空行的Postgres默认聚合值

Question

我写了一个查询，该查询正在执行X的聚合。

它从本周的星期一一直到当前日期。

select 
  extract(dow from time_sent) days
  , count(id)
from messages
where time_sent between date_trunc('week', now()) and now()
group by days
order by days;

我想提示如何在星期一未添加任何行的情况下为星期一设置默认值。

即，星期一是空的，所以我得到了这个：

[ anonymous { day: 2, count: '1' },
  anonymous { day: 3, count: '1' },
  anonymous { day: 4, count: '1' },
  anonymous { day: 5, count: '1' },
  anonymous { day: 6, count: '1' } ]

预期结果：

[ anonymous { day: 1, count: '0' },
  anonymous { day: 2, count: '1' },
  anonymous { day: 3, count: '1' },
  anonymous { day: 4, count: '1' },
  anonymous { day: 5, count: '1' },
  anonymous { day: 6, count: '1' } ]

编辑：添加表结构和示例输出。

id | time_sent 
1  | 2018-01-13 15:26:21.443828+08
2  | 2018-01-12 15:26:21.44755+08
3  | 2018-01-11 15:26:21.459208+08
4  | 2018-01-10 15:26:21.440648+08
5  | 2018-01-09 15:26:21.457262+08

查询1：

select 
  coalesce(extract(dow from time_sent), d) days
  , count(message_batch_id)
from generate_series(0,6) d
left join messages 
on d = extract(dow from time_sent)
where time_sent between date_trunc('week', now()) and now()
group by days
order by days;

查询＃1输出：

days | count
2    | 1
3    | 1
4    | 1
5    | 1
6    | 1

编辑：我需要澄清一下， messages表在星期一没有任何行条目。

编辑：

由于某种原因，此查询返回我想要的行结构：

select 
    extract(dow from d.*) days, 
    count(id) 
from GENERATE_SERIES(DATE_TRUNC('WEEK', NOW()), NOW(), '1 
DAY'::INTERVAL) d
left outer join messages m
on m.time_sent::DATE = d
group by days
order by days;

Answer 1

尝试：

select 
  coalesce(extract(dow from time_sent),d) days
  , count(id)
from generate_series(0,6) d
left outer join messages on d = extract(dow from time_sent)
where time_sent between date_trunc('week', now()) and now()
group by days
order by days;

（未测试）

Answer 2

谓词无法扩展行。 您必须使用GENERATE_SERIES生成日期，并将其与messages一起加入：

SELECT extract(dow from time_sent) days, coalesce(count(id), 0)
FROM GENERATE_SERIES(DATE_TRUNC('WEEK', NOW()), NOW(), '1 DAY'::INTERVAL) d 
    LEFT JOIN messages ON time_sent::DATE = d
GROUP BY days
ORDER BY days;

Answer 3

由于某种原因，这会返回我期望的结果。 即

返回一周中一天中的某一天（即使该天没有行条目）

select 
    extract(dow from d.*) days, 
    count(id) 
from GENERATE_SERIES(DATE_TRUNC('WEEK', NOW()), NOW(), '1 
DAY'::INTERVAL) d
left outer join messages m
on m.time_sent::DATE = d
group by days
order by days;