicates that the user needs to change their password. In SQL Server 2005, 2008 R2 and Denali, I found this was raised as error 18488, not 18456; this is because for SQL logins the change password dialog just delays logging in, and is not actually a login failure.  I suspect that, like state 16, this state will go away.
Error: 18456, Severity: 14, State: 18.
???
23 There could be a few reasons for state 23. The most common one is that connections are being attempted while the service is being shut down. However if this error occurs and it is not surrounded in the log by messages about SQL Server shutting down, and there is no companion reason along with the message, I would look at KB #937745, which implies that this could be the result of an overloaded server that can't service any additional logins because of connection pooling issues. Finally, if there *is* a companion reason, it may be the message indicated to the right, indicating that SQL Server was running as a valid domain account and, upon restarting, it can't validate the account because the domain controller is offline or the account is locked or no longer valid. Try changing the service account to LocalSystem until you can sort out the domain issues. Error: 18456, Severity: 14, State: 23.
Login failed for user '<x>'.
Reason: Access to server validation failed while revalidating the login on the connection.
27 State 27, like state 16, only occurs prior to SQL Server 2008. It means that the database specified in the connection string has been removed, renamed, or is offline (though in every case I tried, it was reported as state 16). This state does not indicate a reason in the error log.  In 2008 and onward this is reported as state 38 (see below), with a reason.
Error: 18456, Severity: 14, State: 27.
Login failed for user '<x>'.
28 I have not experienced this issue but I suspect it involves overloaded connection pooling and connection resets. I think you will only see state 28 prioer to SQL Server 2008.
Error: 18456, Severity: 14, State: 28.
Login failed for user '<x>'.
38 The database specified in the connection string, or selected in the Options > Connection Properties tab of the SSMS connection dialog, is no longer valid or online. I came across this once when I typed <default> here instead of picking that option from the list.  This is reported as state 27 or state 16 prior to SQL Server 2008.
Error: 18456, Severity: 14, State: 38.
Login failed for user '<x>'.
Reason: Failed to open the database specified in the login properties.
40 Usually this means the login's default database is offline or no longer exists.  Resolve by fixing the missing database, or changing the login's default database using ALTER LOGIN (for older versions, use sp_defaultdb (deprecated)).  This is reported as state 16 prior to SQL Server 2008.

Error: 18456, Severity: 14, State: 40.
Login failed for user '<x>'.
Reason: Failed to open the explicitly specified database.

46 State 46 may occur when the login (or login mapping to the service account) does not have a valid database selected as their default database. (I am guessing here but I think this may occur when the login in question is attempting to perform log shipping. Again, just a guess based on the few conversations I discovered online.) It can also occur if the classifier function (Resource Governor) Error: 18456, Severity: 14, State: 46.
Login failed for user '<x>'.
Reason: Failed to open the database configured in the login object while revalidating the login on the connection.
50 As the message implies, this can occur if the default collation for the login does not match the collation of their default database (or the database explicitly specified in the connection string). Error: 18456, Severity: 14, State: 50.
Login failed for user '<x>'.
Reason: Current collation did not match the database's collation during connection reset.
51 Like states 11 & 12, this could have to do with UAC, or that the domain controller could not be reached, or that the domain account could not authenticate against the log shipping partner, or that the log shipping partner was down. Try changing the service account for SQL Server to a known domain or local account, rather than the built-in local service accounts, and validating that the partner instance is accessible. Error: 18456, Severity: 14, State: 51.
Login failed for user '<x>'.
Reason: Failed to send an environment change notification to a log shipping partner node while revalidating the login.
56 State 56 is not very common - again, like states 11 & 12, this could have to do with UAC, or that the domain controller could not be reached. Try changing the service account for SQL Server to a known domain or local account, rather than the built-in local service accounts. Error: 18456, Severity: 14, State: 56.
Login failed for user '<x>'.
Reason: Failed attempted retry of a process tokenvalidation.
58 State 58 occurs when SQL Server is set to use Windows Authentication only, and a client attempts to log in using SQL Authentication. Error: 18456, Severity: 14, State: 58.
Login failed for user '<x>'.
Reason: An attempt to login using SQL authentication failed. Server is configured for Windows authentication only.
65
Container user exists, the database is correct, but the password is invalid.
This can also happen if you use a SQL login to connect to a contained database that has a contained user with the same name (one reason why this is not recommended).
Error: 18456, Severity: 14, State: 65.
Login failed for user '<x>'.
Reason: Password did not match that for the user provided. [Database: '<x>']

I am sure I missed some, but I hope that is a helpful summary of most of the 18456 errors you are likely to come across. Please let me know if you spot any inaccuracies or if you know of any states (or reasons) that I missed.

If you are using contained databases in Denali, there will be a little extra complication in solving login failures, especially if you try to create contained users with the same name as server-level logins.  This is a ball of wax you just probably don't want to get into...

Thanks to Jonathan Kehayias (blog | twitter), Bob Ward (CSS blog | twitter), and Rick Byham for helping with sanity checking.

Posted by 보미아빠
, |


누군가 tempdb 가 이상해요 라고 연락이 왔다. 아래 문서와 쿼리를 이용해 문제를 해결한다.



tempdb 를 core 수만큼 만드는 것은 tempdb 유지관리 비용으로 인해 allocation contention 해결 비용보다 더 클 수 도 있다는 설명도 들어있는 좋은 문서이다. workload 에 따라서 튜닝방법은 달라져야 한다. 
 



CREATE database perf_warehouse
GO
USE perf_warehouse
GO

CREATE TABLE tempdb_space_usage (
  -- This represents the time when the particular row was
  -- inserted
  dt datetime DEFAULT CURRENT_TIMESTAMP,
  -- session id of the sessions that were active at the time
  session_id int DEFAULT null,
  -- this represents the source DMV of information. It can be
  -- track instance, session or task based allocation information.
  scope varchar(10), 
  -- instance level unallocated extent pages in tempdb
  Instance_unallocated_extent_pages bigint,
  -- tempdb pages allocated to verstion store
  version_store_pages bigint,
  -- tempdb pages allocated to user objects in the instance
  Instance_userobj_alloc_pages bigint,   
  -- tempdb pages allocated to internal objects in the instance
  Instance_internalobj_alloc_pages bigint,
  -- tempdb pages allocated in mixed extents in the instance
  Instance_mixed_extent_alloc_pages bigint,
  -- tempdb pages allocated to user obejcts within this sesssion or task.
  Sess_task_userobj_alloc_pages bigint,   
  -- tempdb user object pages deallocated within this sesssion
  -- or task.
  Sess_task_userobj_deallocated_pages bigint,
  -- tempdb pages allocated to internal objects within this sesssion
  -- or task
  Sess_task_internalobj_alloc_pages bigint,
  -- tempdb internal object pages deallocated within this sesssion or
  -- task
  Sess_task_internalobj_deallocated_pages bigint,   
  -- query text for the active query for the task 
  query_text nvarchar(max) 
)
go

-- Create a clustered index on time column when the data was collected
CREATE CLUSTERED INDEX cidx ON tempdb_space_usage (dt)
go

 

 

아래 프로시저를 주기적으로 돌려 tempdb 문제를 일으키는 쿼리를 찾는다.

CREATE PROC sp_sampleTempDbSpaceUsage AS
  -- Instance level tempdb File space usage for all files within
  -- tempdb
  INSERT tempdb_space_usage (
    scope,
    Instance_unallocated_extent_pages,
    version_store_pages,
    Instance_userobj_alloc_pages,
    Instance_internalobj_alloc_pages,
    Instance_mixed_extent_alloc_pages)
  SELECT
    'instance',
    SUM(unallocated_extent_page_count),
    SUM(version_store_reserved_page_count),
    SUM(user_object_reserved_page_count),
    SUM(internal_object_reserved_page_count),
    SUM(mixed_extent_page_count)
  FROM sys.dm_db_file_space_usage
   
    -- 2. tempdb space usage per session
    --
  INSERT tempdb_space_usage (
    scope,
    session_id,
    Sess_task_userobj_alloc_pages,
    Sess_task_userobj_deallocated_pages,
    Sess_task_internalobj_alloc_pages,
    Sess_task_internalobj_deallocated_pages)
  SELECT
    'session',
    session_id,
    user_objects_alloc_page_count,
    user_objects_dealloc_page_count,
    internal_objects_alloc_page_count,
    internal_objects_dealloc_page_count
  FROM sys.dm_db_session_space_usage
    WHERE session_id > 50

    -- 3. tempdb space usage per active task
    --
  INSERT tempdb_space_usage (
    scope,
    session_id,
    Sess_task_userobj_alloc_pages,
    Sess_task_userobj_deallocated_pages,
    Sess_task_internalobj_alloc_pages,
    Sess_task_internalobj_deallocated_pages,
    query_text)
  SELECT
    'task',
    R1.session_id,
    R1.user_objects_alloc_page_count,
    R1.user_objects_dealloc_page_count,
    R1.internal_objects_alloc_page_count,
    R1.internal_objects_dealloc_page_count,
    R3.text
  FROM sys.dm_db_task_space_usage AS R1
    LEFT OUTER JOIN
    sys.dm_exec_requests AS R2
    ON R1.session_id = R2.session_id
    OUTER APPLY sys.dm_exec_sql_text(R2.sql_handle) AS R3
  WHERE R1.session_id > 50


Posted by 보미아빠
, |

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