Question

公司每年使用以下命名约定生成三批：YYYY11,YYYY22,YYYY33

在这种情况下batch_id =1所有批次都是顺序的。但是，batches 200933和201022的缺席会导致batch_id=2不顺序。

with batch_sequences as(
  select  1 as batch_sequence, '200911' as batch_date  from dual union all
  select  2 as batch_sequence, '200922' as batch_date from dual  union all
  select  3 as batch_sequence, '200933' as batch_date  from dual  union all
  select  4 as batch_sequence, '201011' as batch_date  from dual union all
  select  5 as batch_sequence, '201022' as batch_date from dual union all
  select  6 as batch_sequence, '201033' as batch_date  from dual),
batch_entries as
 (
   select   1 as batch_id, '200911' as  batch_date from dual union all
   select   1 as batch_id, '200922' as  batch_date from dual union all
   select   1 as batch_id, '200933' as  batch_date from dual union all
   select   1 as batch_id, '201011' as  batch_date from dual union all
   select   1 as batch_id, '201022' as  batch_date from dual union all
   select   1 as batch_id, '201033' as  batch_date from dual union all
   select   2 as batch_id, '200911' as  batch_date from dual union all
   select   2 as batch_id, '200922' as  batch_date from dual union all
   select   2 as batch_id, '201011' as  batch_date from dual union all
   select   2 as batch_id, '201033' as  batch_date from dual
 )
select batch_sequence,
       e.batch_id,
       s.batch_date,
       lead(batch_sequence,1) over (order by batch_sequence) as next_batch
  from batch_entries e
 inner join batch_sequences s on e.batch_date=s.batch_date
 order by e.batch_id,
          e.batch_date;

我想我可以对铅值进行数学运算，但我没有得到所有的数据 batch_sequence值可以正确计算它。

问题

如何编写查询以显示batch_id=1有'完美运行'且batch_id=2错过了一些batch_dates？

我会满足于任何可以突出显示此结果的结果集。

Answer 1

根据batch_id为每个batch_date分配一个序号，并将其与batch_sequence进行比较：

with cte as
 (
   select batch_id, batch_date,
      row_number() -- sequential number
      over (partition by batch_id
            order by batch_date) as rn
   from batch_entries
 )
 select e.batch_id
 from cte e join batch_sequences s 
   on e.batch_date=s.batch_date 
 group by e.batch_id
 -- if there's no missing batch the difference will always be the same  
 having min(s.batch_sequence - e.rn) <> max(s.batch_sequence - e.rn)

请参阅fiddle

第二批数据：

batch_date  rn    batch_sequence  batch_date
'200911'  -> 1           1         '200911'
'200922'  -> 2           2         '200922'
                         3         '200922'
'201011'  -> 3           4         '201011' 
                         5         '201022'
'201033'  -> 4           6         '201033'

如何确定我的批次是否全部连续？

1 个答案: