Question

我“想”我看到了问题，但不知道如何解决它

得到了一些很棒的/快速的帮助让我的查询理顺，但现在我收到了这个错误： 'where子句'中的未知列'caet.entity_id'

错误出现在第三个select语句中。我想说＆gt;＆gt;选择 caev .entity_id字段与 caet .entity_id匹配的值。我已经检查过，拼写是正确的，这些字段都在两个表格中。

如果这样做，我还需要另一次加入吗？

select c.*, 
(
select caet.value 
from customer_address_entity_text 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 = 15

下面的查询显示了我尝试过的内部联接，但仍然出现错误：

select c.*, 
(
select caet.value 
from customer_address_entity_text 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`
INNER JOIN customer_address_entity_text CAET ON CAET.id = CAEV.id 
where store_id = 8

Answer 1

CAEV不在外部环境中。所以问题是“CAET.id = CAEV.id”。您应该在外部上下文中连接所有表，并在内部上下文中选择。但内部联接不起作用。使用OUTER join。也许你可以在下面使用：

SELECT     c.*, caet.value AS test, caev.value AS two
FROM         customer_entity AS c INNER JOIN
                  customer_address_entity AS cae ON c.entity_id = cae.parent_id LEFT OUTER JOIN
                  customer_address_entity_text AS caet ON cae.entity_id = caet.value_id AND caet.attribute_id = 23 LEFT OUTER JOIN
                  customer_address_entity_varchar AS caev ON cae.entity_id = caev.entity_id AND caev.attribute_id = 23
WHERE     (store_id = 15)

如何在sql查询中修复此未知列错误

1 个答案: