Zapytanie SQL Server o indeksy klastrowane

Question

Czy można napisać zapytanie zwracające wszystkie tabele z indeksami klastrowanymi, które nie są oparte na kluczu tożsamości?

Answer 1

Jak o tym:

SELECT 
    TableName = t.name, 
    ClusteredIndexName = i.name, 
    ColumnName = c.Name 
FROM 
    sys.tables t 
INNER JOIN 
    sys.indexes i ON t.object_id = i.object_id 
INNER JOIN 
    sys.index_columns ic ON i.index_id = ic.index_id AND i.object_id = ic.object_id 
INNER JOIN 
    sys.columns c ON ic.column_id = c.column_id AND ic.object_id = c.object_id 
WHERE 
    i.index_id = 1 -- clustered index 
    AND c.is_identity = 0 
    AND EXISTS (SELECT * 
       FROM sys.columns c2 
       WHERE ic.object_id = c2.object_id AND c2.is_identity = 1) 

OK, ta kwerenda będzie lista tych kluczy podstawowych, które mają kolumnę, która jest nie tożsamość, ale gdzie tam także dodatkowo druga kolumna w klucz podstawowy tego IS kolumna .

Answer 2

SELECT s.name AS schema_name, o.name AS object_name, i.name AS index_name 
FROM sys.indexes i 
JOIN sys.objects o ON i.object_id = o.object_id 
JOIN sys.schemas s ON o.schema_id = s.schema_id 
WHERE i.type = 1 -- Clustered index 
--AND  o.is_ms_shipped = 0 -- Uncomment if you want to see only user objects 
AND  NOT EXISTS (
    SELECT * 
    FROM sys.index_columns ic INNER JOIN sys.columns c ON c.object_id = ic.object_id AND c.column_id = ic.column_id 
    WHERE ic.object_id = i.object_id AND ic.index_id = i.index_id 
    AND  c.is_identity = 1 -- Is identity column 
) 
ORDER BY schema_name, object_name, index_name; 

Przykładowe wyjście (AdventureWorks2008R2):

schema_name object_name     index_name 
-------------- --------------------------- -------------------------------------------------------------------- 
HumanResources Employee     PK_Employee_BusinessEntityID 
HumanResources EmployeeDepartmentHistory PK_EmployeeDepartmentHistory_BusinessEntityID_StartDate_DepartmentID 
HumanResources EmployeePayHistory   PK_EmployeePayHistory_BusinessEntityID_RateChangeDate 
Person   BusinessEntityAddress  PK_BusinessEntityAddress_BusinessEntityID_AddressID_AddressTypeID 
Person   BusinessEntityContact  PK_BusinessEntityContact_BusinessEntityID_PersonID_ContactTypeID 

Answer 3

następujące zapytanie daje wszystkie tabele użytkownika, kolumny, typ danych, a jeżeli kolumna jest częścią indeksu klastra powróci sekwencję kolumny/kolejność w skorupie indeksu inaczej zwróci wartość NULL.

SELECT U.name [OWNER],O.name [TABLE_NAME],C.name [COLUMN_NAME],T.name [DATA_TYPE],C.length [DATA_LENGTH], x.keyno [Primary_Key_order] 
FROM syscolumns C 
inner join sysobjects O on O.Id=C.Id and o.xtype='U' -- User Tables 
inner join sysusers U on O.Uid=U.UID 
inner join systypes T on C.xtype=T.xtype 
left outer join (Select O.name [TABLE_NAME] , C.name [COLUMN_NAME], IK.keyno 
      from syscolumns C 
      inner join sysobjects O on O.Id=C.Id and O.xtype='U' -- User Tables 
      join sysindexkeys IK on O.id=IK.ID and C.colid=IK.COLID and Indid=1 -- Only Clustered Index 
      ) x 
      on x.TABLE_NAME=O.name and X.COLUMN_NAME=C.name 
order by U.name