替代SQL count子查询

我有以下查询：

SELECT DISTINCT 
    e.id,folder,subject,in_reply_to,message_id,"references",e.updated_at,(
        select count(*)  
        from emails  
        where 
        (
            select "references"[1] 
            from emails 
            where message_id = e.message_id
        ) = ANY ("references") 
        or message_id = 
        (
            select "references"[1] 
            from emails 
            where message_id = e.message_id
        )
    )
FROM "emails" e
INNER JOIN "email_participants" 
    ON ("email_participants"."email_id" = e."id") 
WHERE (("user_id" = 220) 
AND ("folder" = 'INBox')) 
ORDER BY e."updated_at" DESC 
LIMIT 10 OFFSET 0;

Here是上述查询的explain analyze 输出.

查询执行正常,直到我添加下面的count子查询：

(
    select count(*)  
    from emails  
    where 
    (
        select "references"[1] 
        from emails 
        where message_id = e.message_id
    ) = ANY ("references") 
    or message_id = 
    (
        select "references"[1] 
        from emails 
        where message_id = e.message_id
    )
)

事实上,我已经尝试过更简单的子查询,似乎是聚合函数本身需要时间.

那么我可以将count子查询附加到每个结果上吗？我应该在初始查询运行后更新结果吗？

这是一个pastebin,它将创建表并在最后运行性能不佳的查询以显示输出应该是什么.

解决方法

@H_404_19@ 扩展Paul Guyot的答案,您可以将子查询移动到派生表中,该表应该执行得更快,因为它在一次扫描(加一个连接)中获取消息计数,而不是每行一次扫描.

SELECT DISTINCT 
    e.id,e.folder,e.subject,e.message_id,e."references",t1.message_count
FROM "emails" e
INNER JOIN "email_participants" 
    ON ("email_participants"."email_id" = e."id") 
INNER JOIN (
    SELECT COUNT(e2.id) message_count,e.message_id
    FROM emails e
    LEFT JOIN emails e2 ON (ARRAY[e."references"[1]] <@ e2."references"
    OR e2.message_id = e."references"[1])
    GROUP BY e.message_id
) t1 ON t1.message_id = e.message_id
WHERE (("user_id" = 220) 
AND ("folder" = 'INBox')) 
ORDER BY e."updated_at" DESC 
LIMIT 10 OFFSET 0;

使用pastebin数据的小提琴 – http://www.sqlfiddle.com/#!15/c6298/7

下面是postgres生成的查询计划,用于通过加入派生表来获取相关子查询中的计数与获取计数.我使用了自己的一张桌子,但我认为结果应该是相似的.

相关子查询

"Limit  (cost=0.00..1123641.81 rows=1000 width=8) (actual time=11.237..5395.237 rows=1000 loops=1)"
"  ->  Seq Scan on visit v  (cost=0.00..44996236.24 rows=40045 width=8) (actual time=11.236..5395.014 rows=1000 loops=1)"
"        SubPlan 1"
"          ->  Aggregate  (cost=1123.61..1123.62 rows=1 width=0) (actual time=5.393..5.393 rows=1 loops=1000)"
"                ->  Seq Scan on visit v2  (cost=0.00..1073.56 rows=20018 width=0) (actual time=0.002..4.280 rows=21393 loops=1000)"
"                      Filter: (company_id = v.company_id)"
"                      Rows Removed by Filter: 18653"
"Total runtime: 5395.369 ms"

加入派生表

"Limit  (cost=1173.74..1211.81 rows=1000 width=12) (actual time=21.819..22.629 rows=1000 loops=1)"
"  ->  Hash Join  (cost=1173.74..2697.72 rows=40036 width=12) (actual time=21.817..22.465 rows=1000 loops=1)"
"        Hash Cond: (v.company_id = visit.company_id)"
"        ->  Seq Scan on visit v  (cost=0.00..973.45 rows=40045 width=8) (actual time=0.010..0.198 rows=1000 loops=1)"
"        ->  Hash  (cost=1173.71..1173.71 rows=2 width=12) (actual time=21.787..21.787 rows=2 loops=1)"
"              Buckets: 1024  Batches: 1  Memory Usage: 1kB"
"              ->  HashAggregate  (cost=1173.67..1173.69 rows=2 width=4) (actual time=21.783..21.784 rows=3 loops=1)"
"                    ->  Seq Scan on visit  (cost=0.00..973.45 rows=40045 width=4) (actual time=0.003..6.695 rows=40046 loops=1)"
"Total runtime: 22.806 ms"

替代SQL count子查询

解决方法

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