我需要在SQL Server数据库中删除一个高度引用的表。我如何才能得到所有外键约束的列表,我将需要删除以便删除表?
(SQL比在管理工作室的GUI中点击更可取)
我需要在SQL Server数据库中删除一个高度引用的表。我如何才能得到所有外键约束的列表,我将需要删除以便删除表?
(SQL比在管理工作室的GUI中点击更可取)
当前回答
以下的解决方案对我来说很有效:
--Eliminar las llaves foraneas
declare @query varchar(8000)
declare cursorRecorrerTabla cursor for
SELECT 'ALTER TABLE [PoaComFinH].['+sch.name+'].['+referencingTable.Name+'] DROP CONSTRAINT ['+foreignKey.name+']' 'query'
FROM PoaComFinH.sys.foreign_key_columns fk
JOIN PoaComFinH.sys.tables referencingTable ON fk.parent_object_id = referencingTable.object_id
JOIN PoaComFinH.sys.schemas sch ON referencingTable.schema_id = sch.schema_id
JOIN PoaComFinH.sys.objects foreignKey ON foreignKey.object_id = fk.constraint_object_id
JOIN PoaComFinH.sys.tables referencedTable ON fk.referenced_object_id = referencedTable.object_id
--3ro. abrir el cursor.
open cursorRecorrerTabla
fetch next from cursorRecorrerTabla
into @query
while @@fetch_status = 0
begin
--inicio cuerpo del cursor
print @query
exec(@query)
--fin cuerpo del cursor
fetch next from cursorRecorrerTabla
into @query
end
--cerrar cursor
close cursorRecorrerTabla
deallocate cursorRecorrerTabla
其他回答
我将使用SQL Server Management Studio中的数据库图表功能,但既然你排除了它-这在SQL Server 2008中为我工作(没有2005)。
获取引用表和列名的列表…
select
t.name as TableWithForeignKey,
fk.constraint_column_id as FK_PartNo, c.
name as ForeignKeyColumn
from
sys.foreign_key_columns as fk
inner join
sys.tables as t on fk.parent_object_id = t.object_id
inner join
sys.columns as c on fk.parent_object_id = c.object_id and fk.parent_column_id = c.column_id
where
fk.referenced_object_id = (select object_id
from sys.tables
where name = 'TableOthersForeignKeyInto')
order by
TableWithForeignKey, FK_PartNo
获取外键约束的名称
select distinct name from sys.objects where object_id in
( select fk.constraint_object_id from sys.foreign_key_columns as fk
where fk.referenced_object_id =
(select object_id from sys.tables where name = 'TableOthersForeignKeyInto')
)
SELECT
object_name(parent_object_id),
object_name(referenced_object_id),
name
FROM sys.foreign_keys
WHERE parent_object_id = object_id('Table Name')
第一个
EXEC sp_fkeys 'Table', 'Schema'
然后使用NimbleText处理你的结果
最初的问题要求将所有外键的列表放入一个高度引用的表中,以便可以删除表。
这个小查询返回将所有外键放入特定表所需的“drop foreign key”命令:
SELECT
'ALTER TABLE ['+sch.name+'].['+referencingTable.Name+'] DROP CONSTRAINT ['+foreignKey.name+']' '[DropCommand]'
FROM sys.foreign_key_columns fk
JOIN sys.tables referencingTable ON fk.parent_object_id = referencingTable.object_id
JOIN sys.schemas sch ON referencingTable.schema_id = sch.schema_id
JOIN sys.objects foreignKey ON foreignKey.object_id = fk.constraint_object_id
JOIN sys.tables referencedTable ON fk.referenced_object_id = referencedTable.object_id
WHERE referencedTable.name = 'MyTableName'
示例输出:
[DropCommand]
ALTER TABLE [dbo].[OtherTable1] DROP CONSTRAINT [FK_OtherTable1_MyTable]
ALTER TABLE [dbo].[OtherTable2] DROP CONSTRAINT [FK_OtherTable2_MyTable]
省略where -子句以获取当前数据库中所有外键的删除命令。
SELECT OBJECT_NAME(fk.parent_object_id) as ReferencingTable,
OBJECT_NAME(fk.constraint_object_id) as [FKContraint]
FROM sys.foreign_key_columns as fk
WHERE fk.referenced_object_id = OBJECT_ID('ReferencedTable', 'U')
这只显示了外键约束的关系。我的数据库显然早于FK约束。一些表使用触发器来强制引用完整性,有时除了一个类似命名的列来指示关系之外什么都没有(根本没有引用完整性)。
幸运的是,我们有一个一致的命名场景,所以我能够找到引用表 观点是这样的:
SELECT OBJECT_NAME(object_id) from sys.columns where name like 'client_id'
我使用这个选择作为生成一个脚本的基础,做我需要做的事情 相关的表格。