每小时分组数据并插入Postgres的汇总表

Question

我有一个包含大约50个表的数据库（现在），由于数据量大量涌入几个表，我的任务是创建数据的每小时汇总并将其转储到另一个表中。所以运行关于原始数据的报告需要很长时间，因为新数据库（2周龄）已经在我正在为其提取数据的两个表中的一个表中达到了200k记录。

该查询为客户提供了三种可能的结果 - “cust1”，cust2“和”cust3“，每个都有一张选定的卡片（产品质量问卷的邮件回复和潜在的奖品获奖），这是13种选择之一。字母表示（“A”Ace，“K”King等相关值）

这是一个子查询和相应的结果：

select sp_cust_card_sequence(cards.cust1) as seq, cards.cust1 as card, count(cards.cust1) as card_count, rtt.game_id as game_id, gt.promo_id as promo_id, gt.choice_id as choice_id, extract(hour from header.start_timestamp) as hour, header.start_timestamp::timestamp::date as date
    from game_bac_cards cards
    inner join card_cust_resp header ON (header.id = cards.game_id)
    inner join game_table gt ON (header.promo_id = gt.promo_id)
    inner join ref_table_type rtt ON (gt.table_type_id = rtt.id)
    where result <> 'undef' 
    group by date, hour, card, rtt.game_id, gt.promo_id, gt.choice_id, cards.cust1

基本上，我想将卡片列中的所有“K”值按小时分组。通过下面的查询，我似乎能够几乎实现这一目标，但下面的代码段显示“ K ”的值为小时“ 19 “on” 2014年12月4日“有两个条目，而不是一个条目。我确信有更优雅的方式来做到这一点。

最终查询：

select date, hour, card, seq, sum(card_count) as card_count, game_id, choice_id, promo_id 
from (
select date, hour, card, seq, sum(card_count) as card_count, game_id, choice_id, promo_id from (
    select sp_cust_card_sequence(cards.cust1) as seq, cards.cust1 as card, count(cards.cust1) as card_count, rtt.game_id as game_id, gt.promo_id as promo_id, gt.choice_id as choice_id, extract(hour from header.start_timestamp) as hour, header.start_timestamp::timestamp::date as date
    from game_bac_cards cards
    inner join card_cust_resp header ON (header.id = cards.game_id)
    inner join game_table gt ON (header.promo_id = gt.promo_id)
    inner join ref_table_type rtt ON (gt.table_type_id = rtt.id)
    where result <> 'undef' 
    group by date, hour, card, rtt.game_id, gt.promo_id, gt.choice_id, cards.cust1
) as cust1_table
where cust1_table.card is not null and cust1_table.card <> ''
group by date, hour, card_count, card, seq, game_id, choice_id, promo_id

union all

select date, hour, card, seq, sum(card_count) as card_count, game_id, choice_id, promo_id from (
    select sp_cust_card_sequence(cards.cust2) as seq, cards.cust2 as card, count(cards.cust2) as card_count, rtt.game_id as game_id, gt.promo_id as promo_id, gt.choice_id as choice_id, extract(hour from header.start_timestamp) as hour, header.start_timestamp::timestamp::date as date
    from game_bac_cards cards
    inner join card_cust_resp header ON (header.id = cards.game_id)
    inner join game_table gt ON (header.promo_id = gt.promo_id)
    inner join ref_table_type rtt ON (gt.table_type_id = rtt.id)
    where result <> 'undef' 
    group by date, hour, card, rtt.game_id, gt.promo_id, gt.choice_id, cards.cust2
) as cust2_table
where cust2_table.card is not null and cust2_table.card <> ''
group by date, hour, card_count, card, seq, game_id, choice_id, promo_id

union all

select date, hour, card, seq, sum(card_count) as card_count, game_id, choice_id, promo_id from (
    select sp_cust_card_sequence(cards.cust3) as seq, cards.cust3 as card, count(cards.cust3) as card_count, rtt.game_id as game_id, gt.promo_id as promo_id, gt.choice_id as choice_id, extract(hour from header.start_timestamp) as hour, header.start_timestamp::timestamp::date as date
    from game_bac_cards cards
    inner join card_cust_resp header ON (header.id = cards.game_id)
    inner join game_table gt ON (header.promo_id = gt.promo_id)
    inner join ref_table_type rtt ON (gt.table_type_id = rtt.id)
    where result <> 'undef' 
    group by date, hour, card, rtt.game_id, gt.promo_id, gt.choice_id, cards.cust3
) as cust3_table
where cust3_table.card is not null and cust3_table.card <> ''
group by date, hour, card_count, card, seq, game_id, choice_id, promo_id
) as card_details
and card_details.card is not null and card_details.card <> ''
group by date, hour, card, card_count, seq, game_id, choice_id, promo_id
order by date, hour, seq

结果仅显示卡 K |小时 19 |日期 2014年12月4日。这应该只有一行而不是两行。 （？）

非常感谢任何帮助！

Answer 1

尝试在外部查询

中的card_count上删除该组

分组按日期，小时， card_count ，卡片，seq，game_id，choice_id，promo_id