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

SQL Server 2014工作解决方案。我在这里只包含了少量的输出字段,但您可以随意添加任何您喜欢的字段。

SELECT
    o.object_id AS objectId
    ,o.name AS objectName
    ,i.index_id AS indexId
    ,i.name AS indexName
    ,i.type_desc AS typeDesc
    ,ic.index_column_id AS indexColumnId
    ,ic.key_ordinal AS keyOrdinal
    ,ic.is_included_column AS isIncludedColumn
    ,ic.column_id AS columnId
    ,c.name AS columnName
FROM {database}.sys.objects AS o
    INNER JOIN {database}.sys.columns AS c ON
        c.object_id = o.object_id
        AND o.type = 'U'
    INNER JOIN {database}.sys.indexes AS i ON
        i.object_id = o.object_id
    INNER JOIN {database}.sys.index_columns AS ic ON
        ic.object_id = i.object_id
        AND ic.index_id = i.index_id
        AND ic.column_id = c.column_id
ORDER BY
    o.object_id
    ,i.index_id
    ,ic.index_column_id

其他回答

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])

我没有经过,但是我在原作者发布的查询中得到了我想要的东西。

我使用它(没有条件/过滤器)来满足我的需求,但它给出了不正确的结果

主要问题是在index_id上没有连接条件的情况下得到叉乘

SELECT S.NAME SCHEMA_NAME,T.NAME TABLE_NAME,I.NAME INDEX_NAME,C.NAME COLUMN_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.INDEX_ID    = I.INDEX_ID**
   AND IC.COLUMN_ID = C.COLUMN_ID
 WHERE 1=1

ORDER BY I.NAME,I.INDEX_ID,IC.KEY_ORDINAL

根据Tim Ford的代码,这是正确答案:

  select tab.[name]  as [table_name],
         idx.[name]  as [index_name],
         allc.[name] as [column_name],
         idx.[type_desc],
         idx.[is_unique],
         idx.[data_space_id],
         idx.[ignore_dup_key],
         idx.[is_primary_key],
         idx.[is_unique_constraint],
         idx.[fill_factor],
         idx.[is_padded],
         idx.[is_disabled],
         idx.[is_hypothetical],
         idx.[allow_row_locks],
         idx.[allow_page_locks],
         idxc.[is_descending_key],
         idxc.[is_included_column],
         idxc.[index_column_id]

     from sys.[tables] as tab

    inner join sys.[indexes]       idx  on tab.[object_id] =  idx.[object_id]
    inner join sys.[index_columns] idxc on idx.[object_id] = idxc.[object_id] and  idx.[index_id]  = idxc.[index_id]
    inner join sys.[all_columns]   allc on tab.[object_id] = allc.[object_id] and idxc.[column_id] = allc.[column_id]

    where tab.[name] Like '%table_name%'
      and idx.[name] Like '%index_name%'
    order by tab.[name], idx.[index_id], idxc.[index_column_id]

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

正确的一个在这里(当我们在一个表上有多个索引时,以上所有帖子都会给出笛卡尔积结果)

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 
                                  AND i.index_id = ic.index_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
 AND  t.name = 'DimCustomer'
order by ic.key_ordinal