【发布时间】:2019-03-01 13:02:12
【问题描述】:
这是我的函数的简化版本,其中包含查询(因此任何变量现在都无用)并且此函数不会完成,但如果我单独运行相同的查询,它会在一秒钟内完成。
永远不会完成的函数
select * from test_function_difference(1);
CREATE OR REPLACE FUNCTION test_function_difference (
p_does_nothing int
)
RETURNS TABLE(
t_datum date,
t_capacity numeric,
t_used numeric,
t_category int,
t_category_name text,
t_used_p numeric,
t_unused_p numeric
)
VOLATILE
AS $dbvis$
declare
p_sql text := '';
p_execute text := '';
rec record;
begin
p_sql :=
'
with
vytizeni as (
select
date_trunc(''day'',mcz.datum)::date as datum ,
sum(zd.v_vytizeni)/3600.0 used
from v_ui_cdc_s5_misto_cas_zdroj_aggregace mcz
left join (select * , pul_den as den_noc from v_ui_cdc_s5_misto_cas_zdroj_aggregace_zdrobneni) zd on mcz.id = zd.id
where
datum between ''2018-12-31'' and ''2018-12-31''
and ( zahranicni = 0 or zahranicni is null )
and den_noc = -1
group by
date_trunc(''day'',mcz.datum)::date
)
,kapacita as (
select
date_trunc(''day'',datum)::date as datum ,
sum(obsazeni_g)/3600.0 capacity
from v_ui_cdc_s5_misto_cas_zdroj_aggregace
where
datum between ''2018-12-31'' and ''2018-12-31''
group by
date_trunc(''day'',datum)::date
)
,zdroj as (
select
k.datum,
k.capacity,
v.used,
-1 category
from kapacita k
join vytizeni v on k.datum = v.datum
)
select
c.* ,
kc.nazev::text categeroy_name,
case when sum(capacity)over(partition by datum) = 0 then 1 else used/sum(capacity)over(partition by datum) end as used_p,
greatest(1 - case when sum(capacity)over(partition by datum) = 0 then 1 else sum(used)over(partition by datum)/sum(capacity)over(partition by datum) end,0) as unused_p
from zdroj c
left join v_ui_cdc_s5_kategorie_cinnosti kc on kc.id = c.category
order by c.datum
';
raise notice '% ' , p_sql;
RETURN QUERY
execute p_sql;
END;
$dbvis$ LANGUAGE plpgsql
以及我单独运行的查询(在 533 毫秒内完成)
with
vytizeni as (
select
date_trunc('day',mcz.datum)::date as datum ,
sum(zd.v_vytizeni)/3600.0 used
from v_ui_cdc_s5_misto_cas_zdroj_aggregace mcz
left join (select * , pul_den as den_noc from v_ui_cdc_s5_misto_cas_zdroj_aggregace_zdrobneni) zd on mcz.id = zd.id
where
datum between '2018-12-31' and '2018-12-31'
and ( zahranicni = 0 or zahranicni is null )
and den_noc = -1
group by
date_trunc('day',mcz.datum)::date
)
,kapacita as (
select
date_trunc('day',datum)::date as datum ,
sum(obsazeni_g)/3600.0 capacity
from v_ui_cdc_s5_misto_cas_zdroj_aggregace
where
datum between '2018-12-31' and '2018-12-31'
group by
date_trunc('day',datum)::date
)
,zdroj as (
select
k.datum,
k.capacity,
v.used,
-1 category
from kapacita k
join vytizeni v on k.datum = v.datum
)
select
c.* ,
kc.nazev::text categeroy_name,
case when sum(capacity)over(partition by datum) = 0 then 1 else used/sum(capacity)over(partition by datum) end as used_p,
greatest(1 - case when sum(capacity)over(partition by datum) = 0 then 1 else sum(used)over(partition by datum)/sum(capacity)over(partition by datum) end,0) as unused_p
from zdroj c
left join v_ui_cdc_s5_kategorie_cinnosti kc on kc.id = c.category
order by c.datum
编辑:大约 28 分钟后,我能够从函数中获得结果(我也在周日晚上尝试过,这意味着我拥有整个服务器的资源,因为在正常加载过程中,即使一小时后函数也没有完成)之后我独立运行查询并在 2.1 秒后得到结果这是解释分析
功能:28 分钟 https://explain.depesz.com/s/v9xJ
独立查询:2.1 秒 https://explain.depesz.com/s/aBri
第二次独立运行 430ms https://explain.depesz.com/s/ENva
有趣的说明:如果我将间隔的开始日期编辑为“2018-12-30”或任何其他日期,函数也会完成
意思是
start date = '2018-12-31'
query => finishes under 1 second
function => won't finish
start date = '2018-12-30'
query => finishes under 1 second
function => finishes under 1 second
版本详情:x86_64-pc-linux-gnu 上的 PostgreSQL 10.7,由 gcc (GCC) 4.8.5 20150623 (Red Hat 4.8.5-36) 编译,64 位
【问题讨论】:
-
从您最初发布的执行计划中,很明显
2018-12-30对行数的估计要高得多。给它一些时间——如果你能产生EXPLAIN (ANALYZE, BUFFERS)的输出,分析问题会更容易。 -
我可以为单独运行的查询解释分析,但我不能为该函数执行此操作,因为它没有完成(我在运行 45 分钟后停止)
-
解释分析查询是否单独运行explain.depesz.com/s/Zji6
-
那么对于长时间运行的查询,普通的
EXPLAIN就足够了。请把这些东西放到问题中,而不是评论。 -
@LaurenzAlbe 你知道是否有一个存储计划的系统表实际上是为这个问题执行的,我有点怀疑我之前给你的执行计划可能不正确,因为它看起来完全正确与来自单独查询的相同,我只是放入函数'EXPLAIN VERBOSE'并存储到变量并打印
标签: postgresql function sql-execution-plan