sql - 在 SQLServer 中使用快照隔离级别检测 tempdb 磁盘速度的性能影响

Question

我们使用具有快照隔离级别的 SQL Server。确切地说，它是一个读提交的快照级别。据我所知，在这种情况下，SQLServer 使用 tempdb 中的版本存储。在我们的环境中，tempdb 存储在比主数据库慢的磁盘上。所以问题是如何检测这种 tempdb 放置对性能的影响。以及如何与我的同事争论这一点？是否有任何有用的管理视图或计数器？

Answer 1

每次更新行时，快照隔离级别是绝对正确的，SQL Server 数据库引擎将原始行的副本存储在tempdb. 而且，反过来，效率disk latency也是具有挑战性的。

https://docs.microsoft.com/en-us/dotnet/framework/data/adonet/sql/snapshot-isolation-in-sql-server

• 快照隔离级别不创建临时表，它将行的副本存储在TempDB
• 数据库引擎从Tempdb
• Tempdb当快照隔离级别打开时，读/写最终会增加

您可以使用此脚本来查找数据库的读/写。无论任何隔离级别，您都可以识别每个数据库的总 I/O

SELECT name AS 'Database Name'
      ,SUM(num_of_reads) AS 'Number of Read'
      ,SUM(num_of_writes) AS 'Number of Writes' 
FROM sys.dm_io_virtual_file_stats(NULL, NULL) I
  INNER JOIN sys.databases D  
      ON I.database_id = d.database_id
GROUP BY name ORDER BY 'Number of Read' DESC;

对于按物理驱动器号显示 I/O 统计信息：

SELECT left(f.physical_name, 1) AS DriveLetter, 
    DATEADD(MS,sample_ms * -1, GETDATE()) AS [Start Date],
    SUM(v.num_of_writes) AS total_num_of_writes, 
    SUM(v.num_of_bytes_written) AS total_num_of_bytes_written, 
    SUM(v.num_of_reads) AS total_num_of_reads, 
    SUM(v.num_of_bytes_read) AS total_num_of_bytes_read, 
    SUM(v.size_on_disk_bytes) AS total_size_on_disk_bytes
FROM sys.master_files f
INNER JOIN sys.dm_io_virtual_file_stats(NULL, NULL) v
ON f.database_id=v.database_id and f.file_id=v.file_id
GROUP BY left(f.physical_name, 1),DATEADD(MS,sample_ms * -1, GETDATE());

用于计算不同数据库驱动器的磁盘延迟

SELECT  LEFT(physical_name, 1) AS drive,
        CAST(SUM(io_stall_read_ms) / 
            (1.0 + SUM(num_of_reads)) AS NUMERIC(10,1)) 
                          AS 'avg_read_disk_latency_ms',
        CAST(SUM(io_stall_write_ms) / 
            (1.0 + SUM(num_of_writes) ) AS NUMERIC(10,1)) 
                          AS 'avg_write_disk_latency_ms',
        CAST((SUM(io_stall)) / 
            (1.0 + SUM(num_of_reads + num_of_writes)) AS NUMERIC(10,1)) 
                          AS 'avg_disk_latency_ms'
FROM    sys.dm_io_virtual_file_stats(NULL, NULL) AS divfs
        JOIN sys.master_files AS mf ON mf.database_id = divfs.database_id
                                       AND mf.file_id = divfs.file_id
GROUP BY LEFT(physical_name, 1)
ORDER BY avg_disk_latency_ms DESC;

希望这可以帮助