如果初始平局为 1，则 Postgresql 密集排名从 2 开始

Question

所以我有一个 table 和一个查询，它对项目的成本进行排名，并且不允许与位置 1 有联系，如果在位置 1 有联系，则排名从 2 开始。

这是带有示例数据的架构

  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);

这里是查询

  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。

SQL FIDDLE

Answer 1

您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，请将所有等级值加 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;

示例：http://rextester.com/EKSLJK65530

如果并列数定义了用于 "new" 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;

Answer 2

一开始没有平局，表示不止一个排在第 1 位

将 r.cost_rank+x.c-1 替换为 r.cost_rank+1 如果固定从 2 级开始到不管有多少人处于平局排名

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;

如果初始平局为 1，则 Postgresql 密集排名从 2 开始

Postgresql dense ranking to start at 2 if there is an initial tie at 1

postgresql

postgresql-9.3