由于源表和目标表的文件组不匹配而导致交换分区失败

  • Z时代
  • 2024-01-10
  • 分类：问答

我想在其中一个表上实现交换机分区，并且确保分区函数，方案，文件组正常工作。但是当我运行下面的命令时，我得到文件组错误。有人可以分享你的想法。由于源表和目标表的文件组不匹配而导致交换分区失败

命令： -

ALTER TABLE XYZ SWITCH PARTITION 5 TO ABC PARTITION 5;

错误： -

ALTER TABLE switch语句失败。表'XYZ'在文件组'PRIMARY'中，表'ABC'的分区5在文件组'FG_5'中。

回答：

某些集群索引可能会在现有表上创建。因此，在可能不同的表上创建群集索引期间会提到文件组。或者如果你可以删除ABC表并重新创建并尝试你的查询。

查询检查表和索引名称的文件组。

select f.name,o.name,i.name from sys.indexes i inner join sys.filegroups f on i.data_space_id=f.data_space_id 
inner join sys.all_objects o on o.object_id= i.object_id 
where o.name in ('ABC','XYZ')

我不会说这是解决方案，但这应该有助于解决问题。

回答：

该错误指示源表和目标表不存储对齐。运行下面的查询以确保源文件组和目标文件组在表和索引中都相同：

SELECT 
    OBJECT_NAME(p.object_id) AS ObjectName, 
    i.name AS IndexName, 
    p.index_id AS IndexID, 
    ds.name AS PartitionScheme, 
    p.partition_number AS PartitionNumber, 
    fg.name AS FileGroupName, 
    prv_left.value AS LowerBoundaryValue, 
    prv_right.value AS UpperBoundaryValue, 
    CASE pf.boundary_value_on_right WHEN 1 THEN 'RIGHT' ELSE 'LEFT' END AS PartitionFunctionRange, 
    p.rows AS Rows 
FROM 
    sys.partitions AS p INNER JOIN 
    sys.indexes AS i ON i.object_id = p.object_id AND i.index_id = p.index_id INNER JOIN 
    sys.data_spaces AS ds ON ds.data_space_id = i.data_space_id INNER JOIN 
    sys.partition_schemes AS ps ON ps.data_space_id = ds.data_space_id INNER JOIN 
    sys.partition_functions AS pf ON pf.function_id = ps.function_id INNER JOIN 
    sys.destination_data_spaces AS dds2 ON dds2.partition_scheme_id = ps.data_space_id AND dds2.destination_id = p.partition_number INNER JOIN 
    sys.filegroups AS fg ON fg.data_space_id = dds2.data_space_id LEFT OUTER JOIN 
    sys.partition_range_values AS prv_left ON ps.function_id = prv_left.function_id AND prv_left.boundary_id = p.partition_number - 1 LEFT OUTER JOIN 
    sys.partition_range_values AS prv_right ON ps.function_id = prv_right.function_id AND prv_right.boundary_id = p.partition_number 
WHERE 
    p.object_id IN (
     OBJECT_ID(N'dbo.ABC') 
     , OBJECT_ID(N'dbo.XYZ') 
    ) 
    AND p.partition_number = 5 
ORDER BY 
    ObjectName 
    ,IndexName 
    ,PartitionNumber;

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