SQL Server 2014 Express。SQL Server - 密鑰更新的死鎖
我我的運動簡化的問題如下:
CREATE TABLE [dbo].[foo](
[fooid] [numeric](10, 0) IDENTITY(1,1) NOT NULL,
[fooval] [nvarchar](4),
CONSTRAINT [foo_PK] PRIMARY KEY CLUSTERED
(
[fooid] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
INSERT INTO [dbo].[foo] ([fooval]) VALUES (1)
GO
INSERT INTO [dbo].[foo] ([fooval]) VALUES (2)
GO
CREATE TABLE [dbo].[bar](
[barid] [numeric](10, 0) IDENTITY(1,1) NOT NULL,
[barval] [nvarchar](4),
CONSTRAINT [bar_PK] PRIMARY KEY CLUSTERED
(
[barid] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
INSERT INTO [dbo].[bar] ([barval]) VALUES (1)
GO
INSERT INTO [dbo].[bar] ([barval]) VALUES (2)
GO
所以我必須對fooid和barid聚集主鍵兩個簡單的表。
我在兩個調試器中運行以下兩個查詢。
第一查詢:
BEGIN TRAN
UPDATE dbo.foo SET fooval = 1 WHERE fooid = 1
UPDATE dbo.bar SET barval = 1 WHERE barval = 1
COMMIT
第二查詢:
BEGIN TRAN
UPDATE dbo.bar SET barval = 2 WHERE barid = 2
UPDATE dbo.foo SET fooval = 2 WHERE fooval = 2
COMMIT
調試時,我執行查詢1的第一更新,然後查詢2的第一更新,查詢1和最後的隨後第二更新第二次更新查詢2.
這會導致死鎖。我正在運行快照隔離級別讀取已提交。
圖表顯示:
<deadlock-list>
<deadlock victim="process2f3ed64e8">
<process-list>
<process id="process2f3ed64e8" taskpriority="0" logused="288" waitresource="KEY: 5:72057607973896192 (227b7397de24)" waittime="2067" ownerId="1978563" transactionname="user_transaction" lasttranstarted="2015-08-24T16:24:57.280" XDES="0x2e2ff23b0" lockMode="U" schedulerid="1" kpid="9892" status="suspended" spid="59" sbid="0" ecid="0" priority="0" trancount="2" lastbatchstarted="2015-08-24T16:24:56.997" lastbatchcompleted="2015-08-24T16:24:56.993" lastattention="1900-01-01T00:00:00.993" clientapp="Microsoft SQL Server Management Studio - Abfrage" hostname="VSL53439" hostpid="9124" loginname="x" isolationlevel="read committed (2)" xactid="1978563" currentdb="5" lockTimeout="4294967295" clientoption1="671088672" clientoption2="128056">
<executionStack>
<frame procname="adhoc" line="6" stmtstart="38" stmtend="146" sqlhandle="0x02000000118b7210fc35334336b07155dea42e1470abe8dd0000000000000000000000000000000000000000">
unknown </frame>
<frame procname="adhoc" line="6" stmtstart="336" stmtend="426" sqlhandle="0x02000000bf0a381fd6fec29b6ed330f87409b4e8c47d26f10000000000000000000000000000000000000000">
unknown </frame>
</executionStack>
<inputbuf>
BEGIN TRAN
UPDATE dbo.bar SET barval = 2 WHERE barid = 2
UPDATE dbo.foo SET fooval = 2 WHERE fooval = 2
COMMIT </inputbuf>
</process>
<process id="process2e01b5088" taskpriority="0" logused="432" waitresource="KEY: 5:72057607973830656 (c939eba47c7b)" waittime="2970" ownerId="1978502" transactionname="user_transaction" lasttranstarted="2015-08-24T16:24:54.100" XDES="0x2df783000" lockMode="U" schedulerid="5" kpid="1928" status="suspended" spid="53" sbid="0" ecid="0" priority="0" trancount="2" lastbatchstarted="2015-08-24T16:24:53.730" lastbatchcompleted="2015-08-24T16:24:53.730" lastattention="1900-01-01T00:00:00.730" clientapp="Microsoft SQL Server Management Studio - Abfrage" hostname="VSL53439" hostpid="4348" loginname="x" isolationlevel="read committed (2)" xactid="1978502" currentdb="5" lockTimeout="4294967295" clientoption1="671088672" clientoption2="128056">
<executionStack>
<frame procname="adhoc" line="6" stmtstart="38" stmtend="146" sqlhandle="0x02000000f8c0c134764c79fe77f7cda514cc62eaf1a50cc80000000000000000000000000000000000000000">
unknown </frame>
<frame procname="adhoc" line="6" stmtstart="336" stmtend="426" sqlhandle="0x020000005c75f728d068a9d6386669fb7b8e315b3e484d640000000000000000000000000000000000000000">
unknown </frame>
</executionStack>
<inputbuf>
BEGIN TRAN
UPDATE dbo.foo SET fooval = 1 WHERE fooid = 1
UPDATE dbo.bar SET barval = 1 WHERE barval = 1
COMMIT </inputbuf>
</process>
</process-list>
<resource-list>
<keylock hobtid="72057607973896192" dbid="5" objectname="dbdevelop.dbo.foo" indexname="foo_PK" id="lock2ea279880" mode="X" associatedObjectId="72057607973896192">
<owner-list>
<owner id="process2e01b5088" mode="X"/>
</owner-list>
<waiter-list>
<waiter id="process2f3ed64e8" mode="U" requestType="wait"/>
</waiter-list>
</keylock>
<keylock hobtid="72057607973830656" dbid="5" objectname="dbdevelop.dbo.bar" indexname="bar_PK" id="lock2eb0e6500" mode="X" associatedObjectId="72057607973830656">
<owner-list>
<owner id="process2f3ed64e8" mode="X"/>
</owner-list>
<waiter-list>
<waiter id="process2e01b5088" mode="U" requestType="wait"/>
</waiter-list>
</keylock>
</resource-list>
</deadlock>
</deadlock-list>
當我看看到鎖定取得我看到下面的鎖已經完成
- 收購 - 九 - OBJECT
- 收購 - IX - 收購 - X - 密鑰
- 收購 - X - EXTENT
- 釋放 - X - EXTENT
- 獲取 - U - EXTENT
- 獲取 - X - PAGE
- 釋放 - U - EXTENT
- 釋放 - X - PAGE
- 釋放 - 0 - KEY
- 發佈 - 0 - 頁
所以,一切都被釋放,除了從開始的對象,這似乎是主鍵索引。我想它會一直保存下來,直到交易完成並且不立即發佈。這似乎導致了僵局。
能否請你回答我以下的問題:
- 我是正確的聚集主鍵索引鎖將保留到提交?
- 我是否正確,這將阻止所有其他併發更新嘗試等待?
- 如果是這樣,爲什麼整個索引在使用where子句中的給定主鍵進行更新時會被鎖定?這意味着主鍵where子句上的每個更新都會鎖定事務的整個表。我無法相信這一點。
- 是在fooval和barval上添加索引的最佳解決方案?
- sql server與sql server express的行爲不同嗎?
這是很多問題。既然你使用兩個調試器來運行單獨的事務,我假設你正在同時運行它們? –
是的,如上所述: 「調試時,我執行查詢1的第一次更新,然後執行查詢2的第一次更新,然後查詢1的第二次更新,最後執行查詢2的第二次更新。」 –
如果你想避免死鎖,你可以嘗試使用「while @@ Trancount <> 0,開始打印'等待'結束」,然後你的交易?這會讓它等到任何交易完成後再嘗試運行另一個交易。我不確定它是否適用 –