Question

我有3行数据： -

Score | Grp
1     | A
10    | B
100   | C

我想要所有组合，就像这样： -

Score | Grp
1     | A
10    | B
100   | C
11    | A/B
101   | A/C
110   | B/C
111   | A/B/C

（将来可能会有超过3组。）

在Teradata上，这段代码让我有很多方法（rn是行号，每行的连续整数）： -

with recursive cte(ev_grp, score, rn, new_grp, new_score) as
 ( select l.ev_grp, l.score, l.rn, ev_grp as new_grp, score as new_score 
   from in_tab as l

   union all

   select inn.ev_grp, inn.score, inn.rn, 
          inn.ev_grp||'/'||cte.ev_grp as new_grp, (inn.score + cte.score) as new_score
   from in_tab as inn  
   join cte
   on inn.rn > cte.rn
 )    
select f.*
from cte f 
order by f.new_score

然而，最后一行 - 所有3组 - 结果都是错误的，我无法思考如何解决它。任何帮助都会很棒。

Answer 1

这不是一个完整的笛卡儿吗？我这样做非常小心，但是：

select
*
from
(select distinct score from <yourtable>) scores
cross join (select distinct grp from <yourtable>) grps
order by grp,score

我无法想象为什么有人会想要这样做，而且我觉得这个解决方案有点脏。

Answer 2

您只需使用new_grp＆amp; new_score代替ev_grp＆amp; score。请注意new_grp不会被截断：

with recursive cte(ev_grp, score, rn, new_grp, new_score) as
 ( select l.ev_grp, l.score, l.rn, 
      CAST(ev_grp AS VARCHAR(1000)) as new_grp, -- must be large enough to store all concatenated groups 
      score as new_score 
   from in_tab as l

   union all

   select inn.ev_grp, inn.score, inn.rn, 
          inn.ev_grp||'/'||cte.new_grp as new_grp, (inn.score + cte.new_score) as new_score
   from in_tab as inn  
   join cte
   on inn.rn > cte.rn
 )    
select f.*
from cte f 
order by f.new_score

将所有SQL行排列加在一起，包括高阶排列

2 个答案: