在 SQLServer 中使用快照隔离级别检测 tempdb 磁盘速度对性能的影响

Detecting performance impact of tempdb disk speed with snapshot isolation level in SQLServer

我们使用 SQL 具有快照隔离级别的服务器。确切地说,它是一个读取提交的快照级别。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 放置的性能影响。以及如何与我的同事争论这一点?有没有什么有用的管理观点或反击?

每次更新行时快照隔离级别是绝对正确的,SQL 服务器数据库引擎会在 tempdb 中存储原始行的副本。而且,反过来 disk latency 的效率具有挑战性。

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,它在 TempDB
  • 中存储行的副本
  • 数据库引擎从 Tempdb
  • 检索行版本
  • Read/Writes 最终会在 Tempdb 开启快照隔离级别

您可以使用此脚本按数据库查找 read/writes。无论任何隔离级别如何,您都可以确定每个数据库的总数 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;

用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());

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;

希望对您有所帮助