如何找出 SQL Server 中哪个 FOREIGN KEY 约束引用了表？

Question

我正在尝试删除表格，但收到以下消息：

消息 3726，第 16 级，状态 1，第 3 行
无法删除对象“dbo.UserProfile”，因为它由外键约束引用。
消息 2714，第 16 级，状态 6，第 2 行
数据库中已有一个名为“UserProfile”的对象。

我用 SQL Server Management Studio 环顾四周，但找不到约束。如何找出外键约束？

Answer 1

这是：

SELECT 
   OBJECT_NAME(f.parent_object_id) TableName,
   COL_NAME(fc.parent_object_id,fc.parent_column_id) ColName
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) = 'YourTableName'

这样，您将获得引用表和列名称。

根据评论建议，编辑为使用 sys.tables 而不是通用 sys.objects。 谢谢，马克_s

Answer 2

另一种方法是检查

的结果

sp_help 'TableName'

（或者仅突出显示引用的表名称并按 ALT+F1）

随着时间的流逝，我决定完善我的答案。下面是

sp_help

提供的结果的屏幕截图。 A 在本示例中使用了 AdventureWorksDW2012 DB。那里有很多好的信息，我们正在寻找的信息在最后 - 以绿色突出显示：

Answer 3

试试这个

SELECT
  object_name(parent_object_id) ParentTableName,
  object_name(referenced_object_id) RefTableName,
  name 
FROM sys.foreign_keys
WHERE parent_object_id = object_id('Tablename')

Answer 4

我发现这个答案非常简单，并且满足了我的需要：https://stackoverflow.com/a/12956348/652519

链接中的摘要，使用此查询：

EXEC sp_fkeys 'TableName'

快速简单。我能够很快找到 15 个表的所有外键表、各自的列和外键名称。

正如 @mdisibio 下面指出的，这里有一个文档链接，详细介绍了可以使用的不同参数：https://learn.microsoft.com/en-us/sql/relational-databases/system-stored-procedures/ sp-fkeys-transact-sql

Answer 5

这是找出所有数据库中外键关系的最佳方法。

exec sp_helpconstraint 'Table Name'

还有一种方法

select * from INFORMATION_SCHEMA.KEY_COLUMN_USAGE where TABLE_NAME='Table Name'
--and left(CONSTRAINT_NAME,2)='FK'(If you want single key)

Answer 6

我正在使用此脚本查找与外键相关的所有详细信息。 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

Answer 7

如果您想在对象资源管理器窗口上通过 SSMS，请右键单击要删除的对象，查看依赖关系。

Answer 8

SELECT 
    obj.name      AS FK_NAME,
    sch.name      AS [schema_name],
    tab1.name     AS [table],
    col1.name     AS [column],
    tab2.name     AS [referenced_table],
    col2.name     AS [referenced_column]
FROM 
     sys.foreign_key_columns fkc
INNER JOIN sys.objects obj
    ON obj.object_id = fkc.constraint_object_id
INNER JOIN sys.tables tab1
    ON tab1.object_id = fkc.parent_object_id
INNER JOIN sys.schemas sch
    ON tab1.schema_id = sch.schema_id
INNER JOIN sys.columns col1
    ON col1.column_id = parent_column_id AND col1.object_id = tab1.object_id
INNER JOIN sys.tables tab2
    ON tab2.object_id = fkc.referenced_object_id
INNER JOIN sys.columns col2
    ON col2.column_id = referenced_column_id 
        AND col2.object_id =  tab2.object_id;

Answer 9

在 SQL Server Management Studio 中，您只需右键单击 对象资源管理器并选择“查看依赖关系”。这会给你一个 良好的起点。它显示引用的表、视图和过程 桌子。

Answer 10

在对象资源管理器中，展开表，然后展开键：

Answer 11

获得桌子

Primary Key

和

Foreign Key

的最简单方法是：

/*  Get primary key and foreign key for a table */
USE DatabaseName;

SELECT CONSTRAINT_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE CONSTRAINT_NAME LIKE 'PK%' AND
TABLE_NAME = 'TableName'

SELECT CONSTRAINT_NAME FROM INFORMATION_SCHEMA.KEY_COLUMN_USAGE
WHERE CONSTRAINT_NAME LIKE 'FK%' AND
TABLE_NAME = 'TableName'

Answer 12

--以下内容可能会为您提供更多您正在寻找的内容：

create Procedure spShowRelationShips 
( 
    @Table varchar(250) = null,
    @RelatedTable varchar(250) = null
)
as
begin
    if @Table is null and @RelatedTable is null
        select  object_name(k.constraint_object_id) ForeginKeyName, 
                object_name(k.Parent_Object_id) TableName, 
                object_name(k.referenced_object_id) RelatedTable, 
                c.Name RelatedColumnName,  
                object_name(rc.object_id) + '.' + rc.name RelatedKeyField
        from sys.foreign_key_columns k
        left join sys.columns c on object_name(c.object_id) = object_name(k.Parent_Object_id) and c.column_id = k.parent_column_id
        left join sys.columns rc on object_name(rc.object_id) = object_name(k.referenced_object_id) and rc.column_id = k.referenced_column_id
        order by 2,3

    if @Table is not null and @RelatedTable is null
        select  object_name(k.constraint_object_id) ForeginKeyName, 
                object_name(k.Parent_Object_id) TableName, 
                object_name(k.referenced_object_id) RelatedTable, 
                c.Name RelatedColumnName,  
                object_name(rc.object_id) + '.' + rc.name RelatedKeyField
        from sys.foreign_key_columns k
        left join sys.columns c on object_name(c.object_id) = object_name(k.Parent_Object_id) and c.column_id = k.parent_column_id
        left join sys.columns rc on object_name(rc.object_id) = object_name(k.referenced_object_id) and rc.column_id = k.referenced_column_id
        where object_name(k.Parent_Object_id) =@Table
        order by 2,3

    if @Table is null and @RelatedTable is not null
        select  object_name(k.constraint_object_id) ForeginKeyName, 
                object_name(k.Parent_Object_id) TableName, 
                object_name(k.referenced_object_id) RelatedTable, 
                c.Name RelatedColumnName,  
                object_name(rc.object_id) + '.' + rc.name RelatedKeyField
        from sys.foreign_key_columns k
        left join sys.columns c on object_name(c.object_id) = object_name(k.Parent_Object_id) and c.column_id = k.parent_column_id
        left join sys.columns rc on object_name(rc.object_id) = object_name(k.referenced_object_id) and rc.column_id = k.referenced_column_id
        where object_name(k.referenced_object_id) =@RelatedTable
        order by 2,3



end

Answer 13

您可以使用此查询来显示

Foreign key

约束：

SELECT
K_Table = FK.TABLE_NAME,
FK_Column = CU.COLUMN_NAME,
PK_Table = PK.TABLE_NAME,
PK_Column = PT.COLUMN_NAME,
Constraint_Name = C.CONSTRAINT_NAME
FROM INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS C
INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS FK ON C.CONSTRAINT_NAME = FK.CONSTRAINT_NAME
INNER JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS PK ON C.UNIQUE_CONSTRAINT_NAME = PK.CONSTRAINT_NAME
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE CU ON C.CONSTRAINT_NAME = CU.CONSTRAINT_NAME
INNER JOIN (
SELECT i1.TABLE_NAME, i2.COLUMN_NAME
FROM INFORMATION_SCHEMA.TABLE_CONSTRAINTS i1
INNER JOIN INFORMATION_SCHEMA.KEY_COLUMN_USAGE i2 ON i1.CONSTRAINT_NAME = i2.CONSTRAINT_NAME
WHERE i1.CONSTRAINT_TYPE = 'PRIMARY KEY'
) PT ON PT.TABLE_NAME = PK.TABLE_NAME
---- optional:
ORDER BY
1,2,3,4
WHERE PK.TABLE_NAME='YourTable'

取自http://blog.sqlauthority.com/2006/11/01/sql-server-query-to-display-foreign-key-relationships-and-name-of-the-constraint-for-each-数据库中的表/

Answer 14

您还可以通过改编@LittleSweetSeas答案来返回有关

Foreign Keys

的所有信息：

SELECT 
   OBJECT_NAME(f.parent_object_id) ConsTable,
   OBJECT_NAME (f.referenced_object_id) refTable,
   COL_NAME(fc.parent_object_id,fc.parent_column_id) ColName
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
order by
ConsTable

Answer 15

流程

_sp_help 'tbl_name'_ 

确实提供了很多信息，但我找到了程序

_sp_fkeys 'tbl_name'_ and 
_sp_pkeys 'tbl_name'_ 

更易于使用，并且可能具有更面向未来的结果。

（他们确实完美地回答了OP）

Answer 16

尝试以下查询。

select object_name(sfc.constraint_object_id) AS constraint_name,
       OBJECT_Name(parent_object_id) AS table_name ,
       ac1.name as table_column_name,
       OBJECT_name(referenced_object_id) as reference_table_name,      
       ac2.name as reference_column_name
from  sys.foreign_key_columns sfc
join sys.all_columns ac1 on (ac1.object_id=sfc.parent_object_id and ac1.column_id=sfc.parent_column_id)
join sys.all_columns ac2 on (ac2.object_id=sfc.referenced_object_id and ac2.column_id=sfc.referenced_column_id) 
where sfc.parent_object_id=OBJECT_ID(<main table name>);

这将给出constraint_name、将要引用的column_names以及将取决于约束的表。

Answer 17

Littlesweetsea 拥有我需要的东西，但通常了解架构是件好事。

SELECT schema_NAME(f.schema_id) schemaname,
   OBJECT_NAME(f.parent_object_id) TableName,
   COL_NAME(fc.parent_object_id,fc.parent_column_id) ColName
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) = 'soldiers'