Question

是否有人碰巧有一个通用的SQL语句，它会列出数据库中的所有表和索引，以及每个分区的当前压缩设置？

感谢。

编辑：这是我尝试查询表时所得到的，但我不确定连接是否正确（我得到重复，这似乎是由索引的存在引起的）

SELECT [t].[name], [p].[partition_number], [p].[data_compression_desc]
FROM [sys].[partitions] AS [p]
INNER JOIN sys.tables AS [t] ON [t].[object_id] = [p].[object_id]

Answer 1

我以为我会分享我的最终查询。这将给出两个结果集，第一个是堆和聚簇索引的数据压缩，第二个是非聚簇索引的索引压缩。

SELECT [t].[name] AS [Table], [p].[partition_number] AS [Partition],
    [p].[data_compression_desc] AS [Compression]
FROM [sys].[partitions] AS [p]
INNER JOIN sys.tables AS [t] ON [t].[object_id] = [p].[object_id]
WHERE [p].[index_id] in (0,1)

SELECT [t].[name] AS [Table], [i].[name] AS [Index],  
    [p].[partition_number] AS [Partition],
    [p].[data_compression_desc] AS [Compression]
FROM [sys].[partitions] AS [p]
INNER JOIN sys.tables AS [t] ON [t].[object_id] = [p].[object_id]
INNER JOIN sys.indexes AS [i] ON [i].[object_id] = [p].[object_id] AND [i].[index_id] = [p].[index_id]
WHERE [p].[index_id] > 1

Answer 2

虽然我认为虽然Barguast发布的最终查询可能会有效，但仍然存在问题/未解释的问题。

基本上index_id 0 是堆， 1 是聚集索引， 2 是其他所有内容（非聚集索引）。

上述查询的问题是，如果表是堆（即使表中有数据），对数据的查询将不起作用。此外，对索引的查询也起作用，因为您指定了index_Id = 2，并且由于未加入index_id和sys.indexes之间的sys.partitions而存在欺骗行为。如果你加入那些，那么结果集中就不会有重复项，你可以做更容易理解的index_id not in (0,1)。

无论如何固定查询如下。我还在第一个查询中添加了索引名称（请注意，如果表是堆，则此字段将为null）。另请注意，您不必在第一个查询中指定index_id的联接，因为where指定(0,1)并且只能有其中一个（换句话说，您可以如果你愿意，可以添加它，但它没有区别。）

-- Data (table) compression (heap or clustered index)
SELECT [t].[name] AS [Table], 
       [i].[name] AS [Index],
       [p].[partition_number] AS [Partition],
       [p].[data_compression_desc] AS [Compression]
FROM [sys].[partitions] AS [p]
INNER JOIN sys.tables AS [t] 
     ON [t].[object_id] = [p].[object_id]
INNER JOIN sys.indexes AS [i] 
     ON [i].[object_id] = [p].[object_id]
WHERE [p].[index_id] in (0,1)

-- Index compression (non-clustered index)
SELECT [t].[name] AS [Table], 
       [i].[name] AS [Index],  
       [p].[partition_number] AS [Partition],
       [p].[data_compression_desc] AS [Compression]
FROM [sys].[partitions] AS [p]
INNER JOIN sys.tables AS [t] 
     ON [t].[object_id] = [p].[object_id]
INNER JOIN sys.indexes AS [i] 
     ON [i].[object_id] = [p].[object_id] AND i.index_id = p.index_id
WHERE [p].[index_id] not in (0,1)

Answer 3

这些答案都很体面，也很有效。由于我为我的工作点缀了一点，我觉得是时候回馈一下了。这个查询添加了Jason的答案（我需要的）。它还解决了一些连接问题，并将结果组合成一个非常简单的摘要。

-- Returns user tables and indexes in a DB and their Compression state
select s.name [Schema], t.name [Table], i.name [Index], p.data_compression_desc Compression
     , case when p.index_id in (0, 1) then 'Table' else 'Index' end CompressionObject
  from sys.tables t
  join sys.schemas s on t.schema_id = s.schema_id
  join sys.indexes i on t.object_id = i.object_id
  join sys.partitions p on (i.object_id = p.object_id and i.index_id = p.index_id)
where t.type = 'U'
order by 1, 2, p.index_id, 3

我使用它作为“工作列表”来生成压缩所有内容的脚本，因为我只是将数据库提升到Azure VM并希望降低IOPS以提高性能。希望这有助于那里的人。

Answer 4

这应该完成这项工作，测试一小部分以确保它能满足您的需求

SELECT DISTINCT s.name [Schema], t.name [Table], i.name [Index Name], p.partition_number, p.data_compression_desc
-- uncommenting the below line will give you dupes
--, p.index_id
FROM sys.schemas s
INNER JOIN sys.tables t
    ON s.schema_id = t.schema_id
    INNER JOIN sys.indexes i
        ON t.object_id = i.object_id
    INNER JOIN sys.partitions p
        ON t.object_id = p.object_id
ORDER BY s.name, t.name

您可能获得欺骗的原因是因为每个表有多个分区记录，例如多个index_id，请参阅此MSDN文章，以了解index_id的含义。添加 DISTINCT 应该可以解决欺骗问题

查询所有表数据和索引压缩

4 个答案: