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
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。
其他回答
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
上面有一些不错的答案。但我更喜欢一个问题就能得到答案。 这段代码来自sys. .Sp_helpconstraint (sys proc)
这是微软查找是否有与tbl关联的外键的方法。
--setup variables. Just change 'Customer' to tbl you want
declare @objid int,
@objname nvarchar(776)
select @objname = 'Customer'
select @objid = object_id(@objname)
if exists (select * from sys.foreign_keys where referenced_object_id = @objid)
select 'Table is referenced by foreign key' =
db_name() + '.'
+ rtrim(schema_name(ObjectProperty(parent_object_id,'schemaid')))
+ '.' + object_name(parent_object_id)
+ ': ' + object_name(object_id)
from sys.foreign_keys
where referenced_object_id = @objid
order by 1
答案看起来像这样:test_db_name.dbo。账户:FK_Account_Customer
with tab_list as (
select t.name AS Table_Name, t.object_id, s.name AS Table_Schema from sys.tables t, sys.schemas s
where t.schema_id = s.schema_id
and s.name = 'your schema')
select IIF(col.column_id = 1, tab.TABLE_SCHEMA + '.' + tab.TABLE_NAME, NULL) Table_Name,
col.Name AS Column_Name, IIF(col.IS_NULLABLE= 0, 'NOT NULL', '') Nullable, st.name Type,
CASE WHEN st.name = 'decimal' THEN CONVERT(NVARCHAR(4000), col.Precision) + ',' + CONVERT(NVARCHAR(4000), col.Scale)
WHEN col.max_length = -1 THEN 'max'
WHEN st.name in ('int', 'bit', 'bigint', 'datetime2') THEN NULL
ELSE CONVERT(NVARCHAR(4000), col.max_length / 2)
END
AS Length,
ss.name + '.' + stab.name Referenced_Table, scol.name Referenced_Column
from sys.COLUMNS col
INNER JOIN tab_list tab ON col.object_id = tab.object_id
INNER JOIN sys.types st ON col.system_type_id = st.system_type_id AND col.user_type_id = st.user_type_id
LEFT JOIN [sys].[foreign_key_columns] sfkc ON col.object_id = sfkc.parent_object_id AND col.column_id = sfkc.parent_column_id
LEFT JOIN sys.tables stab ON sfkc.referenced_object_id = stab.object_id
LEFT JOIN sys.columns scol ON sfkc.referenced_object_id = scol.object_id AND sfkc.referenced_column_id = scol.column_id
LEFT JOIN sys.schemas ss ON ss.schema_id = stab.schema_id
我正在使用这个脚本来查找与外键相关的所有细节。 我正在使用INFORMATION.SCHEMA。 下面是一个SQL脚本:
SELECT
ccu.table_name AS SourceTable
,ccu.constraint_name AS SourceConstraint
,ccu.column_name AS SourceColumn
,kcu.table_name AS TargetTable
,kcu.column_name AS TargetColumn
FROM INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE ccu
INNER JOIN INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS rc
ON ccu.CONSTRAINT_NAME = rc.CONSTRAINT_NAME
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE kcu
ON kcu.CONSTRAINT_NAME = rc.UNIQUE_CONSTRAINT_NAME
ORDER BY ccu.table_name
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。