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

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

其他回答

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

对于每个索引的唯一列:

select s.name, t.name, i.name, i.index_id,c.name,c.column_id
 from sys.schemas s
inner join sys.tables t 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
    and ic.index_id=i.index_id
inner join sys.columns c on c.object_id = t.object_id 
    and ic.column_id = c.column_id
where i.object_id = object_id('previous.account_1')  
order by index_id,column_id

——简短而甜蜜:

SELECT OBJECT_SCHEMA_NAME(T.[object_id],DB_ID()) AS [Schema],  
  T.[name] AS [table_name], I.[name] AS [index_name], AC.[name] AS [column_name],  
  I.[type_desc], I.[is_unique], I.[data_space_id], I.[ignore_dup_key], I.[is_primary_key], 
  I.[is_unique_constraint], I.[fill_factor],    I.[is_padded], I.[is_disabled], I.[is_hypothetical], 
  I.[allow_row_locks], I.[allow_page_locks], IC.[is_descending_key], IC.[is_included_column] 
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] 
  INNER JOIN sys.[all_columns] AC ON T.[object_id] = AC.[object_id] AND IC.[column_id] = AC.[column_id] 
WHERE T.[is_ms_shipped] = 0 AND I.[type_desc] <> 'HEAP' 
ORDER BY T.[name], I.[index_id], IC.[key_ordinal]   

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
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