Corruption Challenge Week 4 – The Winning Solution

Congratulations to Randolph West who won the corruption challenge this week with the following solution which restored all of the data.

First he restored the database to get started. Note some of his code and comments have been reformatted to better fit in the blog format.

Restore database. I use KEEP_CDC and KEEP_REPLICATION because of the hints you dropped in your blog.
The hint is that there is 100% chance of data recovery, so I will look for Change Data Capture tables.

USE master;
GO
--DROP DATABASE [CorruptionChallenge4] ;
RESTORE DATABASE [CorruptionChallenge4]
   FROM DISK = N'C:\DBBackups\CorruptionChallenge4_Corrupt.bak'
   WITH FILE = 1,
        MOVE N'CorruptionChallenge4'
		  TO N'C:\SQL_DATA\CorruptionChallenge4.mdf',
        MOVE N'UserObjects'
		  TO N'C:\SQL_DATA\CorruptionChallenge4_UserObjects.ndf',
        MOVE N'CorruptionChallenge4_log'
		  TO N'C:\SQL_DATA\CorruptionChallenge4_log.ldf',
        NOUNLOAD,
    REPLACE,
    STATS = 5,
    KEEP_CDC,
    KEEP_REPLICATION;

Next he ran CheckDB to see what is wrong with this database.

USE [CorruptionChallenge4];
-- Run DBCC CHECKDB to see what's wrong
DBCC CHECKDB
WITH ALL_ERRORMSGS,
	NO_INFOMSGS,
	DATA_PURITY

CheckDB returned 17,515 lines of errors.

Ouch, lots of errors. But there’s one object ID that keeps showing up: 2105058535

SELECT *
FROM sys.objects
WHERE [object_id] = 2105058535

dbo.Customers seems to be the problem. Let’s cheat a little and look for a CDC table using this name

SELECT *
FROM sys.tables

dbo_Customers_CT … 373576369 … 2015-05-01 08:19:32.570 … 2015-05-01 15:24:30.177
This looks promising. Let’s get the CDC output for net changes on this table, and dump them into a temp table.
We get the Start LSN from sys.fn_cdc_get_min_lsn(‘Customers’)

SELECT id, FirstName, MiddleName, LastName
INTO tempdb.dbo.fff
FROM [cdc].[fn_cdc_get_net_changes_dbo_Customers](0x0000009A000009700267, sys.fn_cdc_get_max_lsn(), 'all')

Which inserted 511740 rows into the tempdb.dbo.fff table.

Hmmm … how many records is that compared to the original corrupt table?

SELECT COUNT(*) FROM dbo.Customers
SELECT COUNT(*) FROM tempdb.dbo.fff

This looks very promising. Now let’s query the rows at ID 510900 and 510901

SELECT * FROM dbo.Customers -- oops, page 3:22 is corrupt

SELECT * FROM dbo.Customers WHERE id IN (510900, 510901) -- oops, page 3:2150 is corrupt too

During my playing around earlier, I discovered that the two NC indexes were affected by corruption in page 3:88 as well.

SELECT * FROM tempdb.dbo.fff WHERE id IN (510900, 510901)

This is a mess – I guess we’ll have to turf the entire Customers table
But first we should double-check that the Orders table is ok

DBCC CHECKTABLE ('[dbo].[Orders]') WITH ALL_ERRORMSGS, NO_INFOMSGS

[Orders] is good.

Set database in single user mode to enable repair

ALTER DATABASE [CorruptionChallenge4]
SET SINGLE_USER
WITH ROLLBACK IMMEDIATE;

The corrupt table is dbo.Customer
Run repair of dbo.Customer

DBCC CheckTable('dbo.Customers', REPAIR_ALLOW_DATA_LOSS)
WITH NO_INFOMSGS;

EXEC sp_changedbowner 'sa'
GO

-- This sometimes shows up, but should be dropped if CDC is off
DISABLE TRIGGER [tr_MScdc_ddl_event]
 ON DATABASE
GO

DISABLE TRIGGER [noDropTables]
 ON DATABASE
GO

DISABLE TRIGGER [noNewTables]
 ON DATABASE
GO

EXEC sys.sp_cdc_disable_db
GO

Based on the DBCC CheckTable removing all rows from the Customers table, this next section could be skipped.

ALTER TABLE [dbo].[Orders]

DROP CONSTRAINT [FK_Orders_People]
GO

ALTER TABLE [dbo].[Customers]

DROP CONSTRAINT [PK_Customers]
GO

TRUNCATE TABLE [dbo].[Customers]

ALTER TABLE [dbo].[Customers] ADD CONSTRAINT [PK_Customers] PRIMARY KEY CLUSTERED ([id] ASC) ON [UserObjects]
GO

Now to put everything back into the Customers table.

SET IDENTITY_INSERT [dbo].[Customers] ON

INSERT INTO [dbo].[Customers] (
 [id],
 [FirstName],
 [MiddleName],
 [LastName]
 )
SELECT [id],
 [FirstName],
 [MiddleName],
 [LastName]
FROM tempdb.dbo.fff

SET IDENTITY_INSERT [dbo].[Customers] OFF

Turn the triggers back on

ALTER TABLE [dbo].[Orders]
 WITH CHECK ADD CONSTRAINT [FK_Orders_People] FOREIGN KEY ([customerId]) REFERENCES [dbo].[Customers]([id])
GO

ALTER TABLE [dbo].[Orders] CHECK CONSTRAINT [FK_Orders_People]
GO

EXEC sys.sp_cdc_enable_db
GO

EXECUTE sys.sp_cdc_enable_table
 @source_schema = N'dbo',
 @source_name = N'Customers',
 @role_name = NULL;

EXECUTE sys.sp_cdc_enable_table
 @source_schema = N'dbo',
 @source_name = N'Orders',
 @role_name = NULL;

ENABLE TRIGGER [noDropTables]
 ON DATABASE
GO

ENABLE TRIGGER [noNewTables]
 ON DATABASE
GO

ALTER DATABASE [CorruptionChallenge4]
SET MULTI_USER

Not for the checks.

SELECT *
FROM [dbo].[Customers]

SELECT * FROM dbo.Customers WHERE id IN (510900, 510901)

SELECT COUNT(*)
FROM sys.objects
WHERE is_ms_shipped = 0;

And that wraps it up, that is how Randolph West won the Week 4 Database Corruption Challenge.

Related Links:

• Week 4 Building the Corrupt Database
• Week 4 Results
• Week 4 Database Corruption Challenge
• Database Corruption Challenge Scores
• Week 4 Extra Clue
• Is Week 4 Too Diabolical?

The post Corruption Challenge Week 4 – The Winning Solution appeared first on Steve Stedman.