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

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])

其他回答

——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]   

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

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
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

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