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

你可以通过以下查询找到:

 SELECT OBJECT_NAME (FK.referenced_object_id) 'Referenced Table', 
      OBJECT_NAME(FK.parent_object_id) 'Referring Table', FK.name 'Foreign Key', 
      COL_NAME(FK.referenced_object_id, FKC.referenced_column_id) 'Referenced Column',
      COL_NAME(FK.parent_object_id,FKC.parent_column_id) 'Referring Column'
     FROM sys.foreign_keys AS FK
             INNER JOIN sys.foreign_key_columns AS FKC 
                 ON FKC.constraint_object_id = FK.OBJECT_ID
     WHERE OBJECT_NAME (FK.referenced_object_id) = 'YourTableName'
     AND COL_NAME(FK.referenced_object_id, FKC.referenced_column_id) = 'YourColumnName'
     order by  OBJECT_NAME(FK.parent_object_id)

其他回答

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')
)

@BankZ的最好回答

sp_help 'TableName'   

另外,对于不同的模式

sp_help 'schemaName.TableName'   

试试这个:

sp_help 'TableName'

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

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