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')
)

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。

 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。

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

以下是我认为在SQL Server 2016中处理这种情况的最佳实践。

你必须列出外键使用:

EXEC sp_fkeys 'TableName'

在这里你可以看到fk的全部信息。注意列FKTABLE_NAME, FKCOLUMN_NAME, FK_NAME, UPDATE_RULE, DELETE_RULE是你需要删除外键并在截断后再次实现它们的信息。

你可以组织一个脚本如下:

-- EXEC sp_fkeys 'TableName'

-- DROP CONSTRAINTS: I drop one, here drop every constraint you desire. 
BEGIN TRANSACTION
GO
ALTER TABLE dbo.TableName
    DROP CONSTRAINT IF EXISTS FK_TableName_OtherTable
GO
ALTER TABLE dbo.TableName SET (LOCK_ESCALATION = TABLE)
GO
COMMIT

-- TRUNCATE 
BEGIN TRANSACTION
TRUNCATE TABLE TableName
GO
COMMIT


-- RECREATE CONSTRAINTS: I recreate 1, here recreate every fk you desire
BEGIN TRANSACTION
GO
ALTER TABLE dbo.TableName SET (LOCK_ESCALATION = TABLE)
GO

ALTER TABLE dbo.TableName ADD CONSTRAINT
    FK_TableName_OtherTable FOREIGN KEY
    (
    Id_FK
    ) REFERENCES dbo.OtherTable
    (
    Id
    ) ON UPDATE  NO ACTION 
     ON DELETE  NO ACTION 
GO
COMMIT

** UPDATE_RULE和DELETE_RULE的值可以在sp_fkeys的文档中看到:

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