这会给你:

FK本身 FK所属的Schema “引用表”或者有FK的表 “引用列”或引用表中指向FK的列 “引用表”或具有FK指向的键列的表 “引用列”或者是FK指向的键的列

下面的代码:

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

 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。

也试一试。

EXEC sp_fkeys 'tableName', 'schemaName'

使用sp_fkeys，不仅可以通过pk表名和模式过滤结果，还可以通过fk表名和模式过滤结果。链接

这会给你:

FK本身 FK所属的Schema “引用表”或者有FK的表 “引用列”或引用表中指向FK的列 “引用表”或具有FK指向的键列的表 “引用列”或者是FK指向的键的列

下面的代码:

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

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