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

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

SELECT
  object_name(parent_object_id),
  object_name(referenced_object_id),
  name 
FROM sys.foreign_keys
WHERE parent_object_id = object_id('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。: -)

Oracle SQL

select *
from
    all_constraints
where
    r_constraint_name in
    (select       constraint_name
    from
       all_constraints
    where
       table_name='PUT_THE_TABLE_NAME_HERE');

all_constraints是Oracle DB中的一个固有表名。

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