ROW_NUMBER 或其他序列取决于日期 (SQL)

Question

我在使用

row_number

创建序列时遇到问题，但仍然无法处理它。

我有桌子

公元前	io	约会
1a	11	2022-01-01
1a	11	2022-01-02
1a	12	2022-01-03
1a	11	2022-01-04

当我使用由

row_number

和

bc

划分的简单

io

并由

date

排序时，我得到了这个结果

公元前	io	约会	rn
1a	11	2022-01-01	1
1a	11	2022-01-02	2
1a	12	2022-01-03	1
1a	11	2022-01-04	3

但是我需要这个结果，当

io

变化时，下一个

io

，已经遇到过，应该从1

开始

公元前	io	约会	rn
1a	11	2022-01-01	1
1a	11	2022-01-02	2
1a	12	2022-01-03	1
1a	11	2022-01-04	1

我试过用这个sql，但是不正确

select tt.*,row_number() over(partition by tt.bc,tt.io order by tt.date ) as rn
from (
    select '1a' as bc, 11 as io, '2021-01-01' as date
    union all
    select '1a' as bc, 11 as io, '2021-01-02' as date
    union all
    select '1a' as bc, 12 as io, '2021-01-03' as date
    union all
    select '1a' as bc, 11 as io, '2021-01-04' as date
) as tt

Answer 1

这是一个常见的间隙和孤岛问题：将每个键的连续属性值分组（给定一些“类似时间”的维度）。方法是这样的：

• 计算按时间维度排序的每个键

  row_number

  。
• 计算

  row_number

  每个键和按时间维度排序的感兴趣的属性。
• 找到它们的差异 - 这会将连续的相同属性值分组（第二个

  row_number

  在某些属性发生变化时重置为1并且差异增加）。

以下是查询：

with src as (
    select inline(array(
      struct('1a', 11, date '2022-01-01'),
      struct('1a', 11, date '2022-01-02'),
      struct('1a', 12, date '2022-01-03'),
      struct('1a', 11, date '2022-01-04')
    )) as (bc, io, dt)
)
, prepared as (
  select
    src.*
    /*Partition by keys*/
    , row_number() over(partition by bc order by dt asc)
        /*Partition by keys AND attributes to track changes and create groups*/
      - row_number() over(partition by bc, io order by dt asc) as rn_diff
  from src
)
select
  bc, io, dt
  /*Partition by keys AND attributes to track changes AND group number*/
  , row_number() over(partition by bc, io, rn_diff order by dt asc) as rn
from prepared
order by dt asc

公元前	io	dt	rn
1a	11	2022-01-01	1
1a	11	2022-01-02	2
1a	12	2022-01-03	1
1a	11	2022-01-04	1

dbfiddle 基于 Postgres（添加了更多属性）。

Answer 2

您可以手动将行与“先前”行进行比较，并基于它创建变化指标。稍后对该指标求和将为您提供分区号，识别不间断的块。

这个分区号可以用在你的分区子句中

row_number

.

with tt as (
    select '1a' as bc, 11 as io, '2021-01-01' as date
    union all
    select '1a' as bc, 11 as io, '2021-01-02' as date
    union all
    select '1a' as bc, 12 as io, '2021-01-03' as date
    union all
    select '1a' as bc, 11 as io, '2021-01-04' as date
), t2 as(
  select 
    tt.*,
    case when bc = lag(bc) over (order by date) and io = lag(io) over (order by date) then 0 else 1 end ind
  from tt
), t3 as (
  select 
    t2.*, 
    sum(ind) over ( order by date) pid 
  from t2
)
select 
  bc,
  io, 
  date, 
  row_number() over (partition by pid order by date) rn
from t3

演示（在 MySQL 中）这里.

---

编辑：忽略

bc

的变化忽略部分案例条件提及

bc

：

with tt as (
    select '1a' as bc, 11 as io, '2021-01-01' as date
    union all
    select '1b' as bc, 11 as io, '2021-01-02' as date
    union all
    select '1a' as bc, 12 as io, '2021-01-03' as date
    union all
    select '1a' as bc, 11 as io, '2021-01-04' as date
), t2 as(
  select 
    tt.*,
    case when io = lag(io) over (order by date) then 0 else 1 end ind
  from tt
), t3 as (
  select 
    t2.*, 
    sum(ind) over ( order by date) pid 
  from t2
)
select 
  bc,
  io, 
  date, 
  row_number() over (partition by pid order by date) rn
from t3

MySQL 中的演示在这里.