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.
32 ответа
Решение
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_helpindex просмотреть все индексы одной таблицы.
EXEC sys.sp_helpindex @objname = N'User' -- nvarchar(77)
И по всем показателям вы можете пройти sys.objects чтобы 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 содержат 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.
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
- Имя таблицы
- Имя индекса
- Индекс Описание
- Индексированные столбцы - в порядке
- 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;
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]
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])
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 соединения 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
Ниже приводится то, что похоже на имя таблицы sp_helpindex
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
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
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 в большой базе данных она может быть длительной. 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
Это мое, работает с одной схемой по умолчанию, но ее 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
Выход
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
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
Я придумал этот, который дает мне точный обзор, который мне нужен. Помогает то, что вы получаете по одной строке 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
Запрос ниже включает в себя 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 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
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]
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
Вышеупомянутое решение элегантно, но согласно MS, INDEXKEY_PROPERTY устарела. Смотрите: http://msdn.microsoft.com/en-us/library/ms186773.aspx
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.
Используя SQL Server 2016, это дает полный список всех индексов с включенным дампом каждой таблицы, чтобы вы могли увидеть, как эти таблицы связаны. Также показаны столбцы, включенные в индексы покрытия:
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
Для уникальных столбцов по индексу:
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
Во-первых, обратите внимание, что все вышеперечисленные запросы могут пропустить или ошибочно включить столбцы INCLUDE 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.
Я использовал следующий запрос, когда у меня было это требование...
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
Рабочее решение для SQL Server 2014. Я включил здесь только несколько выходных полей, но не стесняйтесь добавлять сколько угодно.
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
Вот лучший способ сделать это:
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'Я предпочитаю использовать неявные объединения, так как мне гораздо легче понять. Вы можете удалить ссылку на object_id, так как она может вам не понадобиться.
Приветствия.
Я дал ответ KFD9 обновлением.
Я адаптировал их версию для поддержки спецификации include и не использовал indexkey_property, который устарел.
Это дает вам инструкции create и drop для индексов и ограничений.
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
Правильный здесь (все вышеперечисленные сообщения дадут декартово произведение, когда у нас есть более одного индекса в таблице)
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