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问题:
How to I determine the maximum row size for a table? I'm looking for a tool or script that does this so I don't have to add up each column's size by hand.
My goal is to produce a report of tables that are too wide so we can look into restructing them. I know we have several that are so wide that only 1 row fits on each 8K page, but I want to find the rest.
回答1:
Another way, run this then look at MaximumRecordsize
dbcc showcontig ('YourTableNameHere') with tableresults
回答2:
As the DBCC SHOWCONTIG is going to be deprecated, Microsoft
advises to use the sys.dm_db_index_physical_stats instead. You can get the maximum, minimum and average row size for each index and other stats, too:
SELECT * FROM sys.dm_db_index_physical_stats (DB_ID(N'AdventureWorks2014'), OBJECT_ID(N'[sales].[SalesOrderHeader]'), NULL, NULL , 'DETAILED')
Output:
database_id object_id index_id partition_number index_type_desc alloc_unit_type_desc index_depth index_level avg_fragmentation_in_percent fragment_count avg_fragment_size_in_pages page_count avg_page_space_used_in_percent record_count ghost_record_count version_ghost_record_count min_record_size_in_bytes max_record_size_in_bytes avg_record_size_in_bytes forwarded_record_count compressed_page_count
----------- ----------- ----------- ---------------- ------------------------------------------------------------ ------------------------------------------------------------ ----------- ----------- ---------------------------- -------------------- -------------------------- -------------------- ------------------------------ -------------------- -------------------- -------------------------- ------------------------ ------------------------ ------------------------ ---------------------- ---------------------
6 1266103551 1 1 CLUSTERED INDEX IN_ROW_DATA 3 0 0.439882697947214 28 24.3571428571429 682 98.8644304423029 31465 0 0 154 195 171.487 NULL 0
6 1266103551 1 1 CLUSTERED INDEX IN_ROW_DATA 3 1 0 2 1 2 54.7442550037064 682 0 0 11 11 11 NULL 0
6 1266103551 1 1 CLUSTERED INDEX IN_ROW_DATA 3 2 0 1 1 1 0.296515937731653 2 0 0 11 11 11 NULL 0
6 1266103551 2 1 NONCLUSTERED INDEX IN_ROW_DATA 2 0 0 4 25.5 102 99.0670990857425 31465 0 0 24 24 24 NULL 0
6 1266103551 2 1 NONCLUSTERED INDEX IN_ROW_DATA 2 1 0 1 1 1 31.4801087225105 102 0 0 23 23 23 NULL 0
6 1266103551 3 1 NONCLUSTERED INDEX IN_ROW_DATA 2 0 0 3 36.3333333333333 109 99.8363355572029 31465 0 0 26 26 26 NULL 0
6 1266103551 3 1 NONCLUSTERED INDEX IN_ROW_DATA 2 1 0 1 1 1 36.3355572028663 109 0 0 25 25 25 NULL 0
6 1266103551 4 1 NONCLUSTERED INDEX IN_ROW_DATA 2 0 0 2 27.5 55 98.9285025945145 31465 0 0 12 12 12 NULL 0
6 1266103551 4 1 NONCLUSTERED INDEX IN_ROW_DATA 2 1 0 1 1 1 11.527057079318 55 0 0 15 15 15 NULL 0
6 1266103551 5 1 NONCLUSTERED INDEX IN_ROW_DATA 2 0 0 2 27.5 55 98.9285025945145 31465 0 0 12 12 12 NULL 0
6 1266103551 5 1 NONCLUSTERED INDEX IN_ROW_DATA 2 1 0 1 1 1 13.5656041512231 55 0 0 18 18 18 NULL 0
回答3:
Here is another query I got, but like Tomalak's it is broken because it doesn't take into account things like the overhead for variable length columns.
SELECT OBJECT_NAME (id) tablename
, COUNT (1) nr_columns
, SUM (length) maxrowlength
FROM syscolumns
GROUP BY OBJECT_NAME (id)
ORDER BY OBJECT_NAME (id)
回答4:
Have a look at INFORMATION_SCHEMA.COLUMNS
. You can add the size of each column in a table to give you an idea which tables have the potential to overshoot the 8k/row rule.
SELECT * FROM INFORMATION_SCHEMA.COLUMNS
You will figure out the rest when you see the query result.
回答5:
Might not the best but this is showing the number of rows + for columns with datatype date,time and so on
SELECT
t.NAME AS TableName,
s.Name AS SchemaName,
p.rows AS RowCounts,
SUM(a.total_pages) * 8 AS TotalSpaceKB,
CAST(ROUND(((SUM(a.total_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS TotalSpaceMB,
SUM(a.used_pages) * 8 AS UsedSpaceKB,
CAST(ROUND(((SUM(a.used_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS UsedSpaceMB,
(SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB,
CAST(ROUND(((SUM(a.total_pages) - SUM(a.used_pages)) * 8) / 1024.00, 2) AS NUMERIC(36, 2)) AS UnusedSpaceMB
FROM
sys.tables t
INNER JOIN
sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN
sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
INNER JOIN
sys.allocation_units a ON p.partition_id = a.container_id
LEFT OUTER JOIN
sys.schemas s ON t.schema_id = s.schema_id
INNER JOIN sysobjects so on t.object_id = so.id
INNER JOIN syscolumns SC on (so.id = sc.id)
INNER JOIN systypes st on (st.type = sc.type)
WHERE
t.NAME NOT LIKE 'dt%'
AND t.is_ms_shipped = 0
AND i.OBJECT_ID > 255
AND so.type = 'U'
and st.name IN ('DATETIME', 'DATE', 'TIME')
GROUP BY
t.Name, s.Name, p.Rows
ORDER BY
p.rows DESC
回答6:
Better option to get table size
SELECT
t.NAME AS TableName,
s.Name AS SchemaName,
p.rows AS RowCounts,
SUM(a.total_pages) * 8 AS TotalSpaceKB,
CAST(ROUND(((SUM(a.total_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS TotalSpaceMB,
SUM(a.used_pages) * 8 AS UsedSpaceKB,
CAST(ROUND(((SUM(a.used_pages) * 8) / 1024.00), 2) AS NUMERIC(36, 2)) AS UsedSpaceMB,
(SUM(a.total_pages) - SUM(a.used_pages)) * 8 AS UnusedSpaceKB,
CAST(ROUND(((SUM(a.total_pages) - SUM(a.used_pages)) * 8) / 1024.00, 2) AS NUMERIC(36, 2)) AS UnusedSpaceMB
FROM
sys.tables t
INNER JOIN
sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN
sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
INNER JOIN
sys.allocation_units a ON p.partition_id = a.container_id
LEFT OUTER JOIN
sys.schemas s ON t.schema_id = s.schema_id
WHERE
t.NAME NOT LIKE 'dt%'
AND t.is_ms_shipped = 0
AND i.OBJECT_ID > 255
GROUP BY
t.Name, s.Name, p.Rows
ORDER BY
t.Name