考虑：
with days as (select day::date
from generate_series(date '2013-01-01', date '2013-01-01' + 365, interval '1 day' day) day
)
select 'Inspections'::text as data_label,
count(i.reporting_id) as daily_count, d.day as date_column
from days d
left join inspection i on i.close_case_date = d.day
group by d.day
order by d.day
这将返回一个如下所示的集合：
data_label | daily_count | date_column
Inspections 1 01/01/13
Inspections 2 01/02/13
Inspections 4 01/04/13
Inspections 8 01/06/13
请注意记录集中的 1 天和 2 天间隔。我需要生成一个用 0 填充的值的集合，如下所示：
data_label | daily_count | date_column
Inspections 1 01/01/13
Inspections …