SELECT conrelid::regclass AS table_name, 
       conname AS foreign_key, 
       pg_get_constraintdef(oid) 
FROM   pg_constraint 
WHERE  contype = 'f' 
AND    connamespace = 'public'::regnamespace   
ORDER  BY conrelid::regclass::text, contype DESC;

https://soft-builder.com/how-to-list-all-foreign-keys-in-postgresql-database

以下是Andreas Joseph Krogh从PostgreSQL邮件列表提供的解决方案:http://www.postgresql.org/message-id/200811072134.44750.andreak@officenet.no

SELECT source_table::regclass, source_attr.attname AS source_column,
    target_table::regclass, target_attr.attname AS target_column
FROM pg_attribute target_attr, pg_attribute source_attr,
  (SELECT source_table, target_table, source_constraints[i] source_constraints, target_constraints[i] AS target_constraints
   FROM
     (SELECT conrelid as source_table, confrelid AS target_table, conkey AS source_constraints, confkey AS target_constraints,
       generate_series(1, array_upper(conkey, 1)) AS i
      FROM pg_constraint
      WHERE contype = 'f'
     ) query1
  ) query2
WHERE target_attr.attnum = target_constraints AND target_attr.attrelid = target_table AND
      source_attr.attnum = source_constraints AND source_attr.attrelid = source_table;

这个解决方案处理引用多个列的外键，并避免重复(其他一些答案无法做到这一点)。我唯一改变的是变量名。

下面是一个示例，返回所有引用权限表的雇员列:

SELECT source_column
FROM foreign_keys
WHERE source_table = 'employee'::regclass AND target_table = 'permission'::regclass;

PSQL就是这样做的，如果你用:

psql -E

它将准确地显示执行了哪些查询。在查找外键的情况下，它是:

SELECT conname,
  pg_catalog.pg_get_constraintdef(r.oid, true) as condef
FROM pg_catalog.pg_constraint r
WHERE r.conrelid = '16485' AND r.contype = 'f' ORDER BY 1

在这种情况下，16485是我正在寻找的表的oid -你可以通过将你的表名转换为regclass来获得它:

WHERE r.conrelid = 'mytable'::regclass

如果表名不是唯一的(或者是搜索路径中的第一个)，则对表名进行模式限定:

WHERE r.conrelid = 'myschema.mytable'::regclass

在PostgreSQL提示符上发出\d+ tablename，除了显示表列的数据类型外，它还会显示索引和外键。

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'

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