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。


当前回答

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。

其他回答

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;

我想到了这个,它给了我我需要的准确的概述。有帮助的是,每个索引得到一行,索引列被聚合到其中。

select 
    o.name as ObjectName, 
    i.name as IndexName, 
    i.is_primary_key as [PrimaryKey],
    SUBSTRING(i.[type_desc],0,6) as IndexType,
    i.is_unique as [Unique],
    Columns.[Normal] as IndexColumns,
    Columns.[Included] as IncludedColumns
from sys.indexes i 
join sys.objects o on i.object_id = o.object_id
cross apply
(
    select
        substring
        (
            (
                select ', ' + co.[name]
                from sys.index_columns ic
                join sys.columns co on co.object_id = i.object_id and co.column_id = ic.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('')
            )
            , 3
            , 10000
        )    as [Normal]    
        , substring
        (
            (
                select ', ' + co.[name]
                from sys.index_columns ic
                join sys.columns co on co.object_id = i.object_id and co.column_id = ic.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.key_ordinal
                for xml path('')
            )
            , 3
            , 10000
        )    as [Included]    

) Columns
where o.[type] = 'U' --USER_TABLE
order by o.[name], i.[name], i.is_primary_key desc

当我有这个需求时,我使用了以下查询…

SELECT 
    TableName = t.name,
    ColumnId = col.column_id, 
    ColumnName = col.name,
    DataType = ty.name,
    MaxSize = ty.max_length,
    IsNullable = CASE WHEN (col.is_nullable = 1) THEN 'Y' END,
    IsIdentity = CASE WHEN (col.is_identity = 1) THEN 'Y' END,
    IsPrimaryKey = CASE WHEN (ic.column_id = col.column_id) THEN 'Y' END,
    IsForeignKey = CASE WHEN (fkc.parent_column_id = col.column_id) THEN 'Y' END,
    IsDefault = CASE WHEN (dc.parent_column_id = col.column_id) THEN 'Y' END
FROM 
    sys.tables t
INNER JOIN 
     sys.columns col ON t.object_id = col.object_id 
LEFT JOIN
    sys.indexes ind ON t.object_id = ind.object_id 
LEFT JOIN 
     sys.index_columns ic ON ic.index_id=ind.index_id AND ic.object_id = col.object_id and ic.column_id = col.column_id
LEFT JOIN sys.foreign_key_columns fkc
                ON fkc.parent_object_id = col.object_id AND fkc.parent_column_id=col.column_id
LEFT JOIN sys.default_constraints dc
                ON dc.parent_object_id = col.object_id AND dc.parent_column_id=col.column_id
LEFT JOIN
     sys.types ty on ty.user_type_id = col.user_type_id

WHERE
    --t.name='<TABLENAME>'
    t.schema_id = 10    --SCHEMA ID
    AND ind.is_primary_key=1    
ORDER BY
    t.name, ColumnId

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

我可以大胆回答这个饱和的问题吗?

这是@marc_s答案的自由重做,混合了来自@Tim Ford的一些东西,目标是有一个更干净和更简单的结果集和最终显示和排序,以满足我当前的需要。

SELECT 
    OBJECT_SCHEMA_NAME(t.[object_id],DB_ID()) AS [Schema],
    t.[name] AS [TableName], 
    ind.[name] AS [IndexName], 
    col.[name] AS [ColumnName],
    ic.column_id AS [ColumnId],
    ind.[type_desc] AS [IndexTypeDesc], 
    col.is_identity AS [IsIdentity],
    ind.[is_unique] AS [IsUnique],
    ind.[is_primary_key] AS [IsPrimaryKey],
    ic.[is_descending_key] AS [IsDescendingKey],
    ic.[is_included_column] AS [IsIncludedColumn]
FROM 
    sys.indexes ind 
INNER JOIN 
    sys.index_columns ic 
    ON ind.object_id = ic.object_id AND ind.index_id = ic.index_id 
INNER JOIN 
    sys.columns col 
    ON ic.object_id = col.object_id and ic.column_id = col.column_id 
INNER JOIN 
    sys.tables t 
    ON ind.object_id = t.object_id 
WHERE 
    t.is_ms_shipped = 0
    --ind.is_primary_key = 1 -- include or not pks, etc
    --AND ind.is_unique = 0
    --AND ind.is_unique_constraint = 0 
ORDER BY 
    [Schema],
    TableName, 
    IndexName,
    [ColumnId],
    ColumnName