Question

CHART

查询后，我需要该结果：

name   type   shape
---------------------
name1  type1  shape1
name1  type2  shape1
name1  type3  shape2

我只能思考：

name    shape1Types     shape2Types
-----------------------------------
name1   type1, type2    type3

但是这里说shape1Types是未知列

Answer 1

尝试以下情况，如下尝试

   select name, group_concat(case when shape='shape1' then type end) as shape1Types, group_concat(case when shape='shape2' then type end) as shape2Types 
    from table 
    group by name

Answer 2

在该WHERE子句中别名名称shape1Types和shape2Types未知。
因此，错误。

您可以IN代替其他两个表中的唯一类型，而不必使用JOIN。

select 
 t.name, 
 group_concat(shape1.type) as shape1Types, 
 group_concat(shape2.type) as shape2Types 
from table1 t
left join (select type from table2 where shape = 'shape1' group by type) shape1 
  on shape1.type = t.type
left join (select type from table3 where shape = 'shape2' group by type) shape2 
  on shape2.type = t.type
where t.name = 'name1'
-- and (shape1.type is not null or shape2.type is not null)
group by t.name

组连接不同行

2 个答案: