如何循环MySql子查询结果？

有4个实体：标记、问题、它们之间的连接表-问题有标记，最后一个是相关标记，除了id之外还有两个字段：主标记和子标记。Bouth是对Tag.id的引用

子查询返回多个结果，我需要循环它，但不知道如何循环

select t.id, t.name, count(question_has_tag.tag_id) as i from tag as t
   join question_has_tag on t.id = question_has_tag.tag_id
   where t.id = 
     (select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1)
   group by t.id order by i desc;

---

无法在SQL中对子查询结果进行循环。。您可以在过程语言中使用循环，例如：在使用触发器时使用服务器端语言

无论如何，在您的例子中，您可以使用联接，而不是与子查询结果进行比较

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
INNER JOIN  (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
) t2 t.id = t2.child_tag 
group by t.id order by i desc;

或者，如果您只需要一个结果，请限制结果

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
where t.id =       (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
     ORDER BY child_tag 
     limit 1
)
group by t.id order by i desc;

或者也使用IN子句代替=

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
where t.id IN      (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
)
group by t.id order by i desc;

---

无法在SQL中对子查询结果进行循环。。您可以在过程语言中使用循环，例如：在使用触发器时使用服务器端语言

无论如何，在您的例子中，您可以使用联接，而不是与子查询结果进行比较

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
INNER JOIN  (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
) t2 t.id = t2.child_tag 
group by t.id order by i desc;

或者，如果您只需要一个结果，请限制结果

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
where t.id =       (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
     ORDER BY child_tag 
     limit 1
)
group by t.id order by i desc;

或者也使用IN子句代替=

select t.id, t.name, count(question_has_tag.tag_id) as i 
from tag as t
join question_has_tag on t.id = question_has_tag.tag_id
where t.id IN      (
    select related_tag.child_tag from related_tag
     join tag as t on related_tag.main_tag = t.id
     where related_tag.main_tag = 1
)
group by t.id order by i desc;

---

不

其中..=（..）

但

其中。。在（..）

中。但是最好的方法是重写这个加入。注：仅限团体成员。欢迎参加SO。请参见：不

WHERE..=（..）

但

其中。。在（..）

中。但是最好的方法是重写这个加入。注：仅限团体成员。欢迎参加SO。请看：注意，没有订单的限制是相当有意义的。没有订单的限制是相当没有意义的