외래 키 및 이들이 참조하는 테이블 목록
테이블의 외래 키 목록과 참조하는 테이블과 열을 반환하는 쿼리를 찾으려고합니다. 나는 반쯤있다
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'
그러나 여전히이 키가 참조하는 테이블과 기본 키를 알아야합니다. 어떻게 알 수 있습니까?
참조 된 기본 키 컬럼에서 설명 r_owner
하고 r_constraint_name
테이블의 ALL_CONSTRAINTS
. 원하는 정보를 제공합니다.
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
이 시도:
select * from all_constraints where r_constraint_name in (select constraint_name
from all_constraints where table_name='YOUR_TABLE_NAME');
우리가 사용하는 다목적 스크립트는 매우 유용합니다.
직접 실행할 수 있도록 저장하십시오 (@ fkeys.sql). 소유자와 부모 또는 자식 테이블로 검색하고 외래 키 관계를 표시 할 수 있습니다. 현재 스크립트는 C : \ SQLRPTS로 명시 적으로 스풀링하므로 해당 행을 사용하려는 것으로 변경 폴더를 작성해야합니다.
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
주어진 테이블과 열에 대한 외래 키의 계층 구조를 이동하고 자식과 손자, 모든 자손 테이블의 열을 반환합니다. 하위 쿼리를 사용하여 r_table_name 및 r_column_name을 user_constraints에 추가 한 다음이를 사용하여 행을 연결합니다.
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;
다른 해결책이 있습니다. sys의 기본보기 사용은 너무 느립니다 (제 상황에서는 약 10 초). 이보다 훨씬 빠릅니다 (약 0.5 초).
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 */;
사용자의 외래 키가 모두 필요하면 다음 스크립트를 사용하십시오.
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'
Vincent Malgrat 코드 기반
답변이 늦었지만 어쨌든 대답하겠습니다. 위의 답변 중 일부는 매우 복잡하므로 여기에 훨씬 더 간단한 방법이 있습니다.
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'
UAT 환경 테이블에서 라이브로 FK 제약 조건을 만들려는 경우 동적 쿼리 아래에서 실행하십시오 .....
SELECT 'ALTER TABLE '||OBJ.NAME||' ADD CONSTRAINT '||CONST.NAME||' FOREIGN KEY ('||COALESCE(ACOL.NAME, COL.NAME)||') REFERENCES '
||ROBJ.NAME ||' ('||COALESCE(RACOL.NAME, RCOL.NAME) ||');'
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
AND OBJ.NAME = <table_name>;
겸손한 견해로 볼 때 내 버전은 더 읽기 쉽습니다.
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
Its a bit late to anwser, but I hope my answer been useful for someone, who needs to select Composite foreign keys.
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'
I used below code and it served my purpose-
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';
select d.table_name,
d.constraint_name "Primary Constraint Name",
b.constraint_name "Referenced Constraint Name"
from user_constraints d,
(select c.constraint_name,
c.r_constraint_name,
c.table_name
from user_constraints c
where table_name='EMPLOYEES' --your table name instead of EMPLOYEES
and constraint_type='R') b
where d.constraint_name=b.r_constraint_name
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 :=TABLE_NAME
AND c.owner :=OWNER_NAME;
WITH reference_view AS
(SELECT a.owner, a.table_name, a.constraint_name, a.constraint_type,
a.r_owner, a.r_constraint_name, b.column_name
FROM dba_constraints a, dba_cons_columns b
WHERE a.owner LIKE UPPER ('SYS') AND
a.owner = b.owner
AND a.constraint_name = b.constraint_name
AND constraint_type = 'R'),
constraint_view AS
(SELECT a.owner a_owner, a.table_name, a.column_name, b.owner b_owner,
b.constraint_name
FROM dba_cons_columns a, dba_constraints b
WHERE a.owner = b.owner
AND a.constraint_name = b.constraint_name
AND b.constraint_type = 'P'
AND a.owner LIKE UPPER ('SYS')
)
SELECT
rv.table_name FK_Table , rv.column_name FK_Column ,
CV.table_name PK_Table , rv.column_name PK_Column , rv.r_constraint_name Constraint_Name
FROM reference_view rv, constraint_view CV
WHERE rv.r_constraint_name = CV.constraint_name AND rv.r_owner = CV.b_owner;
For Load UserTable (List of foreign keys and the tables they reference)
WITH
reference_view AS
(SELECT a.owner, a.table_name, a.constraint_name, a.constraint_type,
a.r_owner, a.r_constraint_name, b.column_name
FROM dba_constraints a, dba_cons_columns b
WHERE
a.owner = b.owner
AND a.constraint_name = b.constraint_name
AND constraint_type = 'R'),
constraint_view AS
(SELECT a.owner a_owner, a.table_name, a.column_name, b.owner b_owner,
b.constraint_name
FROM dba_cons_columns a, dba_constraints b
WHERE a.owner = b.owner
AND a.constraint_name = b.constraint_name
AND b.constraint_type = 'P'
) ,
usertableviewlist AS
(
select TABLE_NAME from user_tables
)
SELECT
rv.table_name FK_Table , rv.column_name FK_Column ,
CV.table_name PK_Table , rv.column_name PK_Column , rv.r_constraint_name Constraint_Name
FROM reference_view rv, constraint_view CV , usertableviewlist UTable
WHERE rv.r_constraint_name = CV.constraint_name AND rv.r_owner = CV.b_owner And UTable.TABLE_NAME = rv.table_name;
참고URL : https://stackoverflow.com/questions/1729996/list-of-foreign-keys-and-the-tables-they-reference
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