我想我正在解决这个问题,但是遇到了另一个障碍,并且不知道如何解决它
此查询有什么问题- 只需获取通用的“您有一个错误”,它指向联接,但我不知道如何解决它。我想要对customer_address_entity进行联接,因为它具有我需要在其中一条select语句中使用的另一个唯一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`;
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 莽aev.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
很明显,您在FROM之前有JOIN。
其次,您有一个带有unicode字符的莽aev。我不知道这是否有意。
可能这将工作:
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”同盟旁边添加了一个“ as”,只是为了使您的代码一致。