oracle 外键列表及其引用的表
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List of foreign keys and the tables they reference
提问by stimms
I'm trying to find a query which will return me a list of the foreign keys for a table and the tables and columns they reference. I am half way there with
我试图找到一个查询,它将返回一个表的外键列表以及它们引用的表和列。我已经到了一半
SELECT a.table_name,
a.column_name,
a.constraint_name,
c.owner
FROM ALL_CONS_COLUMNS A, ALL_CONSTRAINTS C
where A.CONSTRAINT_NAME = C.CONSTRAINT_NAME
and a.table_name=:TableName
and C.CONSTRAINT_TYPE = 'R'
But I still need to know which table and primary key are referenced by this key. How would I get that?
但是我仍然需要知道这个键引用了哪个表和主键。我怎么会得到那个?
回答by Vincent Malgrat
The referenced primary key is described in the columns r_owner
and r_constraint_name
of the table ALL_CONSTRAINTS
. This will give you the info you want:
引用的主键在列r_owner
和r_constraint_name
表中描述ALL_CONSTRAINTS
。这将为您提供所需的信息:
SELECT a.table_name, a.column_name, a.constraint_name, c.owner,
-- referenced pk
c.r_owner, c_pk.table_name r_table_name, c_pk.constraint_name r_pk
FROM all_cons_columns a
JOIN all_constraints c ON a.owner = c.owner
AND a.constraint_name = c.constraint_name
JOIN all_constraints c_pk ON c.r_owner = c_pk.owner
AND c.r_constraint_name = c_pk.constraint_name
WHERE c.constraint_type = 'R'
AND a.table_name = :TableName
回答by Touchstone
Try this:
尝试这个:
select * from all_constraints where r_constraint_name in (select constraint_name
from all_constraints where table_name='YOUR_TABLE_NAME');
回答by Doug Porter
Here is an all-purpose script we use that has been incredibly handy.
这是我们使用的通用脚本,非常方便。
Save it off so you can execute it directly (@fkeys.sql). It will let you search by Owner and either the Parent or Child table and show foreign key relationships. The current script does explicitly spool to C:\SQLRPTS so you will need to create that folder of change that line to something you want to use.
保存它以便您可以直接执行它(@fkeys.sql)。它将让您按所有者和父表或子表进行搜索并显示外键关系。当前脚本确实会显式后台打印到 C:\SQLRPTS,因此您需要创建将该行更改为要使用的内容的文件夹。
REM ########################################################################
REM ##
REM ## fkeys.sql
REM ##
REM ## Displays the foreign key relationships
REM ##
REM #######################################################################
CLEAR BREAK
CLEAR COL
SET LINES 200
SET PAGES 54
SET NEWPAGE 0
SET WRAP OFF
SET VERIFY OFF
SET FEEDBACK OFF
break on table_name skip 2 on constraint_name on r_table_name skip 1
column CHILDCOL format a60 head 'CHILD COLUMN'
column PARENTCOL format a60 head 'PARENT COLUMN'
column constraint_name format a30 head 'FK CONSTRAINT NAME'
column delete_rule format a15
column bt noprint
column bo noprint
TTITLE LEFT _DATE CENTER 'FOREIGN KEY RELATIONSHIPS ON &new_prompt' RIGHT 'PAGE:'FORMAT 999 SQL.PNO SKIP 2
SPOOL C:\SQLRPTS\FKeys_&new_prompt
ACCEPT OWNER_NAME PROMPT 'Enter Table Owner (or blank for all): '
ACCEPT PARENT_TABLE_NAME PROMPT 'Enter Parent Table or leave blank for all: '
ACCEPT CHILD_TABLE_NAME PROMPT 'Enter Child Table or leave blank for all: '
select b.owner || '.' || b.table_name || '.' || b.column_name CHILDCOL,
b.position,
c.owner || '.' || c.table_name || '.' || c.column_name PARENTCOL,
a.constraint_name,
a.delete_rule,
b.table_name bt,
b.owner bo
from all_cons_columns b,
all_cons_columns c,
all_constraints a
where b.constraint_name = a.constraint_name
and a.owner = b.owner
and b.position = c.position
and c.constraint_name = a.r_constraint_name
and c.owner = a.r_owner
and a.constraint_type = 'R'
and c.owner like case when upper('&OWNER_NAME') is null then '%'
else upper('&OWNER_NAME') end
and c.table_name like case when upper('&PARENT_TABLE_NAME') is null then '%'
else upper('&PARENT_TABLE_NAME') end
and b.table_name like case when upper('&CHILD_TABLE_NAME') is null then '%'
else upper('&CHILD_TABLE_NAME') end
order by 7,6,4,2
/
SPOOL OFF
TTITLE OFF
SET FEEDBACK ON
SET VERIFY ON
CLEAR BREAK
CLEAR COL
SET PAGES 24
SET LINES 100
SET NEWPAGE 1
UNDEF OWNER
回答by user2323623
This will travel the hierarchy of foreign keys for a given table and column and return columns from child and grandchild, and all descendant tables. It uses sub-queries to add r_table_name and r_column_name to user_constraints, and then uses them to connect rows.
这将遍历给定表和列的外键层次结构,并从子表和孙表以及所有后代表返回列。它使用子查询将 r_table_name 和 r_column_name 添加到 user_constraints,然后使用它们连接行。
select distinct table_name, constraint_name, column_name, r_table_name, position, constraint_type
from (
SELECT uc.table_name,
uc.constraint_name,
cols.column_name,
(select table_name from user_constraints where constraint_name = uc.r_constraint_name)
r_table_name,
(select column_name from user_cons_columns where constraint_name = uc.r_constraint_name and position = cols.position)
r_column_name,
cols.position,
uc.constraint_type
FROM user_constraints uc
inner join user_cons_columns cols on uc.constraint_name = cols.constraint_name
where constraint_type != 'C'
)
start with table_name = 'MY_TABLE_NAME' and column_name = 'MY_COLUMN_NAME'
connect by nocycle
prior table_name = r_table_name
and prior column_name = r_column_name;
回答by Ganbat Bayarbaatar
Here is an another solution. Using sys's default views are so slow (approx 10s in my situation). This is much faster than that (approx. 0.5s).
这是另一个解决方案。使用 sys 的默认视图非常慢(在我的情况下大约 10 秒)。这比这快得多(大约 0.5 秒)。
SELECT
CONST.NAME AS CONSTRAINT_NAME,
RCONST.NAME AS REF_CONSTRAINT_NAME,
OBJ.NAME AS TABLE_NAME,
COALESCE(ACOL.NAME, COL.NAME) AS COLUMN_NAME,
CCOL.POS# AS POSITION,
ROBJ.NAME AS REF_TABLE_NAME,
COALESCE(RACOL.NAME, RCOL.NAME) AS REF_COLUMN_NAME,
RCCOL.POS# AS REF_POSITION
FROM SYS.CON$ CONST
INNER JOIN SYS.CDEF$ CDEF ON CDEF.CON# = CONST.CON#
INNER JOIN SYS.CCOL$ CCOL ON CCOL.CON# = CONST.CON#
INNER JOIN SYS.COL$ COL ON (CCOL.OBJ# = COL.OBJ#) AND (CCOL.INTCOL# = COL.INTCOL#)
INNER JOIN SYS.OBJ$ OBJ ON CCOL.OBJ# = OBJ.OBJ#
LEFT JOIN SYS.ATTRCOL$ ACOL ON (CCOL.OBJ# = ACOL.OBJ#) AND (CCOL.INTCOL# = ACOL.INTCOL#)
INNER JOIN SYS.CON$ RCONST ON RCONST.CON# = CDEF.RCON#
INNER JOIN SYS.CCOL$ RCCOL ON RCCOL.CON# = RCONST.CON#
INNER JOIN SYS.COL$ RCOL ON (RCCOL.OBJ# = RCOL.OBJ#) AND (RCCOL.INTCOL# = RCOL.INTCOL#)
INNER JOIN SYS.OBJ$ ROBJ ON RCCOL.OBJ# = ROBJ.OBJ#
LEFT JOIN SYS.ATTRCOL$ RACOL ON (RCCOL.OBJ# = RACOL.OBJ#) AND (RCCOL.INTCOL# = RACOL.INTCOL#)
WHERE CONST.OWNER# = userenv('SCHEMAID')
AND RCONST.OWNER# = userenv('SCHEMAID')
AND CDEF.TYPE# = 4 /* 'R' Referential/Foreign Key */;
回答by Menelaos Vergis
If you need all the foreign keys of the user then use the following script
如果您需要用户的所有外键,请使用以下脚本
SELECT a.constraint_name, a.table_name, a.column_name, c.owner,
c_pk.table_name r_table_name, b.column_name r_column_name
FROM user_cons_columns a
JOIN user_constraints c ON a.owner = c.owner
AND a.constraint_name = c.constraint_name
JOIN user_constraints c_pk ON c.r_owner = c_pk.owner
AND c.r_constraint_name = c_pk.constraint_name
JOIN user_cons_columns b ON C_PK.owner = b.owner
AND C_PK.CONSTRAINT_NAME = b.constraint_name AND b.POSITION = a.POSITION
WHERE c.constraint_type = 'R'
based on Vincent Malgrat code
基于文森特马尔格拉特代码
回答by arvinq
I know it's kinda late to answer but let me answer anyway, some of the answers above are quite complicated hence here is a much simpler take.
我知道现在回答有点晚了,但无论如何让我回答,上面的一些答案非常复杂,因此这里有一个简单得多的答案。
SELECT a.table_name child_table, a.column_name child_column, a.constraint_name,
b.table_name parent_table, b.column_name parent_column
FROM all_cons_columns a
JOIN all_constraints c ON a.owner = c.owner AND a.constraint_name = c.constraint_name
join all_cons_columns b on c.owner = b.owner and c.r_constraint_name = b.constraint_name
WHERE c.constraint_type = 'R'
AND a.table_name = 'your table name'
回答by Francisco M
My version, in my humble opinion, more readable:
我的版本,以我的拙见,更具可读性:
SELECT PARENT.TABLE_NAME "PARENT TABLE_NAME"
, PARENT.CONSTRAINT_NAME "PARENT PK CONSTRAINT"
, '->' " "
, CHILD.TABLE_NAME "CHILD TABLE_NAME"
, CHILD.COLUMN_NAME "CHILD COLUMN_NAME"
, CHILD.CONSTRAINT_NAME "CHILD CONSTRAINT_NAME"
FROM ALL_CONS_COLUMNS CHILD
, ALL_CONSTRAINTS CT
, ALL_CONSTRAINTS PARENT
WHERE CHILD.OWNER = CT.OWNER
AND CT.CONSTRAINT_TYPE = 'R'
AND CHILD.CONSTRAINT_NAME = CT.CONSTRAINT_NAME
AND CT.R_OWNER = PARENT.OWNER
AND CT.R_CONSTRAINT_NAME = PARENT.CONSTRAINT_NAME
AND CHILD.TABLE_NAME = ::table -- table name variable
AND CT.OWNER = ::owner; -- schema variable, could not be needed
回答by D.Zotov
Its a bit late to anwser, but I hope my answer been useful for someone, who needs to select Composite foreign keys.
anwser 有点晚了,但我希望我的回答对需要选择复合外键的人有用。
SELECT
"C"."CONSTRAINT_NAME",
"C"."OWNER" AS "SCHEMA_NAME",
"C"."TABLE_NAME",
"COL"."COLUMN_NAME",
"REF_COL"."OWNER" AS "REF_SCHEMA_NAME",
"REF_COL"."TABLE_NAME" AS "REF_TABLE_NAME",
"REF_COL"."COLUMN_NAME" AS "REF_COLUMN_NAME"
FROM
"USER_CONSTRAINTS" "C"
INNER JOIN "USER_CONS_COLUMNS" "COL" ON "COL"."OWNER" = "C"."OWNER"
AND "COL"."CONSTRAINT_NAME" = "C"."CONSTRAINT_NAME"
INNER JOIN "USER_CONS_COLUMNS" "REF_COL" ON "REF_COL"."OWNER" = "C"."R_OWNER"
AND "REF_COL"."CONSTRAINT_NAME" = "C"."R_CONSTRAINT_NAME"
AND "REF_COL"."POSITION" = "COL"."POSITION"
WHERE "C"."TABLE_NAME" = 'TableName' AND "C"."CONSTRAINT_TYPE" = 'R'
回答by Anwarul Kabir Khan
I used below code and it served my purpose-
我使用了下面的代码,它达到了我的目的-
SELECT fk.owner, fk.table_name, col.column_name
FROM dba_constraints pk, dba_constraints fk, dba_cons_columns col
WHERE pk.constraint_name = fk.r_constraint_name
AND fk.constraint_name = col.constraint_name
AND pk.owner = col.owner
AND pk.owner = fk.owner
AND fk.constraint_type = 'R'
AND pk.owner = sys_context('USERENV', 'CURRENT_SCHEMA')
AND pk.table_name = :my_table
AND pk.constraint_type = 'P';