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
;

使用Key所引用的主键的名称并查询information_schema:

select table_name, column_name
from information_schema.key_column_usage
where constraint_name IN (select constraint_name
  from information_schema.referential_constraints 
  where unique_constraint_name = 'TABLE_NAME_pkey')

这里'TABLE_NAME_pkey'是外键引用的主键的名称。

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'

选择的答案不为我工作，所以张贴我的sql工作。

select 
    con.conname as constraint_name,
    src_schema.nspname as source_schema,
    source.relname as source_table,
    source_col.attname as source_column,
    trg_schema.nspname as target_schema,
    target.relname as target_table,
    target_col.attname as target_column
from 
    pg_constraint con
inner join 
    pg_class source on source.oid = con.conrelid
inner join
    pg_attribute source_col on source_col.attrelid = con.conrelid and source_col.attnum = con.conkey[1] and source_col.attisdropped = false
inner join
    pg_namespace src_schema on src_schema.oid = source.relnamespace
inner join 
    pg_class target on target.oid = con.confrelid
inner join
    pg_attribute target_col on target_col.attrelid = con.confrelid and target_col.attnum = con.confkey[1] and source_col.attisdropped = false    
inner join
    pg_namespace trg_schema on trg_schema.oid = target.relnamespace 

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。

 SELECT 
    conrelid::regclass AS table_from,
    conname,
    pg_get_constraintdef(oid) as condef    
    FROM pg_catalog.pg_constraint r

也适用于所有约束条件。例如，使用pysql: