Liste de tous les index et colonnes d'index dans la BD SQL Server

Question

Comment obtenir une liste de tous les index et colonnes d'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

Ce qui n'est pas exactement ce que je veux.
Ce que je veux, c'est lister tous les index définis par l'utilisateur, ( ce qui signifie qu'il n'y a pas d'index qui supportent les contraintes uniques et les clés primaires. ) avec toutes les colonnes (classées par ordre d'apparition dans la définition de l'index) et autant de métadonnées que possible.

Answer 1

Voici la meilleure façon de procéder :

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 c'est beaucoup plus facile à comprendre pour moi. Vous pouvez supprimer la référence object_id car vous n'en aurez peut-être pas besoin.

A la vôtre.

Answer 2

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

Answer 3

Solution fonctionnelle pour SQL Server 2014. Je n'ai inclus qu'une poignée de champs de sortie ici, mais n'hésitez pas à en ajouter autant que vous le souhaitez.

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

Answer 4

J'ai donné une mise à jour à la réponse de KFD9.

J'ai adapté leur version pour supporter la spécification include et ne pas utiliser la propriété indexkey_property qui est déprécié

Cela vous donne une déclaration de création et de suppression pour les index et les contraintes.

with indexes as (
    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)+' ' as 'Type',  -- B=basic, C=Clustered, X=XML
      (select string_agg(CAST('[' + c.name + ']' + case when is_descending_key = 1 then ' DESC' else '' end AS NVARCHAR(MAX)), ',') within group (order by index_column_id) 
         from sys.index_columns ic JOIN sys.columns c on ic.column_id = c.column_id and ic.object_id = c.object_id where ic.index_id = si.index_id and ic.object_id = si.object_id and ic.is_included_column = 0) Cols,
      (select string_agg(CAST('[' + c.name + ']' + case when is_descending_key = 1 then ' DESC' else '' end AS NVARCHAR(MAX)), ',') within group (order by index_column_id) 
         from sys.index_columns ic JOIN sys.columns c on ic.column_id = c.column_id and ic.object_id = c.object_id where ic.index_id = si.index_id and ic.object_id = si.object_id and ic.is_included_column = 1) IncludedCols,
      (select count(*) from sys.index_columns ic where ic.index_id = si.index_id and ic.object_id = si.object_id) IndexColsCount
    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
)
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)+
  ' ('+Cols+')'+
  isnull(' include ('+IncludedCols+')', '')+
  '' as CreateIndex,
    CASE pk
    WHEN 'PK' THEN 'ALTER '+
     'TABLE ['+SchemaName+'].['+TableName+'] DROP CONSTRAINT ['+IndexName+'] '
    ELSE 'DROP INDEX ['+IndexName+'] ON ['+SchemaName+'].['+TableName + ']'
    END AS DropIndex,
    IndexColsCount
FROM indexes
ORDER BY SchemaName,TableName,IndexName

Answer 5

La bonne solution est ici (tous les messages ci-dessus donnent le résultat du produit cartésien lorsque nous avons plus d'un index sur une table).

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