如何在單個查詢中使用多個rank（）？

Question

Oracle 11g 如何編寫查詢以獲得期望的結果？具體而言，如何加入要求表中的最新記錄？

選擇標準：最近 細節和最新的要求

with 
detail as (
select 'TAXI' code, '4920' numb, 1990 dstart, 'BLUE' taxi_name from dual union all 
select 'TAXI' code, '4920' numb, 2000 dstart, 'PINK' taxi_name from dual union all 
select 'TAXI' code, '4920' numb, 2008 dstart, 'CYAN' taxi_name from dual), 

requirement as (
select 'TAXI' code, '4920' numb, 2 seqno, 2000 rstart, 25,min_age,'CDL',permit from dual union all 
select 'TAXI' code, '4920' numb, 4 seqno, 2000 rstart, 27 min_age, 'CAR' permit from dual union all 
select 'TAXI' code, '4920' numb, 6 seqno, 2000 rstart, 29 min_age, 'HAZ' permit from dual union all 
select 'TAXI' code, '4920' numb, 2 seqno, 2007 rstart, 24 min_age, 'CDL' permit from dual union all 
select 'TAXI' code, '4920' numb, 4 seqno, 2007 rstart, 26 min_age, 'VAN' permit from dual union all 
select 'TAXI' code, '4920' numb, 6 seqno, 2007 rstart, 28 min_age, 'HAZ' permit from dual 

）

select code,numb,taxi_name, dstart, ranking from (
select code, numb, taxi_name,dstart, rank() over (partition by code, numb 
              order by dstart desc) ranking 
from detail) 
where ranking =1 

息率

CODE NUMB NAME YEAR  
---- ---- ----- ------- 
TAXI 4920 CYAN 2008 


DESIRED OUTPUT 
-------------------------- 
code numb name  dstart  sqno rstart min_age permit 
TAXI 4920 CYAN  2008  2   2007 24  CDL 
TAXI 4920 CYAN  2008  4   2007 26  VAN 
TAXI 4920 CYAN  2008  6   2007 28  HAZ 

Answer 1

好了，你基本上是有這個想法。你只需將其應用到兩個表：

select code,numb,taxi_name, dstart, ranking 
from (select code, numb, taxi_name,dstart, 
      rank() over (partition by code, numb order by dstart desc) as ranking 
     from detail 
    ) d join 
    (select r.*, 
      rank() over (partition by code, numb order by rstart desc) as ranking 
     from requirement 
    ) r 
    on d.code = r.code and 
     d.ranking = 1 and 
     r.ranking = 1 

Answer 2

完整的查詢可以是這樣的：

with detail as 
(
    select 'TAXI' code, '4920' numb, 1990 dstart, 'BLUE' taxi_name from dual union all 
    select 'TAXI' code, '4920' numb, 2000 dstart, 'PINK' taxi_name from dual union all 
    select 'TAXI' code, '4920' numb, 2008 dstart, 'CYAN' taxi_name from dual 
), 
requirement as 
(
    select 'TAXI' code, '4920' numb, 2 seqno, 2000 rstart, 25 min_age,'CDL' permit from dual union all 
    select 'TAXI' code, '4920' numb, 4 seqno, 2000 rstart, 27 min_age, 'CAR' permit from dual union all 
    select 'TAXI' code, '4920' numb, 6 seqno, 2000 rstart, 29 min_age, 'HAZ' permit from dual union all 
    select 'TAXI' code, '4920' numb, 2 seqno, 2007 rstart, 24 min_age, 'CDL' permit from dual union all 
    select 'TAXI' code, '4920' numb, 4 seqno, 2007 rstart, 26 min_age, 'VAN' permit from dual union all 
    select 'TAXI' code, '4920' numb, 6 seqno, 2007 rstart, 28 min_age, 'HAZ' permit from dual 
) 
select code,numb,taxi_name, dstart, seqno, rstart, min_age, permit 
from 
(
    select d.code, d.numb, d.taxi_name, d.dstart, rank() over (partition by d.code, d.numb order by d.dstart desc) ranking, 
     seqno, rstart, min_age, permit 
    from detail d 
    inner join 
    (
     select code, numb, seqno, rstart, min_age, permit, 
      rank() over (partition by code, numb order by rstart desc) ranking 
     from requirement 
    ) r 
     on d.code = r.code 
     and d.numb = r.numb 
     and r.ranking = 1 
) 
where ranking =1 

見SQL Fiddle with Demo

結果是：

| CODE | NUMB | TAXI_NAME | DSTART | SEQNO | RSTART | MIN_AGE | PERMIT | 
------------------------------------------------------------------------ 
| TAXI | 4920 |  CYAN | 2008 |  2 | 2007 |  24 | CDL | 
| TAXI | 4920 |  CYAN | 2008 |  4 | 2007 |  26 | VAN | 
| TAXI | 4920 |  CYAN | 2008 |  6 | 2007 |  28 | HAZ |