Lista de claves extranjeras y las tablas a las que hacen referencia

Question

Estoy tratando de encontrar una consulta que me devuelva una lista de las claves extranjeras para una tabla y las tablas y columnas a las que hacen referencia. Estoy a mitad de camino con

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'

Pero todavía necesito saber a qué tabla y la clave primaria están referenciadas por esta clave. ¿Cómo conseguiría eso?

Answer 1

La clave primaria referenciada se describe en las columnas r_owner y r_constraint_name de la mesa ALL_CONSTRAINTS. Esto le dará la información que desee:

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

Answer 2

Prueba esto:

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

Answer 3

Aquí hay un script para todo uso que utilizamos que ha sido increíblemente útil.

Guárdelo para que pueda ejecutarlo directamente (@fkeys.sql). Le permitirá buscar por propietario y la mesa de los padres o hijos y mostrar relaciones con la clave extranjera. El script actual se coloca explícitamente a c: sqlrpts, por lo que deberá crear esa carpeta de cambio esa línea a algo que desea usar.

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

Answer 4

Esto viajará por la jerarquía de claves extranjeras para una tabla determinada y columnas y columnas de regreso de niño y nieto, y todas las tablas descendientes. Utiliza subteres para agregar r_table_name y r_column_name a user_constraints, y luego las usa para conectar filas.

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;

Answer 5

Aquí hay otra solución. El uso de las vistas predeterminadas de SYS es muy lenta (aproximadamente 10s en mi situación). Esto es mucho más rápido que eso (aproximadamente 0.5s).

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 */;

Answer 6

Si necesita todas las claves extranjeras del usuario, use el siguiente script

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'

Basado en el código Vincent Malgrat

Answer 7

Sé que es un poco tarde para responder, pero déjame responder de todos modos, algunas de las respuestas anteriores son bastante complicadas, por lo tanto, aquí hay una toma mucho más simple.

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'

Answer 8

En caso de que uno quiera crear restricciones FK de la tabla de entorno UAT para vivir, dispare debajo de la consulta dinámica .....

    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>;

Answer 9

Mi versión, en mi humilde opinión, más legible:

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

Answer 10

Es un poco tarde a Anwser, pero espero que mi respuesta haya sido útil para alguien, que necesita seleccionar claves extranjeras compuestas.

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'

Answer 11

Usé a continuación código y sirvió mi propósito-

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';

Answer 12

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

Answer 13

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;

Answer 14

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;

Answer 15

Para la carga usertable (lista de claves extranjeras y las tablas a las que hacen referencia)

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;