Dev*_*Dog 7 t-sql sql-server identity clustered-index
是否可以编写一个返回所有具有不基于身份密钥的聚簇索引的表的查询?
mar*_*c_s 13
这个怎么样:
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)
Run Code Online (Sandbox Code Playgroud)
OK,这个查询将列出有一列是那些主键不认同,但那里也附加在主键约束第二列IS的IDENTITY列.
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;
Run Code Online (Sandbox Code Playgroud)
示例输出(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
Run Code Online (Sandbox Code Playgroud)