Joh*_*ohn 11 t-sql sql-server locking insert
我们遇到了一些死锁的问题,我发布了这个问题.
通过一些帮助和大量搜索自己,我相信我弄清楚发生了什么.为了在不控制锁升级的情况下解决死锁,我需要理解为什么sql server在插入一行时锁定整个表.
这是我的insert语句(带有重命名的变量):
DECLARE
@Type1 INT = 11,
@Type2 INT = NULL,
@Value1 VARCHAR(20) = '0',
@Value2 VARCHAR(20) = '0',
@Value3 VARCHAR(20) = '0',
@Value4 VARCHAR(20) = '0',
@Date1 DATETIME = '2011-11-25',
@Date2 DATETIME = '2011-11-25',
@Value5 NVARCHAR(50) = '',
@Value6 NVARCHAR(50) = '',
@Type3 INT = NULL,
@Value7 VARCHAR(20) = '4',
@Type4 INT = 4,
@Type5 INT = 15153,
@Type6 INT = 3,
@Type7 INT = 31,
@Type8 INT = 5976,
@Type9 INT = 5044,
@Guid1 UNIQUEIDENTIFIER = 'a8293471-3hb4-442b-844f-44t92f17n67s',
@Value8 VARCHAR(200) = '02jfgg55savolhffr1mkjf45',
@value10 INT = 1,
@Option2 BIT = 0,
@Value9 VARCHAR(20) = null,
@Option1 BIT = 0
insert into dbo.OurTable
(
Type1
,Type2
,Value1
,Value2
,Value3
,Value4
,Date1
,Date2
,Value5
,Value6
,Type3
,Value7
,Type4
,Type5
,Type6
,Type7
,Type8
,Type9
,value10
,Col1
,Col2
,Col3
,Col4
,Value8
,Option2
,Value9
)
values
(
CASE
WHEN [dbo].[GetType](@Type1, null) = 6 AND @Option1 = 1 AND [dbo].[GetType](@Type4, 0) <> 1
THEN 7
ELSE [dbo].[GetType](@Type1, null)
END
,[dbo].[GetType](@Type2, null)
,case when @Value1 = 'null' then null else CAST(@Value1 as numeric(18, 6)) end
,case when @Value2 = 'null' then null else CAST(@Value2 as numeric(18, 6)) end
,case when @Value3 = 'null' then null else CAST(@Value3 as numeric(18, 6)) end
,case when @Value4 = 'null' then null else CAST(@Value4 as numeric(18, 6)) end
,[dbo].[GetDate](@Date1, null)
,[dbo].[GetDate](@Date2, null)
,@Value5
,@Value6
,[dbo].[GetType](@Type3, null)
,case when @Value7 = 'null' then null else CAST(@Value7 as numeric(18, 6)) end
,[dbo].[GetType](@Type4, null)
,@Type6
,case when LOWER(@Type7) = 'null' then null else @Type7 end
,@Type5
,@Type9
,@Type8
,@value10
,GETDATE()
,GETDATE()
,[dbo].[GetGuid](@Guid1)
,[dbo].[GetGuid](@Guid1)
,@Value8
,@Option2
,case when @Value9 = 'null' then null else CAST(@Value9 as int) end
)
Run Code Online (Sandbox Code Playgroud)
如果我在事务中运行此语句,然后在提交之前查询sys.dm_tran_locks,那么我将获得属于该会话的10233行.
SELECT *
FROM sys.dm_tran_locks l
WHERE l.resource_type <> 'DATABASE' AND l.request_session_id = 65
Run Code Online (Sandbox Code Playgroud)
65是测试时当前窗口的会话ID.
另外,如果我查看表锁定(这是我的死锁的原因),我可以看到它在表OurTable上放了一个X锁.
resource_type resource_associated_entity_id Name resource_lock_partition request_mode request_type request_status
OBJECT 290100074 OurTable 0 X LOCK GRANT
OBJECT 290100074 OurTable 1 X LOCK GRANT
OBJECT 290100074 OurTable 2 X LOCK GRANT
OBJECT 290100074 OurTable 3 X LOCK GRANT
OBJECT 290100074 OurTable 4 X LOCK GRANT
OBJECT 290100074 OurTable 5 X LOCK GRANT
OBJECT 290100074 OurTable 6 X LOCK GRANT
OBJECT 290100074 OurTable 7 X LOCK GRANT
OBJECT 290100074 OurTable 8 X LOCK GRANT
OBJECT 290100074 OurTable 9 X LOCK GRANT
OBJECT 290100074 OurTable 10 X LOCK GRANT
OBJECT 290100074 OurTable 11 X LOCK GRANT
OBJECT 290100074 OurTable 12 X LOCK GRANT
OBJECT 290100074 OurTable 13 X LOCK GRANT
OBJECT 290100074 OurTable 14 X LOCK GRANT
OBJECT 290100074 OurTable 15 X LOCK GRANT
Run Code Online (Sandbox Code Playgroud)
我不知道是否由于锁升级或者从开始请求表上的独占锁来完成此操作.无论如何这会导致我遇到麻烦.
单个表上有16个锁定行的原因是锁定分区.
我的问题是,为什么不在桌面上请求一个意图独占锁(IX)?相反,它要求独占锁.我该如何防止这种情况?我在调优顾问中没有任何调整技巧,我已经尝试过了.
编辑 OurTable上有一个插入触发器,用于更新OurTable3上的字段.它看起来像这样:
UPDATE OurTable3 SET Date1 = NULL
FROM OurTable3 as E
JOIN OurTable2 as C on E.Id = C.FKId
JOIN OurTable as ETC on ETC.FKId = C.Id
AND (ETC.Date2 IS NULL OR CAST(ETC.Date2 AS DATE) > E.Date1)
AND ETC.Type1 = 1
Run Code Online (Sandbox Code Playgroud)
正如您所看到的,它不会更新OurTable,而是查询OurTable以更新OurTable3中的正确行.
Joh*_*ohn 14
我找到了答案.我们团队中的开发人员犯了一个小错误(我总是责怪其他人:-).我可能应该已经知道了答案,因为Martin Smith在另一个问题中指出我应该检查ALLOW_ROW_LOCKS和ALLOW_PAGE_LOCKS.但那时我们认为partitionid与索引id相关,我只检查了那个索引.
我所做的是创建一个具有相同数据的新表.效果消失了,我只在新表上有正确的IX锁.然后我创建了每个索引并在每个创建之间进行测试,直到我突然再次产生效果.
我在OurTable上找到了这个索引:
CREATE NONCLUSTERED INDEX [IX_OurTable] ON [dbo].[OurTable]
(
[Col1] ASC,
[Col2] ASC,
[Col3] ASC,
[Col4] ASC,
[Col5] ASC
)
INCLUDE ( [Col6],
[Col7],
[Col8],
[Col9]) WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, SORT_IN_TEMPDB = OFF, IGNORE_DUP_KEY = OFF, DROP_EXISTING = OFF, ONLINE = OFF, ALLOW_ROW_LOCKS = OFF, ALLOW_PAGE_LOCKS = OFF, FILLFACTOR = 90) ON [PRIMARY]
GO
Run Code Online (Sandbox Code Playgroud)
使用ALLOW_ROW_LOCKS = OFF和ALLOW_PAGE_LOCKS = OFF时,很明显我们会对插入和选择产生这种影响.
感谢您的评论,非常感谢Martin真正帮助我解决这些死锁问题.