@BankZ的最好回答

sp_help 'TableName'   

另外，对于不同的模式

sp_help 'schemaName.TableName'   

最初的问题要求将所有外键的列表放入一个高度引用的表中，以便可以删除表。

这个小查询返回将所有外键放入特定表所需的“drop foreign key”命令:

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'

示例输出:

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

省略where -子句以获取当前数据库中所有外键的删除命令。

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

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。: -)

试试这个:

sp_help 'TableName'

@BankZ的最好回答

sp_help 'TableName'   

另外，对于不同的模式

sp_help 'schemaName.TableName'