如何获取嵌套的子查询以识别父查询列 [英] How to get a nested sub-query to recognize parent query column
问题描述
我在尝试计算每个给定ID的值总和时遇到问题。我决定使用子查询来执行此操作(通常我会使用联接,但我还会为每个子查询保留一个计数器,以进行剪切--请参阅此问题以获取更多信息)。出于这个问题,假定我有以下 MySQL 查询:
I have a problem where I'm trying to calculate a sum of values per given id. I decided to do this using sub-queries (typically I'd use a join, but I'm also keeping a counter for each sub-query for clipping purposes - see this question for more info). For the sake of this question, assume I have the following MySQL query:
/* 1. */ SELECT
/* 2. */ t1.experiement_id,
/* 3. */ (SELECT sum(x.size)
/* 4. */ FROM (SELECT size, ( @rownum := @rownum + 1 ) AS `rownum`
/* 5. */ FROM data AS t0
/* 6. */ JOIN ( select @rownum := 0 )
/* 7. */ WHERE t0.experiment_id = t1.experiment_id
/* 8. */ ORDER BY size) AS x
/* 9. */ WHERE x.rownum <= t2.clip_index ) AS `sum`
/* 10. */
/* 11. */ FROM data AS t1
/* 12. */ JOIN data_clip AS t2 USING (experiment_id)
/* 13. */
/* 14. */ GROUP BY t1.experiment_id
问题在行 7
上发生,我试图在子查询中隔离与 experiement_id $ c $匹配的行c>-我收到
t1.experiement_i的错误d
是未知列。只有在嵌套了1个以上级别的查询中才会发生这种情况。正如检查一样,我注意到 t2.clip_index
的处理情况很好。如果我注释掉第7行,则查询返回正确(尽管结果错误)。知道如何让子查询识别在我的情况下要使用的父表的列吗?谢谢。
The problem happens on row 7
, where I'm trying to isolate the rows in the sub-query that matches experiement_id
- I'm getting an error that t1.experiement_id
is an unknown column. This only happens on a query that's nested more than 1 level deep. Just as a check, I've noticed that t2.clip_index
is being processed fine. If I comment out row #7, the query returns fine (albeit with wrong results). Any idea how to make the sub-query recognize the parent table's column to use in my condition? Thanks.
推荐答案
您是否尝试过类似这样的方法来计数行而不是用户定义的变量?
Have you tried something like this for counting rows instead of the user-defined variable ?
(SELECT sum(size) FROM data AS t0
WHERE t0.experiment_id = t1.experiment_id
ORDER BY size HAVING COUNT(*)<=t2.clip_index
) AS `sum`
让我知道是否这项工作有效,这是我们正在研究的有趣问题。
Let me know if this works, it is an interesting issue we are examining here.
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