连接当前行和上一行之间的范围

Question

有人能帮帮我吗？我有两张桌子:价格和间隔：

   Prices:           Intervals:
   Price         Interval_bound  Category
    16                 5         cheap
    11                 10        cheap
    9                  15        median
    26                 20        median
    6                  25        expensive

我需要根据其间隔将类别值加入到价格，其中Interval_bound是类别的最低界限：

Price  Category
 16      median
 11      cheap
 9       cheap
 26      expensive
 6       cheap

我试着用

select Price, Category from Prices 
left join Intervals on Prices.Price interpolate previous value Interval.Interval_bound

但它只给了我一个空的类别。我怎么做才是最简单的方法？我在用Vertica。

Answer 1

您可以使用lead()获得下一个上限，然后使用join

select p.Price, i.Category
from Prices p left join
     (select i.*,
             lead(interval_bound) over (order by interval_bound) as next_interval_bound
      from Intervals i
     ) i
     on p.price >= i.interval_bound and
        (p.price < i.next_interval_bound or i.next_interval_bound is null);

Answer 2

我很困惑-为什么你的版本不起作用？

-- your input ..
WITH
prices(price) AS (
          SELECT 16
UNION ALL SELECT 11
UNION ALL SELECT  9
UNION ALL SELECT 26
UNION ALL SELECT  6
)
, 
-- your other input 
intervals(interval_bound,category) AS (
          SELECT  5,'cheap'
UNION ALL SELECT 10,'cheap'
UNION ALL SELECT 15,'median'
UNION ALL SELECT 20,'median'
UNION ALL SELECT 25,'expensive'
)
-- the way I would write it ...
SELECT
  p.price
, i.category
FROM prices p
LEFT JOIN intervals i
ON p.price INTERPOLATE PREVIOUS VALUE i.interval_bound
;
 price | category
-------+-----------
     6 | cheap
     9 | cheap
    11 | cheap
    16 | median
    26 | expensive

-- the way you wrote it ...
select Price, Category from Prices
left join Intervals on Prices.Price interpolate previous value Intervals.Interval_bound;
 Price | Category
-------+-----------
     6 | cheap
     9 | cheap
    11 | cheap
    16 | median
    26 | expensive

你的案子出了什么问题？