Liste de toutes les colonnes index & index dans la base de données SQL Server

Comment puis-je obtenir une liste de toutes les colonnes index & index dans SQL Server 2005+? Le plus proche que j'ai pu obtenir est: 

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

Qui n'est pas exactement ce que je veux.

Ce que je veux, c'est lister tous les index définis par l'utilisateur, ( qui signifie pas d'index supportant des contraintes uniques et des clés primaires ) avec toutes les colonnes (ordonnées par la façon dont elles apparaissent dans la définition de l'index) plus autant de métadonnées que possible.

278
demandé sur Vikrant 2009-04-19 22:38:24

26 réponses

il y a deux "sys" vues de catalogue que vous pouvez consulter: 

select * from sys.indexes


select * from sys.index_columns

ceux-ci vous donneront à peu près n'importe quelle information que vous pourriez éventuellement vouloir sur les indices et leurs colonnes.

EDIT: cette requête devient assez proche de ce que vous cherchez: 

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;
484
répondu marc_s 2017-01-03 14:03:46

vous pouvez utiliser le sp_helpindex pour voir tous les index d'une table. 

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

Et pour tous les indices, vous pouvez parcourir sys.objects pour obtenir tous les indices pour chaque table.

54
répondu Dragon 2012-10-27 22:54:24

rien de ce qui précède n'a fait le travail pour moi, mais cela fait: 

-- 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)

si vos noms comprennent des espaces, ajoutez des crochets autour d'eux dans les scripts de création.

quand la dernière colonne est nulle, vous savez qu'il n'en manque aucune.

filtrer les clés primaires etc comme dans la requête originale est trivial.

NOTE: faites attention avec cette solution car elle ne distingue pas les indexés et les indexés. des colonnes incluses.

34
répondu KDF9 2014-03-06 13:05:11

-- court et doux: 

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]
30
répondu 2009-06-23 20:09:55

Hé les gars, Je ne suis pas passé par là mais j'ai eu ce que je voulais dans la requête Postée par l'auteur original.

Je l'ai utilisé (sans conditions / filtres) pour mes besoins mais il a donné des résultats incorrects

le problème principal était les résultats obtenir produit croisé sans condition de jointure sur 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
8
répondu Solow Developer 2010-07-13 14:50:32

ce qui suit donne ce qui est similaire à sp_helpindex tablename 

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
8
répondu hshen 2011-10-03 11:07:26

j'ai eu besoin d'obtenir des index particuliers, leurs colonnes d'index et leurs colonnes incluses aussi bien. Voici la requête que j'ai utilisée: 

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])
8
répondu gotqn 2014-07-15 08:56:32

cela va fonctionner: 

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

cela ne renvoie pas le nom de la table et vous obtiendrez des avertissements pour toutes les tables sans index, si c'est un problème, vous pouvez créer une boucle sur les tables qui ont des index comme ceci: 

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

MODIFIER

si vous voulez, vous pouvez filtrer les données, voici quelques exemples (ces travaux pour l'une ou l'autre méthode): 

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%'
6
répondu KM. 2009-04-21 13:23:03 
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
6
répondu jona 2011-06-15 21:39:29

les travaux suivants sur SQL Server 2014/2016 ainsi que toute base de données Microsoft Azure SQL.

produit un jeu complet de résultats qui est facilement exportable à Notepad / Excel pour le tranchage et le découpage et comprend

  1. Nom De La Table
  2. Nom Index
  3. Description De L'Index
  4. colonnes indexées - dans L'ordre
  5. Colonnes Incluses - En ordre 
 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;
5
répondu AeyJey 2017-06-10 00:06:40

C'est une façon de reculer dans l'index. Vous pouvez utiliser SHOWCONTIG pour évaluer la fragmentation. Il dressera la liste de tous les index de la base de données ou du tableau, ainsi que des statistiques. Je tiens à souligner que sur une base de données importante, cela peut être de longue durée. Pour moi, l'un des avantages de cette approche est que vous n'avez pas à être un administrateur de l'utiliser.

--afficher les informations de fragmentation sur tous les index dans une base de données 

SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG WITH ALL_INDEXES
GO

...tour NOCOUNT back OFF lorsque tout est terminé

--afficher les informations de fragmentation sur tous les index sur une table 

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

--fragmentation de l'information sur un index spécifique 

SET NOCOUNT ON
USE pubs
DBCC SHOWCONTIG (authors,aunmind)
GO
4
répondu DOK 2009-04-19 18:54:51

puis-je hasarder une autre réponse à cette question saturée?

il s'agit d'une refonte libérale de @marc_s answer, mélangée avec quelques trucs de @Tim Ford, avec l'objectif d'avoir un peu d'un résultat plus propre et plus simple ensemble et l'affichage final et la commande pour mon besoin actuel. 

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
4
répondu Nicholas Petersen 2014-04-30 21:13:45

basé sur le code de Tim Ford, c'est la bonne réponse: 

  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]
3
répondu Stefan Cantacuz 2012-04-19 15:40:04

j'ai trouvé celui-ci, qui me donne la vue d'ensemble exacte dont j'ai besoin. Ce qui est aide est que vous obtenez une ligne par index dans l'index des colonnes sont regroupées. 

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
3
répondu Danie Kritzinger 2013-12-04 13:58:22

basé sur la réponse acceptée et deux autres questions 1 , 2 j'ai rassemblé la requête suivante: 

SELECT
    sys.indexes.name AS IndexName,
    sys.tables.name AS TableName,
    REPLACE((
        SELECT sys.columns.name + CASE WHEN is_descending_key = 1 THEN ' DESC' ELSE '' END AS [data()]
        FROM sys.index_columns
        INNER JOIN sys.columns ON sys.index_columns.object_id = sys.columns.object_id AND sys.index_columns.column_id = sys.columns.column_id
        WHERE sys.index_columns.object_id = sys.indexes.object_id AND sys.index_columns.index_id = sys.indexes.index_id AND sys.index_columns.is_included_column = 0
        ORDER BY sys.index_columns.key_ordinal
        FOR XML PATH('')
    ), ' ', ', ') AS KeyColumns,
    REPLACE((
        SELECT sys.columns.name AS [data()]
        FROM sys.index_columns
        INNER JOIN sys.columns ON sys.index_columns.object_id = sys.columns.object_id AND sys.index_columns.column_id = sys.columns.column_id
        WHERE sys.index_columns.object_id = sys.indexes.object_id AND sys.index_columns.index_id = sys.indexes.index_id AND sys.index_columns.is_included_column = 1
        ORDER BY sys.index_columns.index_column_id
        FOR XML PATH('')
    ), ' ', ', ') AS IncludedColumns,
    sys.dm_db_index_usage_stats.user_updates,
    sys.dm_db_index_usage_stats.user_seeks,
    sys.dm_db_index_usage_stats.user_scans,
    sys.dm_db_index_usage_stats.user_lookups,
    sys.dm_db_index_usage_stats.user_seeks + sys.dm_db_index_usage_stats.user_scans + sys.dm_db_index_usage_stats.user_lookups AS total_usage
FROM sys.indexes
LEFT JOIN sys.tables ON sys.indexes.object_id = sys.tables.object_id
LEFT JOIN sys.dm_db_index_usage_stats ON sys.indexes.object_id = sys.dm_db_index_usage_stats.object_id AND sys.indexes.index_id = sys.dm_db_index_usage_stats.index_id
WHERE sys.indexes.type <> 0 AND sys.tables.is_ms_shipped = 0

cette requête renvoie des résultats tels que ci-dessous qui montre la liste des index, leurs colonnes et leur utilisation. Très utile pour déterminer quel indice donne de meilleurs résultats que les autres:

index list, columns and usage

3
répondu Salman A 2017-12-06 12:10:47

puisque votre profil indique que vous utilisez .NET vous pourrait utiliser des objets gérés par serveur (Omu) de façon programmatique... autrement, une des réponses ci-dessus sont fantastiques.

2
répondu Kane 2009-07-27 12:10:55

la solution ci-dessus est élégante, mais selon MS, INDEXKEY_PROPERTY est déprécié. Voir: http://msdn.microsoft.com/en-us/library/ms186773.aspx

2
répondu Lisa 2011-03-18 22:01:21

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

dans SQL Server avec 

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
2
répondu jona 2012-04-29 06:10:27

notez juste que si vous allez utiliser l'une des requêtes de travail ci-dessus pour script vos index, vous devez incorporer la colonne filter_definition de sys.table des index dans vos requêtes pour obtenir la définition de filtre des index non-groupés en SQL 2008+

AM

2
répondu user3101273 2013-12-14 01:17:44

tout d'abord, veuillez noter que toutes les requêtes ci-dessus peuvent omettre ou incorporer par erreur les colonnes INCLUDE des indices. Il manque aussi dans certains cas l'option d'ordre et/ou ASC/DESC des colonnes.

a modifié la requête ci-dessus par jona. Comme mise à part, dans beaucoup de la base de données que j'utilise, j'installe ma propre fonction agrégée concaténée CLR, de sorte que le code ci-dessous dépend de quelque chose comme ceci étant présent. Les instructions SQL ci-dessus réduisent à un bien plus maintenable: 

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]

il y a beaucoup de agrégats de concaténation dehors si votre environnement permet des fonctions basées CLR ajoutées à elle.

1
répondu alphadogg 2011-10-25 16:35:38

pour les colonnes uniques par index: 

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
1
répondu user4525484 2015-02-03 18:23:55

la requête ci-dessous inclut toutes les informations pertinentes pour les index définis par l'Utilisateur, (pas d'index pour les contraintes uniques et les clés primaires) avec toutes les colonnes: 

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;

en bonus, la requête ci-dessous est formatée pour écrire les scripts create index et drop index: 

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
1
répondu SteveD 2017-04-19 09:18:18

Voici la meilleure façon de le faire: 

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'

je préfère utiliser des jointures implicites car il est beaucoup plus facile pour moi de comprendre. Vous pouvez supprimer la référence object_id car vous n'en avez peut-être pas besoin.

santé.

0
répondu Obinwanne Hill 2010-04-06 15:00:31

en utilisant SQL Server 2016, cela donne une liste complète de tous les index, avec un dump inclus de chaque table de sorte que vous pouvez voir comment les tables se rapportent. Il montre également les colonnes incluses dans les indices de couverture: 

select t.name TableName, i.name IdxName, c.name ColName
    , ic.index_column_id ColPosition
    , i.type_desc Type
    , case when i.is_primary_key = 1 then 'Yes' else '' end [Primary?]
    , case when i.is_unique = 1 then 'Yes' else '' end [Unique?]
    , case when ic.is_included_column = 0 then '' else 'Yes - Included' end [CoveredColumn?]
    , 'indexes >>>>' [*indexes*], i.*, 'index_columns >>>>' [*index_columns*]
    , ic.*, 'tables >>>>' [*tables*]
    , t.*, 'columns >>>>' [*columns*], c.*
from sys.index_columns ic
join sys.tables t on t.object_id = ic.object_id
join sys.columns c on c.object_id = t.object_id and c.column_id = ic.column_id
join sys.indexes i on i.object_id = t.object_id and i.index_id = ic.index_id
order by TableName, IdxName, ColPosition
0
répondu ejamesr 2017-03-21 18:24:49

j'ai utilisé la requête suivante lorsque j'ai eu cette exigence... 

SELECT 
    TableName = t.name,
    ColumnId = col.column_id, 
    ColumnName = col.name,
    DataType = ty.name,
    MaxSize = ty.max_length,
    IsNullable = CASE WHEN (col.is_nullable = 1) THEN 'Y' END,
    IsIdentity = CASE WHEN (col.is_identity = 1) THEN 'Y' END,
    IsPrimaryKey = CASE WHEN (ic.column_id = col.column_id) THEN 'Y' END,
    IsForeignKey = CASE WHEN (fkc.parent_column_id = col.column_id) THEN 'Y' END,
    IsDefault = CASE WHEN (dc.parent_column_id = col.column_id) THEN 'Y' END
FROM 
    sys.tables t
INNER JOIN 
     sys.columns col ON t.object_id = col.object_id 
LEFT JOIN
    sys.indexes ind ON t.object_id = ind.object_id 
LEFT JOIN 
     sys.index_columns ic ON ic.index_id=ind.index_id AND ic.object_id = col.object_id and ic.column_id = col.column_id
LEFT JOIN sys.foreign_key_columns fkc
                ON fkc.parent_object_id = col.object_id AND fkc.parent_column_id=col.column_id
LEFT JOIN sys.default_constraints dc
                ON dc.parent_object_id = col.object_id AND dc.parent_column_id=col.column_id
LEFT JOIN
     sys.types ty on ty.user_type_id = col.user_type_id

WHERE
    --t.name='<TABLENAME>'
    t.schema_id = 10    --SCHEMA ID
    AND ind.is_primary_key=1    
ORDER BY
    t.name, ColumnId
0
répondu Amit Philips 2018-07-19 05:53:29 
sELECT 
     TableName = t.name,
     IndexName = ind.name,
     --IndexId = ind.index_id,
     ColumnId = ic.index_column_id,
     ColumnName = col.name,
     key_ordinal,
     ind.type_desc
     --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 
     and t.name='CompanyReconciliation' --table name
     and key_ordinal>0
ORDER BY 
     t.name, ind.name, ind.index_id, ic.index_column_id
-3
répondu Vinod G 2017-01-02 03:21:11