Question

我有这张桌子：

// qanda
+----+----------------------------------------+---------+-----------+------+
| Id |                   body                 | related |  user_id  | free |
+----+----------------------------------------+---------+-----------+------+
| 1  | content of question1                   | null    | 2         | null |
| 2  | content of first answer for question1  | 1       | 2         | null |
| 3  | content of question2                   | null    | 6         | 300  |
| 4  | content of second answer for question1 | 1       | 4         | null |
| 5  | content of first answer for question2  | 3       | 2         | null |
| 6  | content of question3                   | NULL    | 8         | null |
| 7  | content of first answer for question3  | 6       | 4         | null |
| 8  | content of second answer for question3 | 6       | 2         | null |
+----+----------------------------------------+---------+-----------+------+

我有这个问题：

SELECT count(1)
 FROM qanda question
 JOIN qanda answer ON question.Id = answer.related
WHERE answer.related IS NOT NULL
  AND answer.user_id = 2
  AND question.free IS NULL;

如您所知，上面的查询会返回属于免费问题（在本例中为it returns 2）的答案数。好的都好。我也有这张桌子：

// votes
+----+---------+-------+
| id | post_id | value |
+----+---------+-------+
| 1  | 2       |  1    |
| 2  | 3       | -1    |
| 3  | 2       |  1    |
| 4  | 8       | -1    |
| 5  | 1       |  1    |
| 6  | 4       |  1    |
| 7  | 2       | -1    |
| 8  | 8       |  1    |
| 9  | 8       | -1    |
| 10 | 8       | -1    |
+----+---------+-------+

我需要消除总得分为负的答案。因此在这种情况下它应该返回1。因为在qanda.id = 8处，该答案的总得票率为-2，所以我不想算数。我怎么能这样做？

Answer 1

排除总票数小于零的帖子的最简单方法是：

SELECT count(1)
FROM qanda question
JOIN qanda answer ON question.Id = answer.related
WHERE answer.related IS NOT NULL
AND answer.user_id = 2
AND question.free IS NULL
AND question.id not in (
  select post_id
  from votes
  group by post_id
  having sum(value) < 0)

此处的关键部分是having sum(value) < 0选择具有净负投票的帖子。

来自评论......

找到有太多＆＃34;坏＆＃34;答案，你应该回报多少＆＃34; good＆＃34;他们做出的答案，并决定他们是否总体上是一个坏的＆＃34;用户。例如，有5个答案都是坏的用户与1000个答案的用户非常不同，其中只有5个是坏的，即使他们都有5个不良答案。

试试这个：

select
    sum(score < 0) bad,
    count(*) total,
    sum(score < 0) / sum(.01) percent_bad
from (
    SELECT coalesce(sum(value), 0) score
    FROM qanda question
    JOIN qanda answer ON question.Id = answer.related
    LEFT JOIN votes ON votes.post_id = answer.id
    WHERE answer.related IS NOT NULL
    AND answer.user_id = 2
    AND question.free IS NULL
    AND answer.timestamp > subdate(now(), 365)
    GROUP BY answer.id
) scores

关于SQL功夫的一些注释：

，true为1，false为0，因此通过对条件求和，可以计算出它的真实次数。这比其他数据库所需的笨拙SUM(CASE ...)表达式更容易编码，更容易阅读
通过SUM(.01)计算潜水（我只想到BTW）是获得百分比的最简单的方法，因为它不仅简化了表达式，而且还可以自动避免整数运算的回答舍入

_{免责声明：代码可能无法编译或工作，因为它在我的手机上被翻阅（但它有合理的可能性）}

Answer 2

你可以放置另一个AND条件，并在其中有select语句（求和值，其中post id = qanda.id），用于计算总投票得分并将其评估为大于或等于零。

Answer 3

请尝试一次：

SELECT v.value val
FROM qanda q
INNER JOIN qanda q1 ON q1.related = q.id
INNER JOIN votes v ON v.id = q.user_id
WHERE q.free IS NULL 
AND q1.user_id = 2
AND v.value > 0

如何计算投票得分为零或正数的帖子数量？

3 个答案: