Updating values in sql

Posted by / 28-Oct-2017 02:14

Updating values in sql

Transact-SQL (T-SQL) supports two important date/time functions -- DATEADD and DATEDIFF -- to enable you to easily perform these sorts of calculations.In this, part four of my series on working with datetime values, I'll explain how to use these two functions and provide examples of how they work.There is no measurable performance penalty from having a default value. The ‘default’ value is pushed into the row, even if the column was not modified.Consider this update: Key Hash Value = (8194443284a0) Slot 1 Offset 0x60 Length 15 Record Type = PRIMARY_RECORD Record Attributes = NULL_BITMAP Record Size = 15 Memory Dump @0x000000000AB8A060 0000000000000000: 10000c00 02000000 83010000 020000†††††††††††††........ƒ......To demonstrate the functions, I used the following T-SQL code to create the Sales.Orders table in the Adventure Works sample database: The table definition contains the Order Date and Deliv Date columns, both configured with the DATETIME data type.

Lets look at a simple example, we’ll create a table with some rows and then add a non-NULL column with default values.First create and populate the table: dbcc traceon (3604,-1) dbcc page(6,1,217,3) Page @0x0000000170D5E000 m_page Id = (7) m_header Version = 1 m_type = 1 m_type Flag Bits = 0x0 m_level = 0 m_flag Bits = 0x200 m_obj Id (Alloc Unit Obj) = 84 m_index Id (Alloc Unit Ind) = 256 Metadata: Alloc Unit Id = 72057594043432960 Metadata: Partition Id = 72057594039042048 Metadata: Index Id = 1 Metadata: Object Id = 245575913 m_prev Page = (0:0) m_next Page = (0) pminlen = 12 m_slot Cnt = 476 m_free Cnt = 4 m_free Data = 7236 m_reserved Cnt = 0 m_xact Reserved = 0 m_xdes Id = (0:0) m_ghost Rec Cnt = 0 m_torn Bits = 2135435720 DB Frag ID = 1 Allocation Status GAM (1:2) = ALLOCATED SGAM (1:3) = ALLOCATED PFS (1:1) = 0x60 MIXED_EXT ALLOCATED 0_PCT_FULL DIFF (1:6) = CHANGED ML (1:7) = NOT MIN_LOGGED Slot 0 dbcc traceon (3604,-1) dbcc page(6,1,217,3) Page @0x0000000170D5E000 m_page Id = (7) m_header Version = 1 m_type = 1 m_type Flag Bits = 0x0 m_level = 0 m_flag Bits = 0x200 m_obj Id (Alloc Unit Obj) = 84 m_index Id (Alloc Unit Ind) = 256 Metadata: Alloc Unit Id = 72057594043432960 Metadata: Partition Id = 72057594039042048 Metadata: Index Id = 1 Metadata: Object Id = 245575913 m_prev Page = (0:0) m_next Page = (0) pminlen = 12 m_slot Cnt = 476 m_free Cnt = 4 m_free Data = 7236 m_reserved Cnt = 0 m_xact Reserved = 0 m_xdes Id = (0:0) m_ghost Rec Cnt = 0 m_torn Bits = 2135435720 DB Frag ID = 1 Allocation Status GAM (1:2) = ALLOCATED SGAM (1:3) = ALLOCATED PFS (1:1) = 0x60 MIXED_EXT ALLOCATED 0_PCT_FULL DIFF (1:6) = CHANGED ML (1:7) = NOT MIN_LOGGED Slot 0 The page header is unchanged, the last LSN is still (), proof that the page was not modified, and the physical record is unchanged and has the same size as before. If you pay attention you’ll notice that the Column 3 though has an Offset 0x0 and a physical length of 0.Column 3 is somehow materialized out of thin air, as it does not physically exists in the record on this page.Only rows existing in the table at the time of running ALTER TABLE statement will have missing ‘default’ values.By contrast the default constraint can be dropped or modified and new rows inserted after the ALTER TABLE will always have a value present in row for the new column.

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There is no penalty from having a missing value in a row.

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