Question

我想要一个查询，它会为我提供ID与其他项目匹配的所有项目，其中匹配项目组大于X.

假设我有两个表，提交和submission_items。每个提交都有一个submission_id，每个submission_item都有一个外键，即父提交的submission_id。提交可以有多个submission_items。

我希望获得包含超过X submission_items的所有提交内容。

我试过了：

select submissions.*, submission_item.*
from submission
join submission_item ON submissions.submission_id = submission_item.submission_id
group by submission.submission_id
having count(*) > 1

它错误地说它想要GROUP BY功能中的其他字段。我怎样才能更好地做到这一点？

Answer 1

当您指定GROUP BY时，SELECT语句中的字段必须是group子句的一部分或聚合的一部分。

您需要找到您感兴趣的ID，然后将其重新加入表格。

SELECT * FROM 
submission si INNER JOIN
   (select submission_id
      from submission
      join submission_item ON submissions.submission_id = submission_item.submission_id
       group by submission.submission_id
       having count(*) > 1) c  
ON si.submission_id = c.submission_id

或者您可以单独对所有字段进行分组

select submissions.submission_id, submissions.x, submissions.y
from submission
join submission_item ON submissions.submission_id = submission_item.submission_id
group by submission.submission_id, submissions.x, submissions.y
having count(*) > 1

SQL Query用于获取与至少有X个元素的另一个元素匹配的所有元素

1 个答案: