InstantForum.NET v2.9.0SQLServerCentralhttp://www.sqlservercentral.com/Forums/notifications@sqlservercentral.comThu, 23 May 2013 21:43:58 GMT20http://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]Perry Whittle (10/10/2012)[/b][hr][quote][b]GilaMonster (10/10/2012)[/b][hr]Nope. Not for almost 2 years now. That was the old form, 3 weeks training and exams in Redmond.[/quote]Ah right, gotcha.But, back to my original quote, no study material on hex, binary, file offsets or file layouts ;-)[/quote]I think that's classified as 'stuff you should already know'. Paul's course, for eg, if you didn't understand hex then you'd really struggle with the page internals and log internals sections (better part of 2 full days). That said, you could probably pass the exams without knowing the binary structure of the mdf file.Wed, 10 Oct 2012 07:55:18 GMTGilaMonsterhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]GilaMonster (10/10/2012)[/b][hr]Nope. Not for almost 2 years now. That was the old form, 3 weeks training and exams in Redmond.[/quote]Ah right, gotcha.But, back to my original quote, no study material on hex, binary, file offsets or file layouts ;-)Wed, 10 Oct 2012 07:49:16 GMTPerry Whittlehttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxNope. Not for almost 2 years now. That was the old form, 3 weeks training and exams in Redmond. Now it's just 2 exams, if you want training you have to organise and locate, or you can self-study.Wed, 10 Oct 2012 07:41:10 GMTGilaMonsterhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxDon't they run a boot camp in Redmond?Wed, 10 Oct 2012 07:24:15 GMTPerry Whittlehttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]Perry Whittle (10/10/2012)[/b][hr][quote][b]sql.com (10/10/2012)[/b][hr]@ Gail - I was waiting for a MCM ;-)[/quote]Why??i could be wrong but i'm sure I'm not. I wouldn't think that the MCM course will teach you hex, binary, file offsets and the like. You either know it or you don't.[/quote]MCM course?It's self-study these days. Videos sure, various training courses if you pay for them, but your choice.Wed, 10 Oct 2012 04:50:48 GMTGilaMonsterhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]Perry Whittle (10/10/2012)[/b][hr][quote][b]sql.com (10/10/2012)[/b][hr]@ Gail - I was waiting for a MCM ;-)[/quote]Why??i could be wrong but i'm sure I'm not. I wouldn't think that the MCM course will teach you hex, binary, file offsets and the like. You either know it or you don't.[/quote]Just got over-excited. Nothing more, nothing less. I meant no disrespect.Wed, 10 Oct 2012 04:46:23 GMTsql.comhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]GilaMonster (10/10/2012)[/b][hr]Err... No offence, but if you don't understand hex representation, you're going to find the raw page dumps near impossible to decode.You might want to first read up on hexadecimal and binary and how they're represented. Wiki should have something decent.0x30 = 00110000 expressed in binary. It's the first byte of the header. So bits 4 and 5 are 1 and all the others are 0. Those two bits been set to 1 tell me (comparing against the structure of the DB header) that the row has a null bitmap (bit 4) and variable length columns (bit 5)[/quote]Looks like an interesting read, I will read it tonight as evening study.thanks.Wed, 10 Oct 2012 04:45:24 GMTsql.comhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]sql.com (10/10/2012)[/b][hr]@ Gail - I was waiting for a MCM ;-)[/quote]Why??i could be wrong but i'm sure I'm not. I wouldn't think that the MCM course will teach you hex, binary, file offsets and the like. You either know it or you don't.Wed, 10 Oct 2012 04:44:50 GMTPerry Whittlehttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]GilaMonster (10/10/2012)[/b][hr]Err... No offence, but if you don't understand hex representation, you're going to find the raw page dumps near impossible to decode.You might want to first read up on hexadecimal and binary and how they're represented. Wiki should have something decent.0x30 = 00110000 expressed in binary. It's the first byte of the header. So bits 4 and 5 are 1 and all the others are 0. Those two bits been set to 1 tell me (comparing against the structure of the DB header) that the row has a null bitmap (bit 4) and variable length columns (bit 5)[/quote]That makes sense, Im just rusty, I will dig up my notes on oct, hex and binary number representation.thanks.Wed, 10 Oct 2012 04:43:49 GMTsql.comhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]GilaMonster (10/10/2012)[/b][hr]TagA and TagB are just the two bytes (first and second) of the record header, nothing fancier than that. You don't need to decode them, DBCC Page does it for you, it contains info lik "Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP VARIABLE_COLUMNS"The "00000000" is the offset, so byte 0 = offset 00000000, so the first byte of the dump, in this case, Byte 0 is 0x30. The record has a null bitmap and variable length columns (but I didn't need to decode the header for that, DBCC PAge did it for me "Record Attributes = NULL_BITMAP VARIABLE_COLUMNS"[/quote]Argghhh bytes, bits, Hex, offsets :-)Gail has already given you a good starter, the 0x30 is the hex value. The offset is the point in the file at which the record starts and then has a corresponding record length The slots are the records which you can see hold the column data. It's probably worth reading through [url=http://www.sqlservercentral.com/articles/CheckDB/88963/][u][b]this[/b][/u][/url] to get more of an understanding. Playing with a Hex editor on a test database should enlighten you more. Just don't run it on your Production systems or you won't be popular :-)Wed, 10 Oct 2012 04:40:27 GMTPerry Whittlehttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxErr... No offence, but if you don't understand hex representation, you're going to find the raw page dumps near impossible to decode.You might want to first read up on hexadecimal and binary and how they're represented. Wiki should have something decent.0x30 = 00110000 expressed in binary. It's the first byte of the header. So bits 4 and 5 are 1 and all the others are 0. Those two bits been set to 1 tell me (comparing against the structure of the DB header) that the row has a null bitmap (bit 4) and variable length columns (bit 5)Wed, 10 Oct 2012 04:27:29 GMTGilaMonsterhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxThis is more complicated than the Artificial Intelligence Module I did at University few years ago.....:ermm:@ Craig - No issue, Im just finding it frustrating that I cant grasp this.@ Gail - I was waiting for a MCM ;-)Ok, so offset = 00000000 as you stated.Then Bits 1-3 of byte 0 is what? 3,0,0 or 0,0,0 ?It will probably help if I knew what you mean by 0x30? Is it related to 30000800?(might sound stupid to you - sorry)Wed, 10 Oct 2012 04:13:41 GMTsql.comhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxTagA and TagB are just the two bytes (first and second) of the record header, nothing fancier than that. You don't need to decode them, DBCC Page does it for you, it contains info lik "Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP VARIABLE_COLUMNS"The "00000000" is the offset, so byte 0 = offset 00000000, so the first byte of the dump, in this case, Byte 0 is 0x30. The record has a null bitmap and variable length columns (but I didn't need to decode the header for that, DBCC PAge did it for me "Record Attributes = NULL_BITMAP VARIABLE_COLUMNS"Wed, 10 Oct 2012 03:39:13 GMTGilaMonsterhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspx[quote][b]Dr.sql.com (10/10/2012)[/b][hr]again, apologies if this seems trivial, just require some assistance.[/quote]Arguing with DBCC Page is [i]anything[/i] but trivial except for a limited number of people (<1000 at best, in my knowledge) on the planet. All of the planet. The rest of us just muddle through.Is this simply exploration or are you trying to resolve a particular issue?Wed, 10 Oct 2012 03:23:37 GMTEvil Kraig Fhttp://www.sqlservercentral.com/Forums/Topic1370784-391-1.aspxHi SSC,Im not an advanced DBA so apologies if this question seems simple but I would like some guidance. I am trying to understand the output of DBCC page using the following example - from sqlskills.com--------------------------------------------------------------------------Slot 0 Offset 0x60 Length 33Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP VARIABLE_COLUMNSMemory Dump @0x5C76C06000000000: 30000800 05000000 0300f802 00160021 †0..............!00000010: 0042616e 66667369 67687473 6565696e †.Banffsightseein00000020: 67†††††††††††††††††††††††††††††††††††gSlot 0 Column 0 Offset 0x11 Length 5destination = BanffSlot 0 Column 1 Offset 0x16 Length 11activity = sightseeingSlot 0 Column 2 Offset 0x4 Length 4duration = 5-----------------------------------------------------------------------Can someone just point out to me where:> Byte 0 is ? (This will help me to understand where bits 1-3 is to get the record type)> what is meant by TagA byte of the record metadata and TagB?again, apologies if this seems trivial, just require some assistance.thanks,A.Wed, 10 Oct 2012 03:19:49 GMTsql.com