我想在PostgreSQL中获得索引上的列。

在MySQL中,您可以使用SHOW INDEXES FOR表并查看Column_name列。

mysql> show indexes from foos;

+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
| Table | Non_unique | Key_name            | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment |
+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
| foos  |          0 | PRIMARY             |            1 | id          | A         |       19710 |     NULL | NULL   |      | BTREE      |         | 
| foos  |          0 | index_foos_on_email |            1 | email       | A         |       19710 |     NULL | NULL   | YES  | BTREE      |         | 
| foos  |          1 | index_foos_on_name  |            1 | name        | A         |       19710 |     NULL | NULL   |      | BTREE      |         | 
+-------+------------+---------------------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+

PostgreSQL中存在类似的东西吗?

我已经在psql命令提示符中尝试了\d(使用-E选项来显示SQL),但它没有显示我正在寻找的信息。

更新:感谢大家的回答。cope360提供了我想要的东西,但也有一些人提供了非常有用的链接。为了将来的参考,请查看pg_index的文档(通过Milen A. Radev)和非常有用的文章从PostgreSQL提取META信息(通过micharov Niklas)。


当前回答

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;

其他回答

@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;

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

有一个简单的解决方案:

SELECT 
  t.relname table_name,
  ix.relname index_name,
  indisunique,
  indisprimary, 
  regexp_replace(pg_get_indexdef(indexrelid), '.*\((.*)\)', '\1') columns
FROM pg_index i
JOIN pg_class t ON t.oid = i.indrelid
JOIN pg_class ix ON ix.oid = i.indexrelid
WHERE t.relname LIKE 'test%'

`

PostgreSQL (pg_indexes):

SELECT * FROM pg_indexes WHERE tablename = 'mytable';

MySQL(显示索引)

SHOW INDEX FROM mytable;