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;

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

试试这个:

sp_help 'TableName'

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。

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

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。: -)