Postgresql密集排名从2开始，如果初始排名为1

因此，我有一个表和一个查询，它对项目的成本进行排序，不允许与位置1有联系，如果位置1有联系，则排名从2开始

下面是带有示例数据的模式

  CREATE TABLE applications
      (id int, name char(10), cost int);

  INSERT INTO applications
      (id, name, cost)
  VALUES
      (1, 'nfhfjs', 10),
      (2, 'oopdld', 20),
      (3, 'Wedass', 14),
      (4, 'djskck', 22),
      (5, 'laookd', 25),
      (6, 'mfjjf', 25),
      (7, 'vfhgg', 28),
      (8, 'nvopq', 29),
      (9, 'nfhfj', 56),
      (10, 'voapp', 56);

这是一个问题

  WITH start_tie AS (
SELECT 
  DENSE_RANK() OVER(ORDER BY cost DESC) cost_rank,
  lead(cost,1) OVER (ORDER BY cost DESC) as next_app_cost
  FROM 
  applications LIMIT 1
)
  SELECT 
  *, 
  DENSE_RANK() OVER(ORDER BY cost DESC) cost_rank,
  (CASE start_tie.cost_rank WHEN start_tie.next_app_cost THEN cost_rank+1 ELSE cost_rank END) AS right_cost_rank
  FROM 
  applications;

我的预期结果是

 id  name    cost  cost_rank
 10  voapp    56     2
 9   nfhfj    56     2
 8   nvopq    29     3
 7   vfhgg    28     4
 6   mfjjf    25     5
 5   laookd   25     5
 4   djskck   22     6
 2   oopdld   20     7
 3   Wedass   14     8
 1   nfhfjs   10     9

请修改查询以获得结果

---

你所需要做的就是检查最高成本是否与第二高成本相同。如果是这种情况，则在所有秩值中添加1：

with start_tie as (
  select case 
           when cost = lead(cost) over (order by cost desc) then 1 
           else 0
         end as tie_offset
  from applications
  order by cost desc
  limit 1
)
select *, 
       dense_rank() over (order by cost desc) + (select tie_offset from start_tie) cost_rank
from applications;

例如：

如果领带数量定义了用于新排名的偏移量，则可以使用以下公式计算偏移量：

with start_tie as (
  select count(*) - 1 as tie_offset
  from applications a1
  where cost = (select max(cost) from applications)
)
select *, 
       dense_rank() over(order by cost desc) + (select tie_offset from start_tie) cost_rank
from applications;

---

一开始没有平局，意味着不止一个排名1

如果固定从2级开始，则将r.cost_rank+x.c-1替换为r.cost_rank+1，无论有多少处于并列状态

WITH r AS (
    SELECT
       *
      ,DENSE_RANK() OVER(ORDER BY cost DESC) cost_rank
    FROM
      applications
), x as (select count(*) as c from r where cost_rank=1)
SELECT
  r.*, (CASE WHEN 1<x.c THEN r.cost_rank+x.c-1 ELSE r.cost_rank END) as fixed
FROM
  r,x;

---

如果10,10,10,20，。。。不应该从3开始排名吗？@cske：根据问题，如果排名1的位置出现平局，排名将从2开始——但你是对的，这有点不公平ambiguous@cske它用于投标服务，我们不在第一个位置打成平局，在其他位置打成平局是可以的。如果你有三个值作为最高成本呢？