只需替换'您的表名'在下面的查询与您的表名。

简短但贴心的赞，如果对你有用的话。

select * from information_schema.key_column_usage
where constraint_catalog=current_catalog and table_name='your_table_name'
and position_in_unique_constraint notnull;

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

我写了一个解决方案，喜欢和经常使用。代码在http://code.google.com/p/pgutils/。看这些小疙瘩。foreign_keys视图。

不幸的是，输出太冗长，这里不包括。但是，你可以在数据库的公共版本上尝试一下，就像这样:

$ psql -h unison-db.org -U PUBLIC -d unison -c 'select * from pgutils.foreign_keys;

这至少适用于8.3版本。如果需要的话，我预计会在未来几个月内对其进行更新。

莉丝

这是我目前使用的，它将列出一个表，它的fkey约束[删除表子句，它将列出当前目录中的所有表]:

SELECT

    current_schema() AS "schema",
    current_catalog AS "database",
    "pg_constraint".conrelid::regclass::text AS "primary_table_name",
    "pg_constraint".confrelid::regclass::text AS "foreign_table_name",

    (
        string_to_array(
            (
                string_to_array(
                    pg_get_constraintdef("pg_constraint".oid),
                    '('
                )
            )[2],
            ')'
        )
    )[1] AS "foreign_column_name",

    "pg_constraint".conindid::regclass::text AS "constraint_name",

    TRIM((
        string_to_array(
            pg_get_constraintdef("pg_constraint".oid),
            '('
        )
    )[1]) AS "constraint_type",

    pg_get_constraintdef("pg_constraint".oid) AS "constraint_definition"

FROM pg_constraint AS "pg_constraint"

JOIN pg_namespace AS "pg_namespace" ON "pg_namespace".oid = "pg_constraint".connamespace

WHERE
    --fkey and pkey constraints
    "pg_constraint".contype IN ( 'f', 'p' )
    AND
    "pg_namespace".nspname = current_schema()
    AND
    "pg_constraint".conrelid::regclass::text IN ('whatever_table_name')

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';

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