是否有一种方法使用SQL列出给定表的所有外键?我知道表名/模式,我可以把它插入。


当前回答

我自己的贡献。目标是备份所有外键的定义:

SELECT
    'ALTER TABLE ' || tc.table_schema || '.' || tc.table_name || E'\n
    ADD FOREIGN KEY (' || kcu.column_name || ')' || E'\n
    REFERENCES ' || ccu.table_schema || '.' || ccu.table_name ||
    ' (' || ccu.column_name || ') ' || E'\n    ' ||
    CASE WHEN rc.match_option <> 'NONE' THEN E'\n
    MATCH ' || rc.match_option ELSE '' END ||
    CASE WHEN rc.update_rule <> 'NO ACTION' THEN E'\n
    ON UPDATE ' || rc.update_rule || ' ' ELSE '' END ||
    CASE WHEN rc.delete_rule <> 'NO ACTION'
    THEN 'ON DELETE ' || rc.delete_rule ELSE '' END || ';'
AS add_constraint
FROM
    information_schema.table_constraints AS tc
    JOIN information_schema.key_column_usage AS kcu
        ON tc.constraint_name = kcu.constraint_name
        AND tc.table_schema = kcu.table_schema
    JOIN information_schema.constraint_column_usage AS ccu
        ON ccu.constraint_name = tc.constraint_name
        AND ccu.table_schema = tc.table_schema
    JOIN information_schema.referential_constraints AS rc
        ON tc.constraint_name=rc.constraint_name
WHERE tc.constraint_type = 'FOREIGN KEY'
\t\a\g\a\ta

其他回答

你可以使用PostgreSQL系统目录。也许您可以查询pg_constraint来请求外键。 您还可以使用信息模式

SELECT r.conname
      ,ct.table_name
      ,pg_catalog.pg_get_constraintdef(r.oid, true) as condef
  FROM pg_catalog.pg_constraint r, information_schema.constraint_table_usage ct
 WHERE r.contype = 'f' 
   AND r.conname = ct.constraint_name
 ORDER BY 1

这个查询也适用于组合键:

select c.constraint_name
    , x.table_schema as schema_name
    , x.table_name
    , x.column_name
    , y.table_schema as foreign_schema_name
    , y.table_name as foreign_table_name
    , y.column_name as foreign_column_name
from information_schema.referential_constraints c
join information_schema.key_column_usage x
    on x.constraint_name = c.constraint_name
join information_schema.key_column_usage y
    on y.ordinal_position = x.position_in_unique_constraint
    and y.constraint_name = c.unique_constraint_name
order by c.constraint_name, x.ordinal_position

从最流行的答案改进查询

因为对于postgresql 12+ information_schema是非常慢的

它帮助了我:

SELECT sh.nspname AS table_schema,
  tbl.relname AS table_name,
  col.attname AS column_name,
  referenced_sh.nspname AS foreign_table_schema,
  referenced_tbl.relname AS foreign_table_name,
  referenced_field.attname AS foreign_column_name
FROM pg_constraint c
    INNER JOIN pg_namespace AS sh ON sh.oid = c.connamespace
    INNER JOIN (SELECT oid, unnest(conkey) as conkey FROM pg_constraint) con ON c.oid = con.oid
    INNER JOIN pg_class tbl ON tbl.oid = c.conrelid
    INNER JOIN pg_attribute col ON (col.attrelid = tbl.oid AND col.attnum = con.conkey)
    INNER JOIN pg_class referenced_tbl ON c.confrelid = referenced_tbl.oid
    INNER JOIN pg_namespace AS referenced_sh ON referenced_sh.oid = referenced_tbl.relnamespace
    INNER JOIN (SELECT oid, unnest(confkey) as confkey FROM pg_constraint) conf ON c.oid = conf.oid
    INNER JOIN pg_attribute referenced_field ON (referenced_field.attrelid = c.confrelid AND referenced_field.attnum = conf.confkey)
WHERE c.contype = 'f'

扩展到ollyc配方:

CREATE VIEW foreign_keys_view AS
SELECT
    tc.table_name, kcu.column_name,
    ccu.table_name AS foreign_table_name,
    ccu.column_name AS foreign_column_name
FROM
    information_schema.table_constraints AS tc
    JOIN information_schema.key_column_usage 
        AS kcu ON tc.constraint_name = kcu.constraint_name
    JOIN information_schema.constraint_column_usage 
        AS ccu ON ccu.constraint_name = tc.constraint_name
WHERE constraint_type = 'FOREIGN KEY';

然后:

SELECT * FROM foreign_keys_view WHERE table_name='YourTableNameHere';