Question

我有一个名为causality的表，它有两个数据类型（varchar，varchar）列（eventA，eventB），如下所述

  ----------------------------------------------
   eventA                |  eventB     
  ------------------------------------------------
  Wait - User            |  Resolved
  Wait                   |  Resolved
  Closed                 |  In Progress
  Wait - Vendor          |  Resolved
  Wait - Customer        |  Resolved
  Wait - Implementation  |  Assigned
  Wait - Implementation  |  Resolved
  Closed                 |  Awaiting Assignment
  Wait                   |  Assigned
  Wait - Customer        |  Awaiting Assignment
  Resolved               |  Unmatched
  Unmatched              |  In Progress
  Wait - Vendor          |  Assigned
  Wait - Customer        |  Assigned
  Wait - Implementation  |  Wait - Customer
  In Progress            |  Cancelled

  -----------------------------------------------

现在我想要eventB的所有子集，特别是eventA，就像Wait一样 - eventA列的实现有3个唯一值，因此将生成8个子集{{Assigned}，{Resolved}，{Wait - Customer}，{Assigned ，已解决}，{已分配，等待 - 客户}，{已解决，等待 - 客户}，{已分配，已解决，等待 - 客户}，{}}

现在类似于eventA中的所有值，例如在eventA列中的Closed将有4个子集，Wait in eventA列将有4个子集，WaitA中的WaitA供应商将有4个子集，eventA列中的Wait -Customer将有8个子集，事件进展中eventA列将有2个子集，已解决的eventA列将有2个子集，eventA列中的Unmatched将有2个子集

因此输出表将有2列与因果关系表相似eventA将包含所有单值，eventB将包含evnetA中的值子集。

Answer 1

让我假设您有一个数字表，存储从0到足够大的数字的整数。然后，你可以用＆＃34; bit-fiddling＆＃34;：

来做到这一点

select idc.id, n.n, group_concat(nc.team)
from numbers n cross join
     (select id, count(*) as cnt
      from not_connected
      group by id
     ) idc
     on n.n < idc.cnt left join
     (select id, team,
             (@rn := if(@id = id, @rn + 1,
                        if(@id := id, 1, 1)
                       )
             ) as rn
      from not_connected cross join
           (select @id := 0, @rn := 0) vars
      order by id, team
     ) nc
     on (n & (1 << nc.rn)) > 0
group by idc.id, n.n;

这首先为所有输出生成一行，给定id具有2 ^ n行，其中＆＃34; n＆＃34;是团队的数量。然后，每个位都映射到原始表中的团队名称。最后，group by汇总了给定数字和ID的值。

您需要一个数字表才能完成这项工作。只需为最大数量的团队生成足够大的一个。这是一种方法：

create table numbers as
    select (@rn := @rn + 1) as n
    from not_connected cross join
         (select @rn := 0);

MySQL-列中的所有值子集

1 个答案: