我有以下兩個queries.Query 1是快速的,因爲它使用索引(使用嵌套循環連接)和查詢2使用散列連接,並且它比較慢。通過左連接排序不使用索引和非常慢
查詢1按表1列排序,查詢2按表2排列排序。
查詢1
learning=# explain analyze
select *
from users left join
access_logs
on users.userid = access_logs.userid
order by users.userid
limit 10 offset 90;
QUERY PLAN
--------------------------------------------------------------------------------------------------------------------------------------------------------------
Limit (cost=14.00..15.46 rows=10 width=104) (actual time=1.330..1.504 rows=10 loops=1)
-> Merge Left Join (cost=0.85..291532.97 rows=1995958 width=104) (actual time=0.037..1.482 rows=100 loops=1)
Merge Cond: (users.userid = access_logs.userid)
-> Index Scan using users_pkey on users (cost=0.43..151132.75 rows=1995958 width=76) (actual time=0.018..1.135 rows=100 loops=1)
-> Index Scan using access_logs_userid_idx on access_logs (cost=0.43..110471.45 rows=1995958 width=28) (actual time=0.012..0.198 rows=100 loops=1)
Planning time: 0.469 ms
Execution time: 1.569 ms
查詢2
learning=# explain analyze
select *
from users left join
access_logs
on users.userid = access_logs.userid
order by access_logs.userid
limit 10 offset 90;
QUERY PLAN
------------------------------------------------------------------------------------------------------------------------------------------------
Limit (cost=293584.20..293584.23 rows=10 width=104) (actual time=3821.432..3821.439 rows=10 loops=1)
-> Sort (cost=293583.98..298573.87 rows=1995958 width=104) (actual time=3821.391..3821.415 rows=100 loops=1)
Sort Key: access_logs.userid
Sort Method: top-N heapsort Memory: 51kB
-> Hash Left Join (cost=73231.06..217299.90 rows=1995958 width=104) (actual time=539.859..3168.754 rows=1995958 loops=1)
Hash Cond: (users.userid = access_logs.userid)
-> Seq Scan on users (cost=0.00..44814.58 rows=1995958 width=76) (actual time=0.009..443.260 rows=1995958 loops=1)
-> Hash (cost=34636.58..34636.58 rows=1995958 width=28) (actual time=539.112..539.112 rows=1995958 loops=1)
Buckets: 262144 Batches: 2 Memory Usage: 58532kB
-> Seq Scan on access_logs (cost=0.00..34636.58 rows=1995958 width=28) (actual time=0.006..170.061 rows=1995958 loops=1)
Planning time: 0.480 ms
Execution time: 3832.245 ms
問題
- 第二個查詢是慢,因爲分揀是不要e在計劃之前加入。
- 爲什麼第二個表中的排序不使用索引?下面有這樣一個計劃。
查詢 - 解釋分析SELECT * FROM access_logs爲了通過用戶ID限制10偏移90;
計劃
Limit (cost=5.41..5.96 rows=10 width=28) (actual time=0.199..0.218 rows=10 loops=1)
-> Index Scan using access_logs_userid_idx on access_logs (cost=0.43..110471.45 rows=1995958 width=28) (actual time=0.029..0.201 rows=100 loops=1)
Planning time: 0.120 ms
Execution time: 0.252 ms
編輯1:
我的目標不是兩個查詢比較,其實我要的結果作爲查詢2,我只提供查詢1,使相比之下,我可以理解。
按順序不限於連接列,用戶也可以按表2中的另一列進行排序,計劃如下。
learning=# explain analyze select * from users left join access_logs on users.userid=access_logs.userid order by access_logs.last_login limit 10;
QUERY PLAN
------------------------------------------------------------------------------------------------------------------------------------------------
Limit (cost=260431.83..260431.86 rows=10 width=104) (actual time=3846.625..3846.627 rows=10 loops=1)
-> Sort (cost=260431.83..265421.73 rows=1995958 width=104) (actual time=3846.623..3846.623 rows=10 loops=1)
Sort Key: access_logs.last_login
Sort Method: top-N heapsort Memory: 27kB
-> Hash Left Join (cost=73231.06..217299.90 rows=1995958 width=104) (actual time=567.104..3174.818 rows=1995958 loops=1)
Hash Cond: (users.userid = access_logs.userid)
-> Seq Scan on users (cost=0.00..44814.58 rows=1995958 width=76) (actual time=0.007..443.364 rows=1995958 loops=1)
-> Hash (cost=34636.58..34636.58 rows=1995958 width=28) (actual time=566.814..566.814 rows=1995958 loops=1)
Buckets: 262144 Batches: 2 Memory Usage: 58532kB
-> Seq Scan on access_logs (cost=0.00..34636.58 rows=1995958 width=28) (actual time=0.004..169.137 rows=1995958 loops=1)
Planning time: 0.490 ms
Execution time: 3857.171 ms
這兩個查詢都返回完全不同的結果集。外部連接可能會在'access_logs.userid'中返回NULL,而您在此列中則返回ORDER。 – dnoeth
dnoeth是對的。如果'access_log.userid'不包含'null'值,那麼'users.userid'的排序與'access_log.userid'的排序相同(因爲它們是連接列是相同的)。 –
@a_horse_with_no_name:我上面做了一個編輯,很抱歉沒有先告訴它 –