Question

这是前一个问题的扩展，Order by depending on 2 col values. 我有一个查询，它返回如下输出，

SELECT DISTINCT  a.rev_date ,fruitname,
 fruit_id , primary_fruit_id
FROM fruits a, fruit_lookup s,fruit_reference r 
WHERE a.id = s.id(+) 
 and primary_fruit_id = r.fruit_id(+) 
AND (fruit_id = 24 or fruit_id = 0) 
ORDER BY case when fruit_id = primary_fruit_id then 0 else 1 end,
         fruit_id desc,
         a.rev_date desc

如何处理这个问题，以便以下面的方式获得所需的输出。因此，只要fruit_id为0，那些行就需要进入10行结果集中的第2行和第5行。如果总计结果大约为50，则每页将有10个结果，而在第2和第5位置，我需要获得如下输出。希望这是有道理的。任何想法都表示赞赏。 THX

 NAME       FRUIT_ID    PRIMARY_FRUIT_ID
--------------------------------------
apple       24          24
apple       24          24
apple       24          24   
apple       24          24
orange      24          12   
pear        24           7
kiwi        24           6
melon       24           2  
grape        0          90
banana       0          45
carrot       0          30    
Desired output
NAME    FRUIT_ID    PRIMARY_FRUIT_ID
--------------------------------------
apple       24          24   
grape        0          90 
apple       24          24
apple       24          24 
banana       0          45   
apple       24          24
kiwi        24           6    
orange      24          12   
melon       24           2      
pear        24           7
carrot       0          30

Answer 1

这比以前的问题复杂得多，所以答案也有点复杂。这是我设法做的事情：

with t as (
  select x.*,
      row_number() over (
        partition by fid order by decode(fid , pfid, 1, 2), rd desc) rbr
    from (
      select distinct a.rev_date rd, s.fruitname fn, 
          a.fruit_id fid, primary_fruit_id pfid
        from fruits a left join fruit_lookup s on a.id = s.id
          left join fruit_reference r 
            on primary_fruit_id = r.fruit_id and r.fruit_id in (0, 24) ) x),
ca as (select count(1) cnt from t),
cx as (
  select row_number() over (partition by cwm order by lvl) rn, cwm, lvl 
    from (
      select level lvl, case when mod(level, 10) in (2, 5) then 0 else 24 end cwm 
        from ca connect by level <= cnt*5))
select rd, fn, fid, pfid 
  from t join cx on cx.rn = t.rbr and cx.cwm = t.fid
  order by lvl

您没有提供数据结构和示例行，所以我尝试重现它们以获取原始输入，这里是带有数据和查询的SQLFiddle。

子查询cx根据您的规则生成数字，然后将这些数字分配给您的原始查询最终选择使用这些数字对数据进行排序。一般逻辑是：插入fruit_id = 0的行，在其他行之间编号为2,5,12,15,22 ....

按查询顺序排列特定行位置

1 个答案: