A checksum value is tagged with every data page in SQL Server. This is used to detect issues on the storage system to ensure what has been read from the disk is what has been written to the disk. When the checksum is detected which is not the one it’s supposed to be, the page will not be read into the buffer pool. SQL Server will return error. If you are sure the data on the page are all correct or the checksum is broken purposely, you do have the option to fix it even you don’t know how to calculate it.
Now let’s create our experimental database to get this fun stuff done.
use master drop database Test1 create database Test1 go use Test1 go create table t1(Field1 char(10)) go insert into t1 values('AAAAAAAAAA') select * from t1 /* Field1 ---------- AAAAAAAAAA (1 row(s) affected) */
Now let’s break it
-- first find out where it is use master go alter database Test1 set single_user with rollback immediate go dbcc traceon(3604) go dbcc ind(test1, t1,0) -- page 118 in my case go -- then, check the record dbcc page (test1, 1, 118, 1) /* Slot 0, Offset 0x60, Length 17, DumpStyle BYTE Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 17 Memory Dump @0x0000000010F0A060 0000000000000000: 10000e00 41414141 41414141 41410100 00 ....AAAAAAAAAA... */ --Now let's modify one of the A to B dbcc writepage(test1, 1, 118, 100, 1, 0x42, 1) --Notice that the last parameter I used is 1 not the default value --this will bypass the the buffer pool and direct write the data to the page on disk --Buffer pool calculates the checksum before writting. If I bypass it --it will mean the checksum is broken. go --Let's check the page again dbcc page (test1, 1, 118, 1) /* Slot 0, Offset 0x60, Length 17, DumpStyle BYTE Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 17 Memory Dump @0x0000000009D7A060 0000000000000000: 10000e00 42414141 41414141 41410100 00 ....BAAAAAAAAA... */ go
Ok, now we successfully break the page. if you run DBCC CheckDB(Test1), you will see following errors
Msg 8939, Level 16, State 98, Line 1
Table error: Object ID 245575913, index ID 0, partition ID 72057594039042048, alloc unit ID 72057594043432960 (type In-row data), page (1:118). Test (IS_OFF (BUF_IOERR, pBUF->bstat)) failed. Values are 2057 and -4.
Msg 8928, Level 16, State 1, Line 1
Object ID 245575913, index ID 0, partition ID 72057594039042048, alloc unit ID 72057594043432960 (type In-row data): Page (1:118) could not be processed. See other errors for details.
It does not give you too much information. What if you run
select * from Test1.dbo.t1. The error you you get is very mouthful
Msg 824, Level 24, State 2, Line 1
SQL Server detected a logical consistency-based I/O error: incorrect checksum (expected: 0x5f1b00a8; actual: 0x5f1a80a8). It occurred during a read of page (1:118) in database ID 5 at offset 0x000000000ec000 in file 'F:\Program Files\Microsoft SQL Server\MSSQL11.SQL2012\MSSQL\DATA\Test1.mdf'. Additional messages in the SQL Server error log or system event log may provide more detail. This is a severe error condition that threatens database integrity and must be corrected immediately. Complete a full database consistency check (DBCC CHECKDB). This error can be caused by many factors; for more information, see SQL Server Books Online.
In this error message, there are very meaningful messages delivered – (expected: 0x5f1b00a8; actual: 0x5f1a80a8). One of those 2 numbers must be correct. Those 2 numbers are integer numbers – least significant digits will be saved at lower address. So the values of storage format of those numbers are a8001b5f and a8801a5f. Now let’s fire up DBCC PAGE to see whether we can find one of them. If so, the other one of those 2 will be correct checksum.
dbcc page (test1, 1, 118, 2) /* 000000000FA3A000: 01010000 00820001 00000000 00000e00 00000000 ................... 000000000FA3A014: 00000100 54000000 8d1f7100 76000000 01000000 ....T.....q.v....... 000000000FA3A028: 21000000 52000000 18000000 00000000 00000000 !...R............... 000000000FA3A03C: a8001b5f 00000000 00000000 00000000 00000000 ¨.._................ */
Look at the last line above from DBCC PAGE result. At offset 3C, you can see A8001B5F – aha, that’s the incorrect checksum. Now let’s correct it with value a8801a5f
dbcc writepage(test1, 1, 118, 60, 4, 0xa8801a5f, 1) go dbcc checkdb(test1) go select * from test1..t1 /* Field1 ---------- BAAAAAAAAA (1 row(s) affected) */
Now the error goes away and you can see the modified record. CheckSum is managed by buffer pool. DBCC WritePage with bufferpool(last parameter set to 0) can’t modify that value. You will have to use 1 for the last parameter under single database mode. Enjoy.
John Huang – SQL MCM & MVP, http://www.sqlnotes.info
8 thoughts on “Fix Page Checksum”
hi sir, how can i convert that 2 integers to a storage format so i can check it in the result of dbcc page
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hi john, how do u find the particular offset from the dbcc page result
No, I don’t know where it is. basically I search it based on the result of dbcc checkdb.
I know this is an old article but this is the most relevant article that could possibly helped me with my current issue. I just need to understand how you manage to convert the following:
0x5f1b00a8 = a8001b5f
0x5f1a80a8 = a8801a5f
This the way how numbers are stored…
Thank you so much for replying. But truthfully, i still can’t figure out. 🙁
Here’s a snippet of the error I’m getting “SQL Server detected a logical consistency-based I/O error: incorrect checksum (expected: 0x00011b5b; actual: 0x7b45003b). It occurred during a read of page (1:50942796) in database ID 12 at offset 0x0000612a698000”.
I followed the steps in your article but can’t figure out how to convert 0x00011b5b and 0x7b45003b the way you did. 🙁
Thank you for the very useful information you have provided.
I would love to learn SQL Server from scratch
It is possible for you to start teaching content from scratch or provide us with resources in PDF format.
thank you so much