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博客园 - 光脚码农

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如何获得数据库中所有用户创建的索引
光脚码农 · 2013-09-11 · via 博客园 - 光脚码农

Requirement

今天遇到一个SQL Performance Tuning需求:

对于用户表的索引进行优化,剔出那些重复的索引,添加必要的索引。

那么,我们必须先把表中的索引给找出来,SQL Server数据库和其他数据库一样,都有强大的元数据表,所以我们也可以从元数据表中得到索引(INDEX)的信息。 要完成这个查询需求,会涉及到master数据库中的几个视图(VIEW):

  • sys.[indexes]
  • sys.[index_columns]
  • sys.[all_columns]
  • sys.[tables]
  • sys.[schemas]

Resolution

SELECT '['+Sch.name+'].['+ Tab.[name]+']' AS TableName, 
       Ind.[name] AS IndexName, 
       SUBSTRING(( SELECT ', ' + AC.name 
                     FROM sys.[tables] AS T 
                    INNER JOIN sys.[indexes] I 
                       ON T.[object_id] = I.[object_id] 
                    INNER JOIN sys.[index_columns] IC 
                       ON I.[object_id] = IC.[object_id] 
                      AND I.[index_id] = IC.[index_id] 
                    INNER JOIN sys.[all_columns] AC 
                       ON T.[object_id] = AC.[object_id] 
                      AND IC.[column_id] = AC.[column_id] 
                    WHERE Ind.[object_id] = I.[object_id] 
                      AND Ind.index_id = I.index_id 
                      AND IC.is_included_column = 0 
                    ORDER BY IC.key_ordinal 
                      FOR XML PATH('') 
                 ), 2, 8000) AS KeyColumns, 
       SUBSTRING(( SELECT ', ' + AC.name 
                     FROM sys.[tables] AS T 
                    INNER JOIN sys.[indexes] I 
                       ON T.[object_id] = I.[object_id] 
                    INNER JOIN sys.[index_columns] IC 
                       ON I.[object_id] = IC.[object_id] 
                      AND I.[index_id] = IC.[index_id] 
                    INNER JOIN sys.[all_columns] AC 
                       ON T.[object_id] = AC.[object_id] 
                      AND IC.[column_id] = AC.[column_id] 
                    WHERE Ind.[object_id] = I.[object_id] 
                      AND Ind.index_id = I.index_id 
                      AND IC.is_included_column = 1 
                    ORDER BY IC.key_ordinal 
                      FOR XML PATH('') 
                 ), 2, 8000) AS IncludedColumns,
       Ind.type_desc AS IndexType,
       CASE Ind.is_primary_key
           WHEN 0 THEN 'N'
           WHEN 1 THEN 'Y'
       END AS IsPrimaryKey,
       CASE Ind.is_unique
           WHEN 0 THEN 'N'
           WHEN 1 THEN 'Y'
       END AS IsUnique
  FROM sys.[indexes] Ind 
 INNER JOIN sys.[tables] AS Tab 
    ON Tab.[object_id] = Ind.[object_id] 
 INNER JOIN sys.[schemas] AS Sch 
    ON Sch.[schema_id] = Tab.[schema_id] 
 WHERE Ind.[name] is not null
 ORDER BY TableName DESC

Explanation

上边这段代码的难点可能就是FOR XML PATH的用法。可以参考下面这篇博文:

Examples

在此我们拿Northwind数据库来演示代码执行后的结果。

TableName IndexName KeyColumns IncludeColumns IndexType IsPrimaryKey IsUnique
[dbo].[Territories] PK_Territories TerritoryID NULL NONCLUSTERED Y Y
[dbo].[Suppliers] PK_Suppliers SupplierID NULL CLUSTERED Y Y
[dbo].[Suppliers] CompanyName CompanyName NULL NONCLUSTERED N N
[dbo].[Suppliers] PostalCode PostalCode NULL NONCLUSTERED N N
[dbo].[Shippers] PK_Shippers ShipperID NULL CLUSTERED Y Y
[dbo].[Region] PK_Region RegionID NULL NONCLUSTERED Y Y
[dbo].[Products] PK_Products ProductID NULL CLUSTERED Y Y
[dbo].[Products] CategoriesProducts CategoryID NULL NONCLUSTERED N N