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)。
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>')
我认为这个版本在这个线程上还不存在:它提供了列名列表和索引的ddl。
CREATE OR REPLACE VIEW V_TABLE_INDEXES AS
SELECT
n.nspname as "schema"
,t.relname as "table"
,c.relname as "index"
,i.indisunique AS "is_unique"
,array_to_string(array_agg(a.attname), ', ') as "columns"
,pg_get_indexdef(i.indexrelid) as "ddl"
FROM pg_catalog.pg_class c
JOIN pg_catalog.pg_namespace n ON n.oid = c.relnamespace
JOIN pg_catalog.pg_index i ON i.indexrelid = c.oid
JOIN pg_catalog.pg_class t ON i.indrelid = t.oid
JOIN pg_attribute a ON a.attrelid = t.oid AND a.attnum = ANY(i.indkey)
WHERE c.relkind = 'i'
and n.nspname not in ('pg_catalog', 'pg_toast')
and pg_catalog.pg_table_is_visible(c.oid)
GROUP BY
n.nspname
,t.relname
,c.relname
,i.indisunique
,i.indexrelid
ORDER BY
n.nspname
,t.relname
,c.relname;
我发现使用函数的索引不链接到列名，所以偶尔你会发现一个索引列表，例如一个列名，而实际上是使用3。
例子:
CREATE INDEX ui1 ON table1 (coalesce(col1,''),coalesce(col2,''),col3)
该查询仅返回'col3'作为索引上的列，但DDL显示了索引中使用的全部列。
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。
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)
