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)。
一些样本数据…
create table test (a int, b int, c int, constraint pk_test primary key(a, b));
create table test2 (a int, b int, c int, constraint uk_test2 unique (b, c));
create table test3 (a int, b int, c int, constraint uk_test3b unique (b), constraint uk_test3c unique (c), constraint uk_test3ab unique (a, b));
使用pg_get_indexdef函数:
select pg_get_indexdef(indexrelid) from pg_index where indrelid = 'test'::regclass;
pg_get_indexdef
--------------------------------------------------------
CREATE UNIQUE INDEX pk_test ON test USING btree (a, b)
(1 row)
select pg_get_indexdef(indexrelid) from pg_index where indrelid = 'test2'::regclass;
pg_get_indexdef
----------------------------------------------------------
CREATE UNIQUE INDEX uk_test2 ON test2 USING btree (b, c)
(1 row)
select pg_get_indexdef(indexrelid) from pg_index where indrelid ='test3'::regclass;
pg_get_indexdef
------------------------------------------------------------
CREATE UNIQUE INDEX uk_test3b ON test3 USING btree (b)
CREATE UNIQUE INDEX uk_test3c ON test3 USING btree (c)
CREATE UNIQUE INDEX uk_test3ab ON test3 USING btree (a, b)
(3 rows)
@cope360接受的答案很好,但我想要一些更像Oracle的DBA_IND_COLUMNS, ALL_IND_COLUMNS和USER_IND_COLUMNS(例如,报告表/索引模式和索引在多列索引中的位置),所以我把接受的答案改编成这样:
with
ind_cols as (
select
n.nspname as schema_name,
t.relname as table_name,
i.relname as index_name,
a.attname as column_name,
1 + array_position(ix.indkey, a.attnum) as column_position
from
pg_catalog.pg_class t
join pg_catalog.pg_attribute a on t.oid = a.attrelid
join pg_catalog.pg_index ix on t.oid = ix.indrelid
join pg_catalog.pg_class i on a.attnum = any(ix.indkey)
and i.oid = ix.indexrelid
join pg_catalog.pg_namespace n on n.oid = t.relnamespace
where t.relkind = 'r'
order by
t.relname,
i.relname,
array_position(ix.indkey, a.attnum)
)
select *
from ind_cols
where schema_name = 'test'
and table_name = 'indextest'
order by schema_name, table_name
;
这将给出如下输出:
schema_name | table_name | index_name | column_name | column_position
-------------+------------+------------+-------------+-----------------
test | indextest | testind1 | singleindex | 1
test | indextest | testind2 | firstoftwo | 1
test | indextest | testind2 | secondoftwo | 2
(3 rows)
在处理索引时,索引中构造列的顺序与列本身同样重要。
下面的查询以排序的方式列出给定表的所有索引及其所有列。
SELECT
table_name,
index_name,
string_agg(column_name, ',')
FROM (
SELECT
t.relname AS table_name,
i.relname AS index_name,
a.attname AS column_name,
(SELECT i
FROM (SELECT
*,
row_number()
OVER () i
FROM unnest(indkey) WITH ORDINALITY AS a(v)) a
WHERE v = attnum)
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 'tablename'
ORDER BY table_name, index_name, i
) raw
GROUP BY table_name, index_name