如何计算工作日的间隔和整个间隔

Question

我有以下数据：

entity_id	阶段	old_phase	时间	下一次
1	'日志'	空	1654781946989	1654781949732
1	'批准'	'日志'	1654781949732	1654781952676
1	'满足'	'批准'	1654781952676	1677506971778
1	'接受'	'满足'	1677506971778	1677518742552
1	'回顾'	'接受'	1677518742552	1678097845979
1	'满足'	'回顾'	1678097845979	1678097847325
1	'接受'	'满足'	1678097847325	1678097977816
1	'回顾'	'接受'	1678097977816	空
2	'日志'	空	1645088036633	1645088043676
2	'批准'	'日志'	1645088043676	1645088047318
2	'满足'	'批准'	1645088047318	1677500808099
2	'关闭'	'满足'	1677500808099	空

time

和

next_time

表示为以毫秒为单位的 unix 时间戳。

目标是计算每个

entity_id

阶段的每个

Fulfill

的持续时间总和。

我的数据库是只读的，所以我不能创建函数。

预期结果应该是这样的：

entity_id	时长（不包括周六和周日）	持续时间满
1	186天22小时30分21秒	{“月”：8，“天”：22，“小时”：22，“分钟”：30，“秒”：21}
2	267天3小时32分41秒	{“年”：1，“天”：15，“小时”：3，“分钟”：32，“秒”：41}

电流输出：

entity_id	时长（不包括周六和周日）	持续时间满
1	186天22小时30分19秒	{“月”：8，“天”：22，“小时”：22，“分钟”：30，“秒”：19}
1	-76天0小时0分2秒	{“秒”：2}
2	267天3小时32分41秒	{“年”：1，“天”：15，“小时”：3，“分钟”：32，“秒”：41}

如果

2 Fulfill

上有

1 entity_id

相，则输出不正确。

我的 SQL 请求：

WITH temp2 AS (
    SELECT
        entity_id,
        old_phase,
        phase,
        time,
        next_time,
        to_timestamp(to_char(to_timestamp("time"/1000.0) at time zone 'Europe/Paris', 'yyyy-mm-dd HH24:MI:SS'), 'yyyy-mm-dd HH24:MI:SS') at time zone 'Europe/Paris' AS "TIME 2",
        to_timestamp(to_char(to_timestamp("next_time"/1000.0) at time zone 'Europe/Paris', 'yyyy-mm-dd HH24:MI:SS'), 'yyyy-mm-dd HH24:MI:SS') at time zone 'Europe/Paris'AS "NEXT TIME 2",
        ((next_time - time)/1000.0) AS DIFF
    FROM 
        tbl_history
    WHERE phase = 'Fulfill'
),parms (entity_id, start_date, end_date) AS
(
    SELECT 
        entity_id,
        "TIME 2"::timestamp,
        "NEXT TIME 2"::timestamp
    FROM 
        temp2
), weekend_days (entity_id, wkend) AS 
(
    SELECT 
        entity_id,
        SUM(case when extract(isodow from d) in (6, 7) then 1 else 0 end) 
    FROM 
        parms
    CROSS JOIN 
        generate_series(start_date, end_date, interval '1 day') dn(d)
    GROUP BY entity_id
)

SELECT 
    entity_id,
    CONCAT(
        extract(day from diff), ' days ', 
        extract( hours from diff)   , ' hours ', 
        extract( minutes from diff) , ' minutes ', 
        extract( seconds from diff)::int , ' seconds '
    ) AS "Duration (excluding saturday and sunday)",
    justify_interval(end_date::timestamp - start_date::timestamp) AS "Duration full"
FROM (
    SELECT 
        start_date,
        end_date,
        entity_id,
        (end_date-start_date) - (wkend * interval '1 day') AS diff
    FROM parms 
    JOIN weekend_days USING(entity_id)
) sq;

演示：https://www.db-fiddle.com/f/ujJe2t9CYLhoRm23RAbSbv/2

如何计算每个 entity_id 的 Fulfill 阶段的完整持续时间（有和没有工作日）？

Answer 1

可以求出每个

sum

的时间间隔（包括

justify_interval

）的

entity_id

s，求出

entity_id

所有间隔之间的周末总数，然后join结果，减去周末数从总和区间：

with cte as (
   select t.*, to_timestamp(to_char(to_timestamp(t.time/1000.0) at time zone 'Europe/Paris', 'yyyy-mm-dd HH24:MI:SS'), 'yyyy-mm-dd HH24:MI:SS') at time zone 'Europe/Paris' s,
    to_timestamp(to_char(to_timestamp(t.next_time/1000.0) at time zone 'Europe/Paris', 'yyyy-mm-dd HH24:MI:SS'), 'yyyy-mm-dd HH24:MI:SS') at time zone 'Europe/Paris' e
   from tbl_history t
),
dt_diff as (
   select t.entity_id, t.phase, sum(t.e - t.s) dr, sum(justify_interval(t.e - t.s)) dt 
   from cte t where t.phase = 'Fulfill' group by t.entity_id, t.phase
),
weekends as (
  select t.entity_id, t.phase, sum(case when extract(dow from v) in (0, 6) then 1 else 0 end) wt 
  from cte t cross join generate_series(t.s, t.e, interval '1 day') v  
  where t.phase = 'Fulfill' group by t.entity_id, t.phase
)
select t1.entity_id, t1.dr - make_interval(days => w.wt::int) "Duration (excluding saturday and sunday)", 
   (select jsonb_object_agg(d.key, d.value) from jsonb_each(
      jsonb_build_object('years', extract(years from t1.dt), 
       'months', extract(months from t1.dt), 
       'days', extract(days from t1.dt), 
       'hours', extract(hours from t1.dt), 
       'minutes', extract(minutes from t1.dt), 
       'seconds', extract(seconds from t1.dt))) d where d.value::int > 0) "Duration full"
from dt_diff t1 join weekends w on w.entity_id = t1.entity_id 

见小提琴