添加缺少的每月行

Question

例如，我想在请求中列出两个日期之间缺少的日期

我的数据：

YEAR_MONTH  | AMOUNT    
202001  |  500    
202001  |  600    
201912  |  100    
201910  |  200
201910  |  100     
201909  |  400
201601  | 5000

我希望请求返回

201912  |  100    
201911  |    0    
201910  |  300
201909  |  400     
201908  |    0
201907  |    0
201906  |    0
....    |    0
201712  |    0

我想要从执行之日起的最后 24 个月

我对日期做了类似的事情，但不是 YEAR MONTH yyyyMM

select date_sub(s.date_order ,nvl(d.i,0)) as date_order, case when d.i > 0 then 0 else s.amount end as amount
from
(--find previous date
select date_order, amount, 
        lag(date_order) over(order by date_order) prev_date,
        datediff(date_order,lag(date_order) over(order by date_order)) datdiff
from
( --aggregate
 select date_order, sum(amount) amount from your_data group by date_order )s
)s
--generate rows
lateral view outer posexplode(split(space(s.datdiff-1),' ')) d as i,x
order by date_order;

我使用带有 Apache Hive 连接器的 Cassandra 数据库

有人可以帮我吗？

Answer 1

因此，如果我理解正确，您想添加当前缺少的所有日期，因为这些天 amount 恰好为 0。

你可以用这个：

select adddate('1970-01-01',t4.i*10000 + t3.i*1000 + t2.i*100 + t1.i*10 + t0.i) base_date from
    (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
    (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
    (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
    (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
    (select 0 i union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4
    having base_date between curdate() - interval 24 month and curdate();

这基本上会创建一个 1970 年到 2200 年之间的日期列表（针对您感兴趣的日期进行过滤）。

想法是从中选择一个子查询并与手头的表连接（在日期字段上）。

示例：return empty rows for not existsting data

至于日期格式（YEAR MONTH YYYYMM）你可以这样运行：

DATE_FORMAT(your_date,'%Y%m')

Answer 2

date_range 子查询从当前日期返回 24 个月（如果您想要 24 个月以外的范围，请调整）。将其与您的数据集左连接，请参阅此演示代码中的 cmets：

with date_range as 
(--this query generates months range, check it's output
select date_format(add_months(concat(date_format(current_date,'yyyy-MM'),'-01'),-s.i),'yyyyMM') as year_month 
  from ( select posexplode(split(space(24),' ')) as (i,x) ) s --24 months
),

your_data as (--use your table instead of this example
select stack(7,
202001, 500,    
202001, 600,    
201912, 100,    
201910, 200,
201910, 100,     
201909, 400,
201601,5000 -----this date is beyond 24 months, hence it is not in the output
) as (YEAR_MONTH, AMOUNT )
)

select d.year_month, sum(nvl(s.amount,0)) as amount --aggregate
  from date_range d 
       left join your_data s on d.year_month=s.year_month
  group by d.year_month;

结果：

d.year_month    amount
201801  0
201802  0
201803  0
201804  0
201805  0
201806  0
201807  0
201808  0
201809  0
201810  0
201811  0
201812  0
201901  0
201902  0
201903  0
201904  0
201905  0
201906  0
201907  0
201908  0
201909  400
201910  300
201911  0
201912  100
202001  1100

使用您的表代替 your_data 子查询。必要时添加order by。