Question

让我们说有两个表：must_have_products和products_buyed 我想为每个客户展示他们已经拥有的产品以col A（GROUP_CONCAT）购买，他们仍然需要购买哪些产品col B（GROUP_CONCAT）..

Table A(must_have_products)
|id_a| name  |
|1a  | TV    |
|2a  | House |
|3a  | Car   |

Table B(People)
|id_b| name  |
|1b  | Mr. A |
|2b  | Mr. B |

Table C
|id_c|id_b|id_a|
|1   |1b  |1a  |

预期结果：

|id_b  | buyed  | left       |
|1b    | 1a     | 2a, 3a     |
|2b    |        | 1a, 2a, 3a |

PS：对不起，如果标题错了，我只是不知道该说什么

Answer 1

这是一种方法。使用group_concat购买的物品很容易，但不能直接购买。

select
t2.id_b,
group_concat(t1.id_a) as buyed,
case 
when group_concat(t1.id_a) is null 
then (select group_concat(id_a order by id_a) from tableA) 
else substring_index((select group_concat(id_a order by id_a) from tableA),
concat(group_concat(t1.id_a order by t1.id_a),','),-1)
end as `left`
from tableB t2
left join tableC t3 on t3.id_b = t2.id_b
left join tableA t1 on t1.id_a = t3.id_a
group by t2.id_b

<强> DEMO

Answer 2

left是关键字，您无法使用它mysql-left-function

这是解决方案：

select B.id_b, COALESCE(R1.buyer, '') buyer, COALESCE(R2.lft, '') lft
from
B, 
(select B.id_b , GROUP_CONCAT(DISTINCT A2.id_a SEPARATOR ', ') as   buyer
from B
left join A A2
on Exists(select * from C where A2.id_a = C.id_a and C.id_b = B.id_b)
group by B.id_b
) R1,
(
select B.id_b , GROUP_CONCAT(DISTINCT A2.id_a SEPARATOR ', ') as  `left` 
from B 
left join A A2 on  
NOT Exists(select * from C where A2.id_a = C.id_a and C.id_b = B.id_b)
group by B.id_b
) R2
where
B.id_b = R1.id_b and B.id_b = R2.id_b

DEMO

MySQL - 显示两列，其中表C中的第一个col值和第二个col值不在表C中

2 个答案: