通过 id 识别在另一列中具有顺序值的行数，其中最大值等于特定值

Question

https://www.db-fiddle.com/f/2bzoKxbU2gznwwmQpMmjp5/0

（实际数据库为 Microsoft SQL Server 2014）

以上是我正在尝试做的 fiddle。

CREATE TABLE IF NOT EXISTS table1 (
  id nvarchar(5) NOT NULL,
  year int(4) NOT NULL,
  PRIMARY KEY (id,year)
  );
INSERT INTO table1 (id, year) VALUES
  ('A', '2013'),
  ('A', '2014'),
  ('A', '2017'),
  ('A', '2018'),
  ('B', '2016'),
  ('B', '2017'),
  ('B', '2018'),
  ('C', '2016'),
  ('D', '2014'),
  ('D', '2016'),
  ('D', '2018');

这大致是我正在使用的数据，我想在其中找到每个 id 的 consecutive/sequential 记录数，这些 id 在年份列中也包含“2018”。

到目前为止我的思考过程是这样的：

select id, count(*)
from table1
group by id;

select main.id,
    case when in_2018.id is not null
        then count(*) 
    else 0
    end
from table1 as main
left join table1 as in_2018
    on in_2018.id = main.id
    and
        in_2018.year = 2018
group by main.id;

/*
    Want a table:
    A | 2
    B | 3
    C | 0
    D | 1

    Count of records that are in a single-step incremental that include 2018     by id
*/

显然，这些不是 return 连续行，只是满足“2018”标准的计数。

我尝试了另一种检查方法：

case when count(*) = max(year) - min(year) +1,

在我的数据示例中，这适用于 ID B，因为 B 的所有数据都是顺序的，但它没有解决其他 ID 的损坏模式。

Answer 1

在 SQL 服务器中，您可以使用 row_number():

来解决这个问题

select top (1) id, count(*)
from (select t.*, row_number() over (partition by id order by year) as seqnum
      from table1 t
     ) t
group by id, (year - seqnum)
having sum(case when year = 2018 then 1 else 0 end) > 0
order by count(*) desc;

这使用的观察结果是 year - seqnum 当年份在一个序列中时是常数。

在不支持 window 函数的数据库中，最简单的解决方案可能是执行相同计算的相关子查询：

select id, count(*)
from (select t.*,
             (select count(*)
              from table1 tt
              where tt.id = t.id and tt.year <= t.year
             ) as seqnum
      from table1 t
     ) t
group by id, (year - seqnum)
having sum(case when year = 2018 then 1 else 0 end) > 0
order by count(*) desc
fetch first 1 year only;

Here 是 db<>fiddle.

Answer 2

我看到 Gordon 在这方面领先于我，而且查询要短得多。但我已经走到这一步了，我还是 post 了。我认为总体思路或多或少是相同的，但我的不依赖于任何非标准功能（我认为），我希望我通过添加一些注释来弥补额外的代码，使其更长。 ;-)

并且每个子查询都可以运行分开，因此您可以看到如何逐步 'zoom in' 结果。

select
  id,
  max(span) as nr_of_years
from
  ( -- This inner query gives all the valid ranges, but they have to be deduplicates
    -- For instance, it can give B 2017-2018 while there is also B 2016-2018, which has precedence.
    -- That's why the outer query uses max, to get the longest range
    select
      s.id,
      s.year,
      s.otheryear,
      s.span,
      s.rows_in_span
    from
      ( -- Find all possible 'spans' of years between two rows with the same id.
        -- also find how much rows are in that span. They should match.
        select 
          a.id, 
          a.year,
          b.year as otheryear,
          a.year - b.year + 1 as span,
          ( select count(*) from table1 c
            where
              c.id = a.id and
              c.year >= b.year and
              c.year <= a.year) as rows_in_span
        from
          table1 a
          join table1 b on b.ID = a.ID and b.year <= a.year -- like a cross join, but per ID
      ) s
    where
      -- if they are not equal, it means one year is missing between the lowest and highest year in the span
      s.span = s.rows_in_span and
      -- If the difference between the year and 2018 is more than this, this is a range, but it's out of scope
      abs(s.year - 2018) < s.span
  ) f
group by
  f.id

在 fiddle 中，您可以看到它也适用于 Postgres（您可以在数据库之间切换，我简化了 create 语句以允许这样做）：

DB Fiddle

通过 id 识别在另一列中具有顺序值的行数，其中最大值等于特定值

Identify number of rows by id that have sequential values in another column, with the greatest of those values equaling a specific value

sql

sql-server

sql-server-2014