【问题标题】:How do I use rank() with an inner join?如何将 rank() 与内部连接一起使用?
【发布时间】:2012-04-03 20:35:32
【问题描述】:

我想将我的 CHEESE 表与 FRESHNESS 连接起来以获取 CHEESE 和 FRESHNESS 代码 每个奶酪 ID 的 max(seq_no) 为 MOLD 的奶酪。

使用 rank() 时,我在哪里加入 FRESHNESS?

CHEESE                                    FRESHNESS
CHEESE_ID  SEQ_NO  FRESH_CODE             FRESH_CODE   FRESH_DESC      
=================================         ========================
  1         1        MOLD                 MOLD         MOLDY CHEESE    
  1        23        FRSH                 FRSH         EDIBLE
  1        34        FRSH
  2         2        FRSH
  2        18        MOLD
  3         3        MOLD
  3         5        MOLD
  3         7        MOLD


 DESIRED RESULT
 ==========================
 CHEESE_ID  SEQ_NO  FRESH_CODE  FRESH_DESC      SEQ_RANK
 2           18     MOLD        MOLDY CHEESE    1
 3            7     MOLD        MOLDY CHEESE    1

这是我用来获取所需序列号的代码。

select 
       cheese_id,seq_no,fresh_code,seq_rank
  from ( select 
         cheese_id,seq_no, fresh_code, 
         rank() over (partition by cheese_id
                          order by seq_no desc) seq_rank
from cheese
where seq_rank = 1
 and  fresh_code = 'MOLD'

【问题讨论】:

    标签: oracle11g inner-join oracle-analytics


    【解决方案1】:

    您可以在子查询中进行连接

    select cheese_id,seq_no,fresh_code,fresh_desc,seq_rank
      from ( select cheese_id,
                    seq_no, 
                    fresh_code, 
                    fresh_desc,
                    rank() over (partition by cheese_id
                                     order by seq_no desc) seq_rank
              from cheese
                   join freshness using (fresh_code) )
     where seq_rank = 1
       and fresh_code = 'MOLD'
    

    或者你可以加入你的子查询

    select cheese_id,seq_no,fresh_code,fresh_desc,seq_rank
      from ( select cheese_id,
                    seq_no, 
                    fresh_code, 
                    fresh_desc,
                    rank() over (partition by cheese_id
                                     order by seq_no desc) seq_rank
              from cheese ) cheese_outer
           join freshness using (fresh_code)
     where seq_rank = 1
       and fresh_code = 'MOLD'
    

    【讨论】:

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