How can I list all foreign keys referencing a give

2018-12-31 14:24发布

I need to remove a highly referenced table in a SQL Server database. How can I get a list of all the foreign key constraints I will need to remove in order to drop the table?

(SQL answers preferable over clicking about in the GUI of the management studio.)

21条回答
查无此人
2楼-- · 2018-12-31 15:05

This gets any foreign key that involves the chosen table. *Assumes a _FIRSTABLENAME_SECONDTABLENAME format.

 declare @tablename as varchar(MAX)
 SET @tablename = 'yourtablename'
 SELECT name
 FROM YOURDATABASE.sys.objects
 WHERE type_desc = 'FOREIGN_KEY_CONSTRAINT' and (name LIKE '%_' + @tablename + 'empdb_%' or name LIKE '%_' + @tablename )

This is a more general form:

 SELECT name
 FROM YOURDATABASE_PROD.sys.objects
 WHERE type_desc = 'FOREIGN_KEY_CONSTRAINT' and name LIKE '%' + @tablename + '%' and
 name NOT LIKE '[a-zA-Z0-9]' + @tablename + '%' and name NOT LIKE '%' + @tablename + '[a-zA-Z0-9]' 
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永恒的永恒
3楼-- · 2018-12-31 15:05

The following solution work for me:

--Eliminar las llaves foraneas
declare @query varchar(8000)
declare cursorRecorrerTabla cursor for

SELECT  'ALTER TABLE [PoaComFinH].['+sch.name+'].['+referencingTable.Name+'] DROP CONSTRAINT ['+foreignKey.name+']' 'query'
FROM PoaComFinH.sys.foreign_key_columns fk
JOIN PoaComFinH.sys.tables referencingTable ON fk.parent_object_id = referencingTable.object_id
JOIN PoaComFinH.sys.schemas sch ON referencingTable.schema_id = sch.schema_id
JOIN PoaComFinH.sys.objects foreignKey ON foreignKey.object_id = fk.constraint_object_id
JOIN PoaComFinH.sys.tables referencedTable ON fk.referenced_object_id = referencedTable.object_id


--3ro. abrir el cursor.
open cursorRecorrerTabla
fetch next from cursorRecorrerTabla
into @query
while @@fetch_status = 0
begin
--inicio cuerpo del cursor
    print @query
    exec(@query)
--fin cuerpo del cursor
fetch next from cursorRecorrerTabla
into @query
end
--cerrar cursor
close cursorRecorrerTabla
deallocate cursorRecorrerTabla
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千与千寻千般痛.
4楼-- · 2018-12-31 15:07

Not sure why no one suggested but I use sp_fkeys to query foreign keys for a given table:

EXEC sp_fkeys 'TableName'
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后来的你喜欢了谁
5楼-- · 2018-12-31 15:07
SELECT
OBJECT_NAME(parent_object_id) 'Parent table',
c.NAME 'Parent column name',
OBJECT_NAME(referenced_object_id) 'Referenced table',
cref.NAME 'Referenced column name'
FROM 
sys.foreign_key_columns fkc 
INNER JOIN 
sys.columns c 
   ON fkc.parent_column_id = c.column_id 
      AND fkc.parent_object_id = c.object_id
INNER JOIN 
sys.columns cref 
   ON fkc.referenced_column_id = cref.column_id 
      AND fkc.referenced_object_id = cref.object_id  where   OBJECT_NAME(parent_object_id) = 'tablename'

If you want to get the foreign key relation of all the tables exclude the where clause else write your tablename instead of tablename

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低头抚发
6楼-- · 2018-12-31 15:09

The original question asked to get a list of all foreign keys into a highly referenced table so that the table can be removed.

This little query returns all the 'drop foreign key' commands needed to drop all foreign keys into a particular table:

SELECT 
   'ALTER TABLE ['+sch.name+'].['+referencingTable.Name+'] DROP CONSTRAINT ['+foreignKey.name+']' '[DropCommand]'
FROM sys.foreign_key_columns fk
    JOIN sys.tables referencingTable ON fk.parent_object_id = referencingTable.object_id
    JOIN sys.schemas sch ON referencingTable.schema_id = sch.schema_id
    JOIN sys.objects foreignKey ON foreignKey.object_id = fk.constraint_object_id
    JOIN sys.tables referencedTable ON fk.referenced_object_id = referencedTable.object_id
WHERE referencedTable.name = 'MyTableName'

Example output:

[DropCommand]
ALTER TABLE [dbo].[OtherTable1] DROP CONSTRAINT [FK_OtherTable1_MyTable]
ALTER TABLE [dbo].[OtherTable2] DROP CONSTRAINT [FK_OtherTable2_MyTable]

Omit the WHERE-clause to get the drop commands for all foreign keys in the current database.

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路过你的时光
7楼-- · 2018-12-31 15:09
SELECT
  object_name(parent_object_id),
  object_name(referenced_object_id),
  name 
FROM sys.foreign_keys
WHERE parent_object_id = object_id('Table Name')
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