我认为你想要的和@ollyc写的很接近的是:

SELECT
tc.constraint_name, 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' AND ccu.table_name='YourTableNameHere';

这将列出所有使用指定表作为外键的表

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'

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

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

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;

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