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

Mysql服务器有information_schema。REFERENTIAL_CONSTRAINTS表供参考,您可以通过表名或引用表名过滤它。

我知道这是一个很晚(非常晚)的回复,但我找到了这些简单的方法来找到所有的foreign_key_references。这是解决方案;

解决方案1:

EXEC SP_FKEYS 'MyTableName';   // It'll show you the all the information(in multiple tables) regarding to the TableName with all ForeignKey_References.

解决方案2:

EXEC SP_HELP 'MyTableName';   // It'll show all ForeignKey references in a single table.

解决方案03:

// It'll show you the Column_Name with Referenced_Table_Name

SELECT 
   COL_NAME(fc.parent_object_id,fc.parent_column_id) Column_Name,
   OBJECT_NAME(f.parent_object_id) Table_Name
FROM 
   sys.foreign_keys AS f
INNER JOIN 
   sys.foreign_key_columns AS fc 
      ON f.OBJECT_ID = fc.constraint_object_id
INNER JOIN 
   sys.tables t 
      ON t.OBJECT_ID = fc.referenced_object_id
WHERE 
   OBJECT_NAME (f.referenced_object_id) = 'MyTableName'

希望这对你有很大帮助。: -)

@BankZ的最好回答

sp_help 'TableName'   

另外,对于不同的模式

sp_help 'schemaName.TableName'   

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

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;