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

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

其他回答

Oracle SQL

select *
from
    all_constraints
where
    r_constraint_name in
    (select       constraint_name
    from
       all_constraints
    where
       table_name='PUT_THE_TABLE_NAME_HERE');

all_constraints是Oracle DB中的一个固有表名。

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

这个答案建立在sp_fkeys的基础上,但格式化类似于sp_fkeys,适用于多个列并列出它们的顺序。

SELECT fk_obj.name    AS FK_NAME,
       pk_schema.name AS PKTABLE_OWNER,
       pk_table.name  AS PKTABLE_NAME,
       pk_column.name AS PKCOLUMN_NAME,
       fk_schema.name AS FKTABLE_OWNER,
       fk_table.name  AS FKTABLE_NAME,
       fk_column.name AS FKCOLUMN_NAME,
       ROW_NUMBER() over (
           PARTITION BY fk_obj.name, fk_schema.name
           ORDER BY fkc.constraint_column_id
           )          AS KEY_SEQ
FROM sys.foreign_key_columns fkc
         INNER JOIN sys.objects fk_obj
                    ON fk_obj.object_id = fkc.constraint_object_id
         INNER JOIN sys.tables fk_table
                    ON fk_table.object_id = fkc.parent_object_id
         INNER JOIN sys.schemas fk_schema
                    ON fk_table.schema_id = fk_schema.schema_id
         INNER JOIN sys.columns fk_column
                    ON fk_column.column_id = parent_column_id
                        AND fk_column.object_id = fk_table.object_id
         INNER JOIN sys.tables pk_table
                    ON pk_table.object_id = fkc.referenced_object_id
         INNER JOIN sys.schemas pk_schema
                    ON pk_table.schema_id = pk_schema.schema_id
         INNER JOIN sys.columns pk_column
                    ON pk_column.column_id = fkc.referenced_column_id
                        AND pk_column.object_id = pk_table.object_id;

下面是我将使用的SQL代码。

SELECT 
   f.name AS 'Name of Foreign Key',
   OBJECT_NAME(f.parent_object_id) AS 'Table name',
   COL_NAME(fc.parent_object_id,fc.parent_column_id) AS 'Fieldname',
   OBJECT_NAME(t.object_id) AS 'References Table name',
   COL_NAME(t.object_id,fc.referenced_column_id) AS 'References fieldname',

   'ALTER TABLE [' + OBJECT_NAME(f.parent_object_id) + ']  DROP CONSTRAINT [' + f.name + ']' AS 'Delete foreign key',

   'ALTER TABLE [' + OBJECT_NAME(f.parent_object_id) + ']  WITH NOCHECK ADD CONSTRAINT [' + 
        f.name + '] FOREIGN KEY([' + COL_NAME(fc.parent_object_id,fc.parent_column_id) + ']) REFERENCES ' + 
        '[' + OBJECT_NAME(t.object_id) + '] ([' +
        COL_NAME(t.object_id,fc.referenced_column_id) + '])' AS 'Create foreign key'
    -- , delete_referential_action_desc AS 'UsesCascadeDelete'
FROM sys.foreign_keys AS f,
     sys.foreign_key_columns AS fc,
     sys.tables t 
WHERE f.OBJECT_ID = fc.constraint_object_id
AND t.OBJECT_ID = fc.referenced_object_id
AND OBJECT_NAME(t.object_id) = 'Employees'      --  Just show the FKs which reference a particular table
ORDER BY 2

这不是特别清晰的SQL,所以让我们看一个例子。

所以,假设我想删除Microsoft心爱的Northwind数据库中的Employees表,但是SQL Server告诉我一个或多个外键阻止我这样做。

上面的SQL命令将返回这些结果…

它显示有3个外键引用雇员表。换句话说,除非先删除这三个外键,否则不允许删除(删除)这个表。

在结果中,第一行是以下外键约束在结果中的显示方式。

ALTER TABLE [dbo].[Employees]  WITH NOCHECK 
ADD CONSTRAINT [FK_Employees_Employees] FOREIGN KEY([ReportsTo])
REFERENCES [dbo].[Employees] ([EmployeeID])

倒数第二列显示了我需要使用的SQL命令来删除这些外键之一,例如:

ALTER TABLE [Employees] DROP CONSTRAINT [FK_Employees_Employees]

...右边一列显示了创建它的SQL…

ALTER TABLE [Employees] WITH NOCHECK 
ADD CONSTRAINT [FK_Employees_Employees] 
FOREIGN KEY([ReportsTo]) REFERENCES [Employees] ([EmployeeID])

使用所有这些命令,您就拥有了删除相关外键所需的一切,从而可以删除表,然后再重新创建它们。

唷。希望这能有所帮助。

 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。