Question

我认为我正在接近这个，但遇到另一个障碍，不知道如何解决它

这个查询出了什么问题 - 只是得到通用的“你有错误”，它指的是连接，但我不知道如何修复它。我想在customer_address_entity上加入，因为它有一个我需要在其中一个select语句中使用的唯一ID。

    select c.*, 
(
select caet.value 
from customer_address_entity_test caet 
where cae.entity_id = caet.value_id 
and caet.attribute_id = 23
) as test,
(
select caev.value 
from customer_address_entity_varchar caev 
where caet.entity_id = caev.entity_id 
and caev.attribute_id = 23
) as two
from customer_entity c where store_id = 15
join customer_address_entity cae on c.`entity_id` = cae.`parent_id`;

Answer 1

首先，您的代码如下所示：

select  c.*, 
        (
            select  caet.value 
            from    customer_address_entity_test caet 
            where   cae.entity_id = caet.value_id 
                    and caet.attribute_id = 23
        ) as test,
        (
            select  çaev.value 
            from    customer_address_entity_varchar caev 
            where   caet.entity_id = caev.entity_id 
                    and caev.attribute_id = 23
        ) two
join customer_address_entity cae on c.`entity_id` = cae.`parent_id`
from customer_entity c where store_id = 8

显然你在FROM之前有JOIN。

其次，你有一个具有unicode字符的çaev。我不知道这是否有意。

可能会有效：

select  c.*, 
        (
            select  caet.value 
            from    customer_address_entity_test caet 
            where   cae.entity_id = caet.value_id 
                    and caet.attribute_id = 23
        ) as test,
        (
            select  caev.value 
            from    customer_address_entity_varchar caev 
            where   caet.entity_id = caev.entity_id 
                    and caev.attribute_id = 23
        ) as two
from customer_entity c
join customer_address_entity cae on c.`entity_id` = cae.`parent_id`
where store_id = 8

我在“两个”别名旁边添加了“as”，只是为了让您的代码保持一致。

如何修复此查询中的联接

1 个答案: