如何在SET子查询中引用外表？

Question

说我有这些表：

create table letter (id_letter bigint, id_group_table bigint, letter char(1));  
create table group_table (id_group_table bigint, id_whatever bigint, champion char(1));
create table whatever (id_whatever bigint);

我想更新group_table，以便我可以在champion列中设置letter表中与{{1}中的每一行相关的字母数最多的字母}}。今天我必须在我的应用程序中迭代group_table中的所有行并对每行运行查询以发现最常用的字母...我想在一次更新中执行此操作，是否可能？

这是我尝试（并且不起作用）的内容：

group_table

MySQL不允许我使用update group_table gt set gt.champion = ( select inner_champ from ( select le.letter as inner_champ, count(*) from letter le where le.id_group_table = gt.id_group_table group by le.letter order by count(*) desc limit 1 ) ) where gt.id_whatever in (1,2,3,4); 引用子查询中的group_table ...是否可以这样做？

谢谢！

Answer 1

您可以在内部查询中单独使用JOIN进行group_table，如下所示

update group_table 
join
        (
            select letter,
            id_group_table, 
            count(distinct id_group_table) as occurences
            from letter
            group by letter
            having max(occurences)
        ) tab on group_table.id_group_table = tab.id_group_table
set champion = tab.letter
where group_table.id_whatever in (1,2,3,4);

Answer 2

您可以尝试使用连接而不是where。尝试类似的东西：

select inner_champ from 
    (
        select le.letter as inner_champ, count(*) from letter le
        left join group_table gt2 on le.id_group_table = gt2.id_group_table
        group by le.letter
        order by count(*) desc
     )

Answer 3

update group_table 
join
  (
        select tab1.letter, tab1.id_group_table from 
        (
          select  letter,
          id_group_table, 
          count(letter) as occurrences
          from letter
          group by id_group_table, letter 
          order by occurrences desc
        ) tab1
        group by tab1.id_group_table having max(tab1.occurrences)    

  ) tab2 on group_table.id_group_table = tab2.id_group_table

set champion = tab2.letter
where group_table.id_whatever in (1,2,3,4);