延伸到@Cope360的好答案。要获取某个表(如果它们是相同的表名但不同的模式)，只需使用表OID。

select
     t.relname as table_name
    ,i.relname as index_name
    ,a.attname as column_name
    ,a.attrelid tableid

from
    pg_class t,
    pg_class i,
    pg_index ix,
    pg_attribute a
where
    t.oid = ix.indrelid
    and i.oid = ix.indexrelid
    and a.attrelid = t.oid
    and a.attnum = ANY(ix.indkey)
    and t.relkind = 'r'
    -- and t.relname like 'tbassettype'
    and a.attrelid = '"dbLegal".tbassettype'::regclass
order by
    t.relname,
    i.relname;

解释:我有表名'tbassettype'在两个模式'dbAsset'和'dbLegal'。要在dbLegal上只获取表，只需让a.attrelid =它的OID。

下面是一个包装cope360答案的函数:

CREATE OR REPLACE FUNCTION getIndices(_table_name varchar)
  RETURNS TABLE(table_name varchar, index_name varchar, column_name varchar) AS $$
  BEGIN
    RETURN QUERY
    select
    t.relname::varchar as table_name,
    i.relname::varchar as index_name,
    a.attname::varchar as column_name
from
    pg_class t,
    pg_class i,
    pg_index ix,
    pg_attribute a
where
    t.oid = ix.indrelid
    and i.oid = ix.indexrelid
    and a.attrelid = t.oid
    and a.attnum = ANY(ix.indkey)
    and t.relkind = 'r'
    and t.relname = _table_name
order by
    t.relname,
    i.relname;
  END;
  $$ LANGUAGE plpgsql;

用法:

select * from getIndices('<my_table>')

稍微修改一下@cope360的回答:

create table test (a int, b int, c int, constraint pk_test primary key(c, a, b));
select i.relname as index_name,
       ix.indisunique as is_unique,
       a.attname as column_name,
from pg_class c
       inner join pg_index ix on c.oid=ix.indrelid
       inner join pg_class i on ix.indexrelid=i.oid
       inner join pg_attribute a on a.attrelid=c.oid and a.attnum=any(ix.indkey)
where c.oid='public.test'::regclass::oid
order by array_position(ix.indkey, a.attnum) asc;

这将显示索引列的正确顺序:

index_name      is_unique  column_name
pk_test         true       c
pk_test         true       a
pk_test         true       b

类似于接受的答案，但有左连接pg_attribute作为正常连接或查询pg_attribute不给出索引，如: 为用户创建唯一索引unique_user_name_index (lower(name))

select 
    row_number() over (order by c.relname),
    c.relname as index, 
    t.relname as table, 
    array_to_string(array_agg(a.attname), ', ') as column_names 
from pg_class c
join pg_index i on c.oid = i.indexrelid and c.relkind='i' and c.relname not like 'pg_%' 
join pg_class t on t.oid = i.indrelid
left join pg_attribute a on a.attrelid = t.oid and a.attnum = ANY(i.indkey) 
group by t.relname, c.relname order by c.relname;

@cope360的精彩回答，转换为使用连接语法。

select t.relname as table_name
     , i.relname as index_name
     , array_to_string(array_agg(a.attname), ', ') as column_names
from pg_class t
join pg_index ix
on t.oid = ix.indrelid
join pg_class i
on i.oid = ix.indexrelid
join pg_attribute a
on a.attrelid = t.oid
and a.attnum = ANY(ix.indkey)
where t.relkind = 'r'
and t.relname like 'test%'
group by t.relname
       , i.relname
order by t.relname
       , i.relname
;

该命令还显示了表变量、索引和约束的视图

=# \d table_name;

例子:

testannie=# \d dv.l_customer_account;