如何在SQL Server 2005+中获得所有索引和索引列的列表?我能想到的最接近的是:
select s.name, t.name, i.name, c.name from sys.tables t
inner join sys.schemas s on t.schema_id = s.schema_id
inner join sys.indexes i on i.object_id = t.object_id
inner join sys.index_columns ic on ic.object_id = t.object_id
inner join sys.columns c on c.object_id = t.object_id and
ic.column_id = c.column_id
where i.index_id > 0
and i.type in (1, 2) -- clustered & nonclustered only
and i.is_primary_key = 0 -- do not include PK indexes
and i.is_unique_constraint = 0 -- do not include UQ
and i.is_disabled = 0
and i.is_hypothetical = 0
and ic.key_ordinal > 0
order by ic.key_ordinal
这可不是我想要的。
我想要的是，列出所有用户定义的索引，(这意味着不支持唯一约束和主键的索引)与所有列(按它们在索引定义中的出现方式排序)以及尽可能多的元数据。
select i.object_id, i.name as [index] , STRING_AGG(c.name,', ') as [column], o.name as [table] from sys.indexes i
INNER join sys.index_columns ic on ic.object_id = i.object_id and ic.index_id = i.index_id
INNER join sys.columns c on c.object_id = ic.object_id and ic.column_id = c.column_id
INNER JOIN sys.objects o on o.object_id = i.object_id
where i.object_id > 100 and i.is_primary_key = 0 and i.is_unique = 0 and o.is_ms_shipped <> 1
group by i.object_id, i.name, o.name
order by i.name
将此用于sql 2016及更高级别，它会显示object_id, indexname，列和表名为非唯一的，没有主键
以下工作在SQL Server 2014/2016以及任何Microsoft Azure SQL数据库。
生成一个全面的结果集，可以很容易地导出到Notepad/Excel中进行切片和切块
表名
索引名称
指数描述
索引列-按顺序
包括列-按顺序
SELECT '[' + s.NAME + '].[' + o.NAME + ']' AS 'table_name'
,+ i.NAME AS 'index_name'
,LOWER(i.type_desc) + CASE
WHEN i.is_unique = 1
THEN ', unique'
ELSE ''
END + CASE
WHEN i.is_primary_key = 1
THEN ', primary key'
ELSE ''
END AS 'index_description'
,STUFF((
SELECT ', [' + sc.NAME + ']' AS "text()"
FROM syscolumns AS sc
INNER JOIN sys.index_columns AS ic ON ic.object_id = sc.id
AND ic.column_id = sc.colid
WHERE sc.id = so.object_id
AND ic.index_id = i1.indid
AND ic.is_included_column = 0
ORDER BY key_ordinal
FOR XML PATH('')
), 1, 2, '') AS 'indexed_columns'
,STUFF((
SELECT ', [' + sc.NAME + ']' AS "text()"
FROM syscolumns AS sc
INNER JOIN sys.index_columns AS ic ON ic.object_id = sc.id
AND ic.column_id = sc.colid
WHERE sc.id = so.object_id
AND ic.index_id = i1.indid
AND ic.is_included_column = 1
FOR XML PATH('')
), 1, 2, '') AS 'included_columns'
FROM sysindexes AS i1
INNER JOIN sys.indexes AS i ON i.object_id = i1.id
AND i.index_id = i1.indid
INNER JOIN sysobjects AS o ON o.id = i1.id
INNER JOIN sys.objects AS so ON so.object_id = o.id
AND is_ms_shipped = 0
INNER JOIN sys.schemas AS s ON s.schema_id = so.schema_id
WHERE so.type = 'U'
AND i1.indid < 255
AND i1.STATUS & 64 = 0 --index with duplicates
AND i1.STATUS & 8388608 = 0 --auto created index
AND i1.STATUS & 16777216 = 0 --stats no recompute
AND i.type_desc <> 'heap'
AND so.NAME <> 'sysdiagrams'
ORDER BY table_name
,index_name;
以下是最好的方法:
SELECT sys.tables.object_id, sys.tables.name as table_name, sys.columns.name as column_name, sys.indexes.name as index_name,
sys.indexes.is_unique, sys.indexes.is_primary_key
FROM sys.tables, sys.indexes, sys.index_columns, sys.columns
WHERE (sys.tables.object_id = sys.indexes.object_id AND sys.tables.object_id = sys.index_columns.object_id AND sys.tables.object_id = sys.columns.object_id
AND sys.indexes.index_id = sys.index_columns.index_id AND sys.index_columns.column_id = sys.columns.column_id)
AND sys.tables.name = 'your_table_name'
我更喜欢使用隐式连接，因为它对我来说更容易理解。您可以删除object_id引用，因为您可能不需要它。
欢呼。
根据公认的答案和另外两个问题1,2，我整理了以下问题:
SELECT
QUOTENAME(t.name) AS TableName,
QUOTENAME(i.name) AS IndexName,
i.is_primary_key,
i.is_unique,
i.is_unique_constraint,
STUFF(REPLACE(REPLACE((
SELECT QUOTENAME(c.name) + CASE WHEN ic.is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
FROM sys.index_columns AS ic
INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.column_id
WHERE ic.object_id = i.object_id AND ic.index_id = i.index_id AND ic.is_included_column = 0
ORDER BY ic.key_ordinal
FOR XML PATH
), '<row>', ', '), '</row>', ''), 1, 2, '') AS KeyColumns,
STUFF(REPLACE(REPLACE((
SELECT QUOTENAME(c.name) AS [data()]
FROM sys.index_columns AS ic
INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.column_id
WHERE ic.object_id = i.object_id AND ic.index_id = i.index_id AND ic.is_included_column = 1
ORDER BY ic.index_column_id
FOR XML PATH
), '<row>', ', '), '</row>', ''), 1, 2, '') AS IncludedColumns,
u.user_seeks,
u.user_scans,
u.user_lookups,
u.user_updates
FROM sys.tables AS t
INNER JOIN sys.indexes AS i ON t.object_id = i.object_id
LEFT JOIN sys.dm_db_index_usage_stats AS u ON i.object_id = u.object_id AND i.index_id = u.index_id
WHERE t.is_ms_shipped = 0
AND i.type <> 0
该查询返回如下所示的结果，其中显示了索引的列表、它们的列和用法。非常有助于确定哪个指数比其他指数表现更好:
这是一种回退到索引的方法。您可以使用SHOWCONTIG来评估碎片。它将列出数据库或表的所有索引，以及统计信息。我要提醒的是，在大型数据库上，它可能是长时间运行的。对我来说，这种方法的好处之一是您不必是管理员就可以使用它。
——显示数据库中所有索引的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG WITH ALL_INDEXES
GO
.．.完成后关闭NOCOUNT
——显示表中所有索引的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors) WITH ALL_INDEXES
GO
——显示特定索引上的碎片信息
SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors,aunmind)
GO
我需要得到特定的索引，它们的索引列和包含的列。以下是我使用的查询:
SELECT INX.[name] AS [Index Name]
,TBL.[name] AS [Table Name]
,DS1.[IndexColumnsNames]
,DS2.[IncludedColumnsNames]
FROM [sys].[indexes] INX
INNER JOIN [sys].[tables] TBL
ON INX.[object_id] = TBL.[object_id]
CROSS APPLY
(
SELECT STUFF
(
(
SELECT ' [' + CLS.[name] + ']'
FROM [sys].[index_columns] INXCLS
INNER JOIN [sys].[columns] CLS
ON INXCLS.[object_id] = CLS.[object_id]
AND INXCLS.[column_id] = CLS.[column_id]
WHERE INX.[object_id] = INXCLS.[object_id]
AND INX.[index_id] = INXCLS.[index_id]
AND INXCLS.[is_included_column] = 0
FOR XML PATH('')
)
,1
,1
,''
)
) DS1 ([IndexColumnsNames])
CROSS APPLY
(
SELECT STUFF
(
(
SELECT ' [' + CLS.[name] + ']'
FROM [sys].[index_columns] INXCLS
INNER JOIN [sys].[columns] CLS
ON INXCLS.[object_id] = CLS.[object_id]
AND INXCLS.[column_id] = CLS.[column_id]
WHERE INX.[object_id] = INXCLS.[object_id]
AND INX.[index_id] = INXCLS.[index_id]
AND INXCLS.[is_included_column] = 1
FOR XML PATH('')
)
,1
,1
,''
)
) DS2 ([IncludedColumnsNames])
