SqlServer中的（分区）表文件组

获取一个表的文件组信息，有如下脚本：

SELECT OBJECT_SCHEMA_NAME(t.object_id) AS schema_name
,t.name AS table_name
,i.index_id
,i.name AS index_name
,ds.name AS filegroup_name
,FORMAT(p.rows, '#,###') AS rows
FROM sys.tables t
INNER JOIN sys.indexes i ON t.object_id=i.object_id
INNER JOIN sys.filegroups ds ON i.data_space_id=ds.data_space_id
INNER JOIN sys.partitions p ON i.object_id=p.object_id AND i.index_id=p.index_id
ORDER BY t.name, i.index_id

但红框圈出的表是做过表分区的，获取的文件组信息是错误的，应考虑表分区问题，用如下脚本获取表文件组：

SELECT OBJECT_SCHEMA_NAME(t.object_id) AS schema_name
,t.name AS table_name
,i.index_id
,i.name AS index_name
,p.partition_number
,fg.name AS filegroup_name
,FORMAT(p.rows, '#,###') AS rows
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
LEFT OUTER JOIN sys.partition_schemes ps ON i.data_space_id=ps.data_space_id
LEFT OUTER JOIN sys.destination_data_spaces dds ON ps.data_space_id=dds.partition_scheme_id AND p.partition_number=dds.destination_id
INNER JOIN sys.filegroups fg ON COALESCE(dds.data_space_id, i.data_space_id)=fg.data_space_id

 这样，获取的文件组才是正确的。学习自：https://jasonstrate.com/2013/01/29/determining-file-group-for-a-table/，摘其最重要的一条信息：

  简言之，“分区索引不再决定数据的位置，而是由分区方案决定”。