Question

我正在写一个SQL查询，但我陷入了困境，无法弄清楚如何解决这个问题。首先看看下面的查询：

SELECT user_id, COUNT(*) as count 
FROM hwc_attend 
WHERE at_id IN 
    (SELECT evdet_id
     FROM eve_detail
     WHERE evdet_id IN (SELECT at_id FROM hwc_attend WHERE attendstate=1 )
       AND location <> ''
       AND evdet_id > 999
       AND location NOT IN (SELECT ASIN FROM pReviews )
    )
GROUP BY user_id

此查询工作正常，但结果不是必需的，因为部分SELECT ASIN FROM pReviews应该像SELECT ASIN FROM pReviews where cID={place current value of "location" field from table eve_detail here。

为了更好地理解，这里是错误的查询：

SELECT user_id, COUNT(*) as count 
FROM hwc_attend 
WHERE at_id IN 
    (SELECT evdet_id **, location**
     FROM eve_detail
     WHERE evdet_id IN (SELECT at_id FROM hwc_attend WHERE attendstate=1)
       AND location <> ''
       AND evdet_id > 999
       AND location NOT IN (SELECT ASIN FROM pReviews where cID=**location**)
    )
GROUP BY user_id

很难解释..简而言之，我必须从表“eve_detail”中提取的结果中删除“location”值，该表也存在于列cID中的表“pReviews”中。

此外，如果有人可以将其转换为联接，那将是很好的。我需要两个查询来学习。

Answer 1

将其翻译为联接会使用类似的东西。使用LEFT OUTER JOIN并检查NULL而不是NOT IN。我假设hwc_attend有一个名为id的唯一列，它在计数中用于获取不同的行。

SELECT ha1.user_id, COUNT(DISTINCT ha1.id) as count 
FROM hwc_attend ha1
INNER JOIN eve_detail ed ON ha1.at_id = ed.evdet_id
INNER JOIN hwc_attend ha2 ON ed.evdet_id = ha2.at_id 
LEFT OUTER JOIN pReviews pr ON ed.location = pr.ASIN AND cID = **location**
WHERE ha2.attendstate = 1
AND ed.location <> ''
AND ed.evdet_id > 999
AND pr.ASIN IS NULL
GROUP BY ha1.user_id

Answer 2

更改

AND location NOT IN (SELECT ASIN FROM pReviews )

要

AND NOT EXISTS (SELECT ASIN FROM pReviews WHERE eve_detail.location = ASIN.cID )

SQL子查询逻辑混乱

2 个答案: