Список всех индексов и столбцов индекса в БД SQL Server

Как получить список всех индексов и столбцов индекса в SQL Server 2005+? Самое близкое, что я мог получить:

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

Что не совсем то, что я хочу.
Я хочу, чтобы список всех пользовательских индексов ( что означает отсутствие индексов, поддерживающих уникальные ограничения и первичные ключи ) со всеми столбцами (упорядоченными по тому, как они отображаются в определении индекса) плюс как можно больше метаданных.

Существует два вида каталога "sys", с которыми вы можете ознакомиться:

select * from sys.indexes

select * from sys.index_columns

Они дадут вам практически любую информацию о индексах и их столбцах.

РЕДАКТИРОВАТЬ: Этот запрос приближается к тому, что вы ищете:

SELECT 
     TableName = t.name,
     IndexName = ind.name,
     IndexId = ind.index_id,
     ColumnId = ic.index_column_id,
     ColumnName = col.name,
     ind.*,
     ic.*,
     col.* 
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 
     ind.is_primary_key = 0 
     AND ind.is_unique = 0 
     AND ind.is_unique_constraint = 0 
     AND t.is_ms_shipped = 0 
ORDER BY 
     t.name, ind.name, ind.index_id, ic.index_column_id;

Вы можете использовать sp_helpindexдля просмотра всех индексов одной таблицы.

EXEC sys.sp_helpindex @objname = N'User' -- nvarchar(77)

И для всех индексов вы можете пройти, sys.objectsчтобы получить все индексы для каждой таблицы.

Ничто из вышеперечисленного не сделало работу для меня, но это делает:

-- KDF9's concise index list for SQL Server 2005+  (see below for 2000)
--   includes schemas and primary keys, in easy to read format
--   with unique, clustered, and all ascending/descendings in a single column
-- Needs simple manual add or delete to change maximum number of key columns
--   but is easy to understand and modify, with no UDFs or complex logic
--
SELECT
  schema_name(schema_id) as SchemaName, OBJECT_NAME(si.object_id) as TableName, si.name as IndexName,
  (CASE is_primary_key WHEN 1 THEN 'PK' ELSE '' END) as PK,
  (CASE is_unique WHEN 1 THEN '1' ELSE '0' END)+' '+
  (CASE si.type WHEN 1 THEN 'C' WHEN 3 THEN 'X' ELSE 'B' END)+' '+  -- B=basic, C=Clustered, X=XML
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,1,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,2,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,3,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,4,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,5,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  (CASE INDEXKEY_PROPERTY(si.object_id,index_id,6,'IsDescending') WHEN 0 THEN 'A' WHEN 1 THEN 'D' ELSE '' END)+
  '' as 'Type',
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,1) as Key1,
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,2) as Key2,
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,3) as Key3,
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,4) as Key4,
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,5) as Key5,
  INDEX_COL(schema_name(schema_id)+'.'+OBJECT_NAME(si.object_id),index_id,6) as Key6
FROM sys.indexes as si
LEFT JOIN sys.objects as so on so.object_id=si.object_id
WHERE index_id>0 -- omit the default heap
  and OBJECTPROPERTY(si.object_id,'IsMsShipped')=0 -- omit system tables
  and not (schema_name(schema_id)='dbo' and OBJECT_NAME(si.object_id)='sysdiagrams') -- omit sysdiagrams
ORDER BY SchemaName,TableName,IndexName

-------------------------------------------------------------------
-- or to generate creation scripts put a simple wrapper around that
SELECT SchemaName, TableName, IndexName,
  (CASE pk
    WHEN 'PK' THEN 'ALTER '+
     'TABLE '+SchemaName+'.'+TableName+' ADD CONSTRAINT '+IndexName+' PRIMARY KEY'+
     (CASE substring(Type,3,1) WHEN 'C' THEN ' CLUSTERED' ELSE '' END)
    ELSE 'CREATE '+
     (CASE substring(Type,1,1) WHEN '1' THEN 'UNIQUE ' ELSE '' END)+
     (CASE substring(Type,3,1) WHEN 'C' THEN 'CLUSTERED ' ELSE '' END)+
     'INDEX '+IndexName+' ON '+SchemaName+'.'+TableName
    END)+
  ' ('+
    (CASE WHEN Key1 is null THEN '' ELSE      Key1+(CASE substring(Type,4+1,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    (CASE WHEN Key2 is null THEN '' ELSE ', '+Key2+(CASE substring(Type,4+2,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    (CASE WHEN Key3 is null THEN '' ELSE ', '+Key3+(CASE substring(Type,4+3,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    (CASE WHEN Key4 is null THEN '' ELSE ', '+Key4+(CASE substring(Type,4+4,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    (CASE WHEN Key5 is null THEN '' ELSE ', '+Key5+(CASE substring(Type,4+5,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    (CASE WHEN Key6 is null THEN '' ELSE ', '+Key6+(CASE substring(Type,4+6,1) WHEN 'D' THEN ' DESC' ELSE '' END) END)+
    ')' as CreateIndex
FROM (
  ...
  ...listing SQL same as above minus the ORDER BY...
  ...
  ) as indexes
ORDER BY SchemaName,TableName,IndexName

----------------------------------------------------------
-- For SQL Server 2000 the following should work
--   change table names to sysindexes and sysobjects (no dots)
--   change object_id => id, index_id => indid,
--   change is_primary_key => (select count(constid) from sysconstraints as sc where sc.id=si.id and sc.status&15=1)
--   change is_unique => INDEXPROPERTY(si.id,si.name,'IsUnique')
--   change si.type => INDEXPROPERTY(si.id,si.name,'IsClustered')
--   remove all references to schemas including schema name qualifiers, and the XML type
--   add select where indid<255 and si.status&64=0 (to omit the text/image index and autostats)

Если ваши имена включают пробелы, добавьте квадратные скобки вокруг них в сценарии создания.

Когда в последнем столбце Key все нули, вы знаете, что ни один из них не пропущен.

Отфильтровывать первичные ключи и т. Д., Как в исходном запросе, тривиально.

ПРИМЕЧАНИЕ. Будьте осторожны с этим решением, поскольку оно не различает проиндексированные и включенные столбцы.

Коротко и сладко

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]   

Следующее работает на SQL Server 2014/2016, а также на любой базе данных Microsoft Azure SQL.

Создает полный набор результатов, который легко экспортировать в Блокнот / Excel для нарезки и нарезки кубиками и включает в себя

1. Имя таблицы
2. Имя индекса
3. Индекс Описание
4. Индексированные столбцы - в порядке
5. Включенные столбцы - в порядке

 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;

Эй, ребята, я не прошел, но я получил то, что хотел в запросе, опубликованном оригинальным автором.

Я использовал его (без условий / фильтров) для моего требования, но это дало неправильные результаты

Основная проблема заключалась в получении кросс-продукта результатов без условия объединения по index_id

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

Мне нужно было получить определенные индексы, их столбцы индекса и включенные столбцы, а также. Вот запрос, который я использовал:

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])

После дает то , что подобно тому, как sp_helpindex имя_таблица

select T.name as TableName, I.name as IndexName, AC.Name as ColumnName, I.type_desc as IndexType 
from sys.tables as T inner join sys.indexes as I on T.[object_id] = I.[object_id] 
   inner join sys.index_columns as IC on IC.[object_id] = I.[object_id] and IC.[index_id] = I.[index_id] 
   inner join sys.all_columns as AC on IC.[object_id] = AC.[object_id] and IC.[column_id] = AC.[column_id] 
order by T.name, I.name

На основании принятого ответа и двух других вопросов ^{1 , 2} я собрал следующий запрос:

SELECT
    QUOTENAME(t.name) AS TableName,
    QUOTENAME(i.name) AS IndexName,
    i.is_primary_key,
    i.is_unique,
    i.is_unique_constraint,
    STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) + CASE WHEN ic.is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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
    ), '<row>', ', '), '</row>', ''), 1, 2, '') AS KeyColumns,
    STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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.index_column_id
        FOR XML PATH
    ), '<row>', ', '), '</row>', ''), 1, 2, '') AS IncludedColumns,
    u.user_seeks,
    u.user_scans,
    u.user_lookups,
    u.user_updates
FROM sys.tables AS t
INNER JOIN sys.indexes AS i ON t.object_id = i.object_id
LEFT JOIN sys.dm_db_index_usage_stats AS u ON i.object_id = u.object_id AND i.index_id = u.index_id
WHERE t.is_ms_shipped = 0
AND i.type <> 0

Этот запрос возвращает результаты, такие как ниже, который показывает список индексов, их столбцы и использование. Очень полезно при определении, какой индекс работает лучше, чем другие:

это будет работать:

DECLARE @IndexInfo  TABLE (index_name         varchar(250)
                          ,index_description  varchar(250)
                          ,index_keys         varchar(250)
                          )

INSERT INTO @IndexInfo
exec sp_msforeachtable 'sp_helpindex ''?'''
select * from @IndexInfo

это не возвращает имя таблицы, и вы получите предупреждения для всех таблиц без индекса; если это проблема, вы можете создать цикл над таблицами, индексы которых выглядят следующим образом:

DECLARE @IndexInfoTemp  TABLE (index_name         varchar(250)
                              ,index_description  varchar(250)
                              ,index_keys         varchar(250)
                              )

DECLARE @IndexInfo  TABLE (table_name         sysname
                          ,index_name         varchar(250)
                          ,index_description  varchar(250)
                          ,index_keys         varchar(250)
                          )

DECLARE @Tables Table (RowID       int not null identity(1,1)
                      ,TableName   sysname 
                      )
DECLARE @MaxRow       int
DECLARE @CurrentRow   int
DECLARE @CurrentTable sysname

INSERT INTO @Tables
    SELECT
        DISTINCT t.name 
        FROM sys.indexes i
            INNER JOIN sys.tables t ON i.object_id = t.object_id
        WHERE i.Name IS NOT NULL
SELECT @MaxRow=@@ROWCOUNT,@CurrentRow=1

WHILE @CurrentRow<=@MaxRow
BEGIN

    SELECT @CurrentTable=TableName FROM @Tables WHERE RowID=@CurrentRow

    INSERT INTO @IndexInfoTemp
    exec sp_helpindex @CurrentTable

    INSERT INTO @IndexInfo
            (table_name   , index_name , index_description , index_keys)
        SELECT
            @CurrentTable , index_name , index_description , index_keys
        FROM @IndexInfoTemp

    DELETE FROM @IndexInfoTemp

    SET @CurrentRow=@CurrentRow+1

END --WHILE
SELECT * from @IndexInfo

РЕДАКТИРОВАТЬ,
если вы хотите, вы можете отфильтровать данные, вот несколько примеров (они работают для любого метода):

SELECT * FROM @IndexInfo WHERE index_description NOT LIKE '%primary key%'
SELECT * FROM @IndexInfo WHERE index_description NOT LIKE '%nonclustered%' AND index_description  LIKE '%clustered%'
SELECT * FROM @IndexInfo WHERE index_description LIKE '%unique%'

with connect(schema_name,table_name,index_name,index_column_id,column_name) as
(   select s.name schema_name, t.name table_name, i.name index_name, index_column_id, cast(c.name as varchar(max)) 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 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 index_column_id=1
union all
select s.name schema_name, t.name table_name, i.name index_name, ic.index_column_id, cast(connect.column_name + ',' + c.name as varchar(max)) 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 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 join connect on
connect.index_column_id+1 = ic.index_column_id
and connect.schema_name = s.name
and connect.table_name = t.name
and connect.index_name = i.name)
select connect.schema_name,connect.table_name,connect.index_name,connect.column_name
from connect join (select schema_name,table_name,index_name,MAX(index_column_id) index_column_id
from connect group by schema_name,table_name,index_name) mx
on connect.schema_name = mx.schema_name
and connect.table_name = mx.table_name
and connect.index_name = mx.index_name
and connect.index_column_id = mx.index_column_id
order by 1,2,3

Это способ возврата к индексам. Вы можете использовать SHOWCONTIG для оценки фрагментации. В нем будут перечислены все индексы для базы данных или таблицы, а также статистика. Я хотел бы предупредить, что на большой базе данных, это может быть длительным. Для меня одним из преимуществ этого подхода является то, что вам не нужно быть администратором, чтобы использовать его.

- Показать информацию фрагментации на все индексы в базе данных

SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG WITH ALL_INDEXES
GO

... выключите NOCOUNT, когда закончите

- Показать информацию фрагментации на все индексы в таблице

SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors) WITH ALL_INDEXES
GO

- Показать информацию фрагментации по конкретному индексу

SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors,aunmind)
GO

Могу ли я дать другой ответ на этот насыщенный вопрос?

Это либеральная доработка ответа @marc_s, смешанного с некоторыми материалами @Tim Ford, с целью получить немного более чистый и простой набор результатов, а также окончательное отображение и порядок для моих текущих потребностей.

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

основываясь на коде Тима Форда, это правильный ответ:

  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]

Я придумал этот, который дает мне точный обзор, который мне нужен. Помогает то, что вы получаете одну строку на индекс, в которую агрегируются столбцы индекса.

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

Поскольку в вашем профиле указано, что вы используете .NET, вы можете программно использовать объекты, управляемые сервером (SMO) ... в противном случае любой из приведенных выше ответов является фантастическим.

Вышеупомянутое решение элегантно, но согласно MS, INDEXKEY_PROPERTY устарела. Смотрите: http://msdn.microsoft.com/en-us/library/ms186773.aspx

В Oracle

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

В SQL Server с

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

Просто отметьте, что если вы собираетесь использовать любой из вышеперечисленных рабочих запросов для сценариев ваших индексов, вам нужно включить столбец filter_definition из таблицы sys.indexes в свои запросы, чтобы получить определение фильтра для некластеризованных индексов в SQL 2008+

AM

Приведенный ниже запрос включает всю необходимую информацию для пользовательских индексов (без индексов для уникальных ограничений и первичных ключей) со всеми столбцами:

SELECT I.name as IndexName, 
        CASE WHEN I.is_unique = 1 THEN 'Yes' ELSE 'No' END as 'Unique',
        I.type_desc COLLATE DATABASE_DEFAULT as Index_Type,
        '[' + SCHEMA_NAME(T.schema_id) + ']' as 'Schema',
        '[' + T.name + ']' as TableName,
        STUFF((SELECT ', [' + C.name + CASE WHEN IC.is_descending_key = 0 THEN '] ASC' ELSE '] DESC' END
            FROM sys.index_columns IC INNER JOIN sys.columns C ON  IC.object_id = C.object_id  AND IC.column_id = C.column_id
            WHERE IC.is_included_column = 0 AND IC.object_id = I.object_id AND IC.index_id = I.Index_id
            FOR XML PATH('')), 1, 2, '') as Key_Columns,
        Included_Columns, 
        I.filter_definition,
        CASE WHEN I.is_padded = 1 THEN 'ON' ELSE 'OFF' END as PAD_INDEX, 
        CASE WHEN ST.no_recompute = 0 THEN 'OFF' ELSE 'ON' END as [Statistics_Norecompute],
        CONVERT(VARCHAR(5), CASE WHEN I.fill_factor = 0 THEN 100 ELSE I.fill_factor END) as [Fillfactor],
        CASE WHEN I.ignore_dup_key = 1 THEN 'ON' ELSE 'OFF' END as [Ignore_Dup_Key],       
        CASE WHEN I.allow_row_locks = 1 THEN 'ON' ELSE 'OFF' END as [Allow_Row_Locks], 
        CASE WHEN I.allow_page_locks = 1 THEN 'ON' ELSE 'OFF' END [Allow_Page_Locks]        
FROM    sys.indexes I INNER JOIN        
        sys.tables T ON  T.object_id = I.object_id INNER JOIN       
        sys.stats ST ON  ST.object_id = I.object_id AND ST.stats_id = I.index_id INNER JOIN 
        sys.data_spaces DS ON  I.data_space_id = DS.data_space_id INNER JOIN 
        sys.filegroups FG ON  I.data_space_id = FG.data_space_id LEFT OUTER JOIN 
        (SELECT * FROM 
            (SELECT IC2.object_id, IC2.index_id,
                STUFF((SELECT ', ' + C.name FROM sys.index_columns IC1 INNER JOIN 
                    sys.columns C ON C.object_id = IC1.object_id
                        AND C.column_id = IC1.column_id
                        AND IC1.is_included_column = 1
                    WHERE  IC1.object_id = IC2.object_id AND IC1.index_id = IC2.index_id
                    GROUP BY IC1.object_id, C.name, index_id  FOR XML PATH('')
                ), 1, 2, '') as Included_Columns
            FROM sys.index_columns IC2
            GROUP BY IC2.object_id, IC2.index_id) tmp1
            WHERE Included_Columns IS NOT NULL
        ) tmp2
        ON tmp2.object_id = I.object_id AND tmp2.index_id = I.index_id
WHERE I.is_primary_key = 0 AND I.is_unique_constraint = 0;

В качестве дополнительного бонуса приведенный ниже запрос отформатирован для записи сценариев создания индекса и удаления индекса:

SELECT I.name as IndexName, 
        -- Uncommnent line below to include checking for index exists as part of the script
        --'IF NOT EXISTS (SELECT name FROM sysindexes WHERE name = '''+ I.name +''') ' +
        'CREATE ' + CASE WHEN I.is_unique = 1 THEN ' UNIQUE ' ELSE '' END +
        I.type_desc COLLATE DATABASE_DEFAULT + ' INDEX [' +
        I.name + '] ON [' + SCHEMA_NAME(T.schema_id) + '].[' + T.name + '] (' + STUFF(
        (SELECT ', [' + C.name + CASE WHEN IC.is_descending_key = 0 THEN '] ASC' ELSE '] DESC' END
            FROM sys.index_columns IC INNER JOIN sys.columns C ON  IC.object_id = C.object_id  AND IC.column_id = C.column_id
            WHERE IC.is_included_column = 0 AND IC.object_id = I.object_id AND IC.index_id = I.Index_id
            FOR XML PATH('')), 1, 2, '')  + ') ' +
        ISNULL(' INCLUDE (' + IncludedColumns + ') ', '') +
        ISNULL(' WHERE ' + I.filter_definition, '') + 
        'WITH (PAD_INDEX = ' + CASE WHEN I.is_padded = 1 THEN 'ON' ELSE 'OFF' END + 
        ', STATISTICS_NORECOMPUTE = ' + CASE WHEN ST.no_recompute = 0 THEN 'OFF' ELSE 'ON' END + 
        ', SORT_IN_TEMPDB = OFF' + 
        ', FILLFACTOR = ' + CONVERT(VARCHAR(5), CASE WHEN I.fill_factor = 0 THEN 100 ELSE I.fill_factor END) +
        ', IGNORE_DUP_KEY = ' + CASE WHEN I.ignore_dup_key = 1 THEN 'ON' ELSE 'OFF' END +      
        ', ONLINE = OFF' + 
        ', ALLOW_ROW_LOCKS = ' + CASE WHEN I.allow_row_locks = 1 THEN 'ON' ELSE 'OFF' END + 
        ', ALLOW_PAGE_LOCKS = ' + CASE WHEN I.allow_page_locks = 1 THEN 'ON' ELSE 'OFF' END + 
        ') ON [' + DS.name + '];' + CHAR(13) + CHAR(10) + 'GO' as [CreateIndex],
        'DROP INDEX ['+ I.name +'] ON ['+ SCHEMA_NAME(T.schema_id) +'].['+ T.name +'];' +
        CHAR(13) + CHAR(10) + 'GO' AS [DropIndex]
FROM    sys.indexes I INNER JOIN        
        sys.tables T ON  T.object_id = I.object_id INNER JOIN       
        sys.stats ST ON  ST.object_id = I.object_id AND ST.stats_id = I.index_id INNER JOIN 
        sys.data_spaces DS ON  I.data_space_id = DS.data_space_id INNER JOIN 
        sys.filegroups FG ON  I.data_space_id = FG.data_space_id LEFT OUTER JOIN 
        (SELECT * FROM 
            (SELECT IC2.object_id, IC2.index_id,
                STUFF((SELECT ', ' + C.name FROM sys.index_columns IC1 INNER JOIN 
                    sys.columns C ON C.object_id = IC1.object_id
                        AND C.column_id = IC1.column_id
                        AND IC1.is_included_column = 1
                    WHERE  IC1.object_id = IC2.object_id AND IC1.index_id = IC2.index_id
                    GROUP BY IC1.object_id, C.name, index_id  FOR XML PATH('')
                ), 1, 2, '') as IncludedColumns
            FROM sys.index_columns IC2
            GROUP BY IC2.object_id, IC2.index_id) tmp1
            WHERE IncludedColumns IS NOT NULL
        ) tmp2
        ON tmp2.object_id = I.object_id AND tmp2.index_id = I.index_id
WHERE I.is_primary_key = 0 AND I.is_unique_constraint = 0 

Это моё, работает на одной схеме по умолчанию, но её можно легко улучшить. Она дает 3 столбца с SQLQueries - Create / Drop / Rebuild (без реорганизации)

Запрос:

SELECT
'CREATE ' + 
CASE WHEN is_primary_key=1 THEN 'CLUSTERED' 
WHEN is_primary_key=0 and is_unique_constraint=0 THEN 'NONCLUSTERED'
WHEN is_primary_key=0 and is_unique_constraint=1 THEN 'UNIQUE' END  
+ ' INDEX ' +
QUOTENAME(i.name) + ' ON ' +
QUOTENAME(t.name) + ' ( '  + 
STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) + CASE WHEN ic.is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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
    ), '<row>', ', '), '</row>', ''), 1, 2, '') + ' ) '  -- keycols
+ COALESCE(' INCLUDE ( ' +
    STUFF(REPLACE(REPLACE((
        SELECT QUOTENAME(c.name) AS [data()]
        FROM sys.index_columns AS ic
        INNER JOIN sys.columns AS c ON ic.object_id = c.object_id AND ic.column_id = c.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.index_column_id
        FOR XML PATH
    ), '<row>', ', '), '</row>', ''), 1, 2, '') + ' ) ',    -- included cols
    '') as [Create],
'DROP INDEX ' + QUOTENAME(i.name) + ' ON ' + QUOTENAME(t.name) as [Drop],
'ALTER INDEX ' + QUOTENAME(i.name)  + ' ON ' +QUOTENAME(t.name) + ' REBUILD ' as [Rebuild]
FROM sys.tables AS t
INNER JOIN sys.indexes AS i ON t.object_id = i.object_id
LEFT JOIN sys.dm_db_index_usage_stats AS u ON i.object_id = u.object_id AND i.index_id = u.index_id
WHERE t.is_ms_shipped = 0
AND i.type <> 0
order by QUOTENAME(t.name), is_primary_key desc

Вывод

Create                                                                                                      Drop                                    Rebuild
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
CREATE CLUSTERED INDEX [PK_Table1] ON [Table1] ( [Tab1_ID] )                                                DROP INDEX [PK_Table1] ON [Table1]      ALTER INDEX [PK_Table1] ON [Table1] REBUILD 
CREATE UNIQUE INDEX [IX_Table1_Name] ON [Table1] ( [Tab1_Name] )                                            DROP INDEX [IX_Table1_Name] ON [Table1] ALTER INDEX [IX_Table1_Name] ON [Table1] REBUILD 
CREATE NONCLUSTERED INDEX [IX_Table2] ON [Table2] ( [Tab2_Name], [Tab2_City] )  INCLUDE ( [Tab2_PhoneNo] )  DROP INDEX [IX_Table2] ON [Table2]      ALTER INDEX [IX_Table2] ON [Table2] REBUILD

Во-первых, обратите внимание, что все вышеперечисленные запросы могут быть пропущены или по ошибке включают INCLUDE. столбцы в индексы. В некоторых также отсутствует правильное упорядочение и / или опция ASC / DESC столбцов.

Модифицированный выше запрос от jona. Кроме того, во многих базах данных, которые я использую, я устанавливаю свою собственную агрегатную функцию CLR CONCATENATE, поэтому приведенный ниже код зависит от наличия чего-то подобного. Вышеприведенные операторы SQL сводятся к гораздо более понятному:

SELECT
  s.[name] AS [schema_name]
, t.[name] AS [table_name]
, i.[name] AS [index_name]
, dbo.Concatenate(CASE WHEN ic.[key_ordinal] > 0 AND ic.[is_descending_key] = 1 THEN c.[name] + ' DESC' WHEN key_ordinal > 0 THEN c.[name] ELSE NULL END,',',1) AS [columns]
, dbo.Concatenate(CASE WHEN ic.[is_included_column] = 1 THEN c.[name] ELSE NULL END,',',1) AS [includes]
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 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
GROUP BY
  s.[name]
, t.[name]
, i.[name]
ORDER BY
  s.[name]
, t.[name]
, i.[name]

Существует множество конкатенационных агрегатов , если ваша среда допускает добавление к ней функций, основанных на CLR.