Список всех индексов и столбцов индекса в БД SQL Server

Question

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, чего я хочу.
Я хочу, чтобы список всех пользовательских индексов ( означал отсутствие индексов, поддерживающих уникальные ограничения и первичные ключи ) со всеми столбцами (упорядоченными по тому, как они отображаются в определении индекса) плюс столько метаданных насколько это возможно.

Answer 1

Существует два вида каталога "sys", к которым вы можете обратиться:

select * from sys.indexes

select * from sys.index_columns

Они предоставят вам практически любую информацию о индексах и их столбцах, которую вы, возможно, захотите.

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;

Answer 2

Вы можете использовать sp_helpindex для просмотра всех индексов одной таблицы.

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

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

Answer 3

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 пустые значения, вы знаете, что ни один из них не пропущен.

Фильтрация первичных ключей и т. Д., Как в исходном запросе, тривиальна.

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

Answer 4

- 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]   

Answer 5

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

Я использовал его (без условий / фильтров) для своих требований, но он дал неверные результаты

Основная проблема заключалась в получении результатов для перекрестного продукта 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

Answer 6

Мне нужно было получить определенные индексы, их столбцы индексов и включенные в них столбцы. 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])

Answer 7

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

1. Название таблицы
2. Индекс
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;

Answer 8

Следующее дает то, что похоже на sp_helpindex имя таблицы

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

Answer 9

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 так:

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

EDIT
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%'

Answer 10

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