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

为了扩展Martin的精彩回答，这里有一个查询，它允许您根据父表进行过滤，并显示每个父表的子表的名称，以便您可以根据父表中的外键约束查看所有依赖的表/列。

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

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')

$1 ('my_schema')是模式，$2 ('my_table')是表名:

SELECT ss.conname constraint_name, a.attname column_name, ss.refnamespace fk_table_schema, ss.reflname fk_table_name, af.attname fk_column_name
    FROM  pg_attribute a, pg_attribute af,
        (SELECT r.oid roid, c.conname, rf.relname reflname, information_schema._pg_expandarray(c.conkey) x,
                nrf.nspname refnamespace, rf.oid rfoid, information_schema._pg_expandarray(cf.confkey) xf
            FROM pg_namespace nr, pg_class r, pg_constraint c,
                pg_namespace nrf, pg_class rf, pg_constraint cf
            WHERE nr.oid = r.relnamespace
                AND r.oid = c.conrelid
                AND rf.oid = cf.confrelid
                AND c.conname = cf.conname
                AND nrf.oid = rf.relnamespace
                AND nr.nspname = $1
                AND r.relname = $2) ss
    WHERE ss.roid = a.attrelid AND a.attnum = (ss.x).x AND NOT a.attisdropped
        AND ss.rfoid = af.attrelid AND af.attnum = (ss.xf).x AND NOT af.attisdropped
    ORDER BY ss.conname, a.attname;

我升级了@ollyc的答案，目前在顶部。 我同意@fionbio，因为key_column_usage和constraint_column_usage在列级上没有相关信息。

如果constraint_column_usage具有像key_column_usage一样的ordinal_position列，则可以将其与该列连接。所以我做了一个ordinal_position到constraint_column_usage如下所示。

我无法确认手动创建的ordinal_position与key_column_usage的顺序完全相同。但我检查了一下，至少在我的箱子里是完全一样的顺序。

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 (select row_number() over (partition by table_schema, table_name, constraint_name order by row_num) ordinal_position,
                 table_schema, table_name, column_name, constraint_name
          from   (select row_number() over (order by 1) row_num, table_schema, table_name, column_name, constraint_name
                  from   information_schema.constraint_column_usage
                 ) t
         ) AS ccu
      ON ccu.constraint_name = tc.constraint_name
      AND ccu.table_schema = tc.table_schema
      AND ccu.ordinal_position = kcu.ordinal_position
WHERE tc.constraint_type = 'FOREIGN KEY' AND tc.table_name = 'mytable'

另一种方式:

WITH foreign_keys AS (
    SELECT
      conname,
      conrelid,
      confrelid,
      unnest(conkey)  AS conkey,
      unnest(confkey) AS confkey
    FROM pg_constraint
    WHERE contype = 'f' -- AND confrelid::regclass = 'your_table'::regclass
)
-- if confrelid, conname pair shows up more than once then it is multicolumn foreign key
SELECT fk.conname as constraint_name,
       fk.confrelid::regclass as referenced_table, af.attname as pkcol,
       fk.conrelid::regclass as referencing_table, a.attname as fkcol
FROM foreign_keys fk
JOIN pg_attribute af ON af.attnum = fk.confkey AND af.attrelid = fk.confrelid
JOIN pg_attribute a ON a.attnum = conkey AND a.attrelid = fk.conrelid
ORDER BY fk.confrelid, fk.conname
;