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

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

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

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]

在Oracle中

select CONNECYBY.SCHEMA_NAME,CONNECYBY.TABLE_NAME,CONNECYBY.INDEX_NAME,CONNECYBY.COLUMN_NAME
from (  select TABLE_OWNER SCHEMA_NAME,TABLE_NAME,INDEX_NAME,COLUMN_POSITION,trim(',' from sys_connect_by_path(COLUMN_NAME,',')) COLUMN_NAME
        from DBA_IND_COLUMNS
        start with COLUMN_POSITION = 1
        connect by TABLE_OWNER = prior TABLE_OWNER
        and TABLE_NAME = prior TABLE_NAME
        and INDEX_NAME = prior INDEX_NAME
        and COLUMN_POSITION = prior COLUMN_POSITION + 1) CONNECYBY
join (  select TABLE_OWNER SCHEMA_NAME,TABLE_NAME,INDEX_NAME,max(COLUMN_POSITION) COLUMN_POSITION
        from DBA_IND_COLUMNS
        group by TABLE_OWNER,TABLE_NAME,INDEX_NAME) MAX_CONNECYBY
on (    CONNECYBY.SCHEMA_NAME = MAX_CONNECYBY.SCHEMA_NAME
        and CONNECYBY.TABLE_NAME = MAX_CONNECYBY.TABLE_NAME
        and CONNECYBY.INDEX_NAME = MAX_CONNECYBY.INDEX_NAME
        and CONNECYBY.COLUMN_POSITION = MAX_CONNECYBY.COLUMN_POSITION)
order by CONNECYBY.SCHEMA_NAME,CONNECYBY.TABLE_NAME,CONNECYBY.INDEX_NAME

SQL Server 与

CONNECTBY(SCHEMA_NAME,TABLE_NAME,INDEX_NAME,INDEX_COLUMN_ID,COLUMN_NAME) 
as 
    (   select SCHEMAS.NAME SCHEMA_NAME
            , TABLES.NAME TABLE_NAME
            , INDEXES.NAME INDEX_NAME
            , INDEX_COLUMNS.INDEX_COLUMN_ID INDEX_COLUMN_ID
            , cast(COLUMNS.NAME AS VARCHAR(MAX)) COLUMN_NAME
        from SYS.INDEXES
        join SYS.TABLES on (INDEXES.OBJECT_ID = TABLES.OBJECT_ID)
        join SYS.SCHEMAS on (TABLES.SCHEMA_ID = SCHEMAS.SCHEMA_ID)
        join SYS.INDEX_COLUMNS on ( INDEXES.OBJECT_ID = INDEX_COLUMNS.OBJECT_ID 
                                    and INDEX_COLUMNS.INDEX_ID = INDEXES.INDEX_ID)
        join SYS.COLUMNS on (   INDEXES.OBJECT_ID = COLUMNS.OBJECT_ID 
                                and INDEX_COLUMNS.COLUMN_ID = COLUMNS.COLUMN_ID)
        where INDEX_COLUMNS.INDEX_COLUMN_ID = 1
        union all
        select SCHEMAS.NAME SCHEMA_NAME
            , TABLES.NAME TABLE_NAME
            , INDEXES.NAME INDEX_NAME
            , INDEX_COLUMNS.INDEX_COLUMN_ID INDEX_COLUMN_ID
            , cast(PRIOR.COLUMN_NAME + ',' + COLUMNS.NAME AS VARCHAR(MAX)) COLUMN_NAME
        from SYS.INDEXES
        join SYS.TABLES on (INDEXES.OBJECT_ID = TABLES.OBJECT_ID)
        join SYS.SCHEMAS on (TABLES.SCHEMA_ID = SCHEMAS.SCHEMA_ID)
        join SYS.INDEX_COLUMNS on ( INDEXES.OBJECT_ID = INDEX_COLUMNS.OBJECT_ID 
                                    and INDEX_COLUMNS.INDEX_ID = INDEXES.INDEX_ID)
        join SYS.COLUMNS on (   INDEXES.OBJECT_ID = COLUMNS.OBJECT_ID 
                                and INDEX_COLUMNS.COLUMN_ID = COLUMNS.COLUMN_ID)
        join CONNECTBY as PRIOR on (SCHEMAS.NAME = PRIOR.SCHEMA_NAME 
                                    and TABLES.NAME = PRIOR.TABLE_NAME 
                                    and INDEXES.NAME = PRIOR.INDEX_NAME 
                                    and INDEX_COLUMNS.INDEX_COLUMN_ID = PRIOR.INDEX_COLUMN_ID + 1))
select CONNECTBY.SCHEMA_NAME,CONNECTBY.TABLE_NAME,CONNECTBY.INDEX_NAME,CONNECTBY.COLUMN_NAME
from CONNECTBY
join (  select  SCHEMA_NAME
                , TABLE_NAME
                , INDEX_NAME
                , MAX(INDEX_COLUMN_ID) INDEX_COLUMN_ID
        from CONNECTBY 
        group by SCHEMA_NAME,TABLE_NAME,INDEX_NAME) MAX_CONNECTBY
        on (CONNECTBY.SCHEMA_NAME = MAX_CONNECTBY.SCHEMA_NAME
            and CONNECTBY.TABLE_NAME = MAX_CONNECTBY.TABLE_NAME
            and CONNECTBY.INDEX_NAME = MAX_CONNECTBY.INDEX_NAME
            and CONNECTBY.INDEX_COLUMN_ID = MAX_CONNECTBY.INDEX_COLUMN_ID)
order by CONNECTBY.SCHEMA_NAME,CONNECTBY.TABLE_NAME,CONNECTBY.INDEX_NAME