我正在使用这个Mysql查询,它有效:
SELECT
nom
,prenom
,(SELECT GROUP_CONCAT(category_en) FROM
(SELECT DISTINCT category_en FROM categories c WHERE id IN
(SELECT DISTINCT category_id FROM m3allems_to_categories m2c WHERE m3allem_id = 37)
) cS
) categories
,(SELECT GROUP_CONCAT(area_en) FROM
(SELECT DISTINCT area_en FROM areas c WHERE id IN
(SELECT DISTINCT area_id FROM m3allems_to_areas m2a WHERE m3allem_id = 37)
) aSq
) areas
FROM m3allems m
WHERE m.id = 37
结果是:
nom prenom categories areas
Man Multi Carpentry,Paint,Walls Beirut,Baalbak,Saida
它工作正确,但只有当我将查询硬编码到我想要的ID时(37)。 我想让它适用于m3allem表中的所有条目,所以我试试这个:
SELECT
nom
,prenom
,(SELECT GROUP_CONCAT(category_en) FROM
(SELECT DISTINCT category_en FROM categories c WHERE id IN
(SELECT DISTINCT category_id FROM m3allems_to_categories m2c WHERE m3allem_id = m.id)
) cS
) categories
,(SELECT GROUP_CONCAT(area_en) FROM
(SELECT DISTINCT area_en FROM areas c WHERE id IN
(SELECT DISTINCT area_id FROM m3allems_to_areas m2a WHERE m3allem_id = m.id)
) aSq
) areas
FROM m3allems m
我收到错误:
'where'中的未知列'm.id' 条款“
为什么呢? 从MySql手册:
13.2.8.7. Correlated Subqueries
[...]
Scoping rule: MySQL evaluates from inside to outside.
所以......当子查询在SELECT部分时,这不起作用吗?我没有读到任何关于此的内容。
有谁知道吗?我该怎么办?我花了很长时间来构建这个查询...我知道它是一个怪物查询,但它在单个查询中得到了我想要的东西,而且我非常接近它的工作!
有人可以帮忙吗?
答案 0 :(得分:6)
您只能将一个级别关联起来。
使用:
SELECT m.nom,
m.prenom,
x.categories,
y.areas
FROM m3allens m
LEFT JOIN (SELECT m2c.m3allem_id,
GROUP_CONCAT(DISTINCT c.category_en) AS categories
FROM CATEGORIES c
JOIN m3allems_to_categories m2c ON m2c.category_id = c.id
GROUP BY m2c.m3allem_id) x ON x.m3allem_id = m.id
LEFT JOIN (SELECT m2a.m3allem_id,
GROUP_CONCAT(DISTINCT a.area_en) AS areas
FROM AREAS a
JOIN m3allems_to_areas m2a ON m2a.area_id = a.id
GROUP BY m2a.m3allem_id) y ON y.m3allem_id = m.id
WHERE m.id = ?
答案 1 :(得分:1)
错误的原因是未定义子查询m
。它稍后在外部查询中定义。