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')

正确的解决方案，使用information_schema，使用多列键，在两个表中正确地连接不同名称的列，并且与ms sqlsever兼容:

select fks.TABLE_NAME as foreign_key_table_name
, fks.CONSTRAINT_NAME as foreign_key_constraint_name
, kcu_foreign.COLUMN_NAME as foreign_key_column_name
, rc.UNIQUE_CONSTRAINT_NAME as primary_key_constraint_name
, pks.TABLE_NAME as primary_key_table_name
, kcu_primary.COLUMN_NAME as primary_key_column_name
from INFORMATION_SCHEMA.TABLE_CONSTRAINTS fks -- foreign keys
inner join INFORMATION_SCHEMA.KEY_COLUMN_USAGE kcu_foreign -- the columns of the above keys
    on fks.TABLE_CATALOG = kcu_foreign.TABLE_CATALOG
    and fks.TABLE_SCHEMA = kcu_foreign.TABLE_SCHEMA
    and fks.TABLE_NAME = kcu_foreign.TABLE_NAME
    and fks.CONSTRAINT_NAME = kcu_foreign.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.REFERENTIAL_CONSTRAINTS rc -- referenced constraints
    on rc.CONSTRAINT_CATALOG = fks.CONSTRAINT_CATALOG
    and rc.CONSTRAINT_SCHEMA = fks.CONSTRAINT_SCHEMA
    and rc.CONSTRAINT_NAME = fks.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.TABLE_CONSTRAINTS pks -- primary keys (referenced by fks)
    on rc.UNIQUE_CONSTRAINT_CATALOG = pks.CONSTRAINT_CATALOG
    and rc.UNIQUE_CONSTRAINT_SCHEMA = pks.CONSTRAINT_SCHEMA
    and rc.UNIQUE_CONSTRAINT_NAME = pks.CONSTRAINT_NAME
inner join INFORMATION_SCHEMA.KEY_COLUMN_USAGE kcu_primary
    on pks.TABLE_CATALOG = kcu_primary.TABLE_CATALOG
    and pks.TABLE_SCHEMA = kcu_primary.TABLE_SCHEMA
    and pks.TABLE_NAME = kcu_primary.TABLE_NAME
    and pks.CONSTRAINT_NAME = kcu_primary.CONSTRAINT_NAME
    and kcu_foreign.ORDINAL_POSITION = kcu_primary.ORDINAL_POSITION -- this joins the columns
where fks.TABLE_SCHEMA = 'dbo' -- replace with schema name
and fks.TABLE_NAME = 'your_table_name' -- replace with table name
and fks.CONSTRAINT_TYPE = 'FOREIGN KEY'
and pks.CONSTRAINT_TYPE = 'PRIMARY KEY'
order by fks.constraint_name, kcu_foreign.ORDINAL_POSITION

注意:在potgresql和sqlserver的information_schema实现之间存在一些差异，这使得顶部答案在两个系统上给出不同的结果——一个显示外键表的列名，另一个显示主键表的列名。出于这个原因，我决定使用KEY_COLUMN_USAGE视图。

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';

Ollyc的答案很好，因为它不是特定于postgres的，但是，当外键引用多个列时，它就会崩溃。以下查询适用于任意数量的列，但它严重依赖于Postgres扩展:

select 
    att2.attname as "child_column", 
    cl.relname as "parent_table", 
    att.attname as "parent_column",
    conname
from
   (select 
        unnest(con1.conkey) as "parent", 
        unnest(con1.confkey) as "child", 
        con1.confrelid, 
        con1.conrelid,
        con1.conname
    from 
        pg_class cl
        join pg_namespace ns on cl.relnamespace = ns.oid
        join pg_constraint con1 on con1.conrelid = cl.oid
    where
        cl.relname = 'child_table'
        and ns.nspname = 'child_schema'
        and con1.contype = 'f'
   ) con
   join pg_attribute att on
       att.attrelid = con.confrelid and att.attnum = con.child
   join pg_class cl on
       cl.oid = con.confrelid
   join pg_attribute att2 on
       att2.attrelid = con.conrelid and att2.attnum = con.parent

您可以通过information_schema表来实现这一点。例如:

SELECT
    tc.table_schema, 
    tc.constraint_name, 
    tc.table_name, 
    kcu.column_name, 
    ccu.table_schema AS foreign_table_schema,
    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
      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
WHERE tc.constraint_type = 'FOREIGN KEY' AND tc.table_name='mytable';

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'