Question

LEFT JOIN
(
    SELECT user_id, review, COUNT(user_id) totalCount
    FROM reviews
    GROUP BY user_id
) b ON b.user_id= b.user_id

我试图让WHERE LENGTH(review) > 100适合这个地方，但每次我说它都会给我带来麻烦。

上面的子查询按user_id计算所有总评论。我只想添加一个资格。只计算超过100长度的评论。

另一方面，我看过函数CHAR_LENGTH - 不确定我是否需要。

修改

以下是完整的查询，可以按照我的需要完美地运行：

    static public $top_users = "
            SELECT  u.username, u.score,
            (COALESCE(a.totalCount, 0) * 4) +
            (COALESCE(b.totalCount, 0) * 5) +
            (COALESCE(c.totalCount, 0) * 1)  +
            (COALESCE(d.totalCount, 0) * 2)  +
            (COALESCE(u.friend_points, 0)) AS totalScore
            FROM users u 
            LEFT JOIN
                (
                    SELECT user_id, COUNT(user_id) totalCount
                    FROM items 
                    GROUP BY user_id
                ) a ON a.user_id= u.user_id
            LEFT JOIN
                (
                    SELECT user_id, COUNT(user_id) totalCount
                    FROM reviews
                    GROUP BY user_id
                ) b ON b.user_id= u.user_id
            LEFT JOIN
                (
                    SELECT user_id, COUNT(user_id) totalCount
                    FROM ratings
                    GROUP BY user_id
                ) c ON c.user_id = u.user_id
             LEFT JOIN
                (
                    SELECT user_id, COUNT(user_id) totalCount
                    FROM comments
                    GROUP BY user_id
                ) d ON d.user_id = u.user_id

                ORDER BY totalScore DESC LIMIT 25;";

Answer 1

LENGTH()返回以字节为单位的字符串长度。你可能想要CHAR_LENGTH()因为它会给你实际的字符。

SELECT user_id, review, COUNT(user_id) totalCount
FROM reviews
WHERE CHAR_LENGTH(review) > 100
GROUP BY user_id, review

您还没有正确使用GROUP BY。

See the documentation

Answer 2

您想要的查询是：

LEFT JOIN
                (
                    SELECT user_id, COUNT(user_id) totalCount,
                           sum(case when length(review) > 100 then 1 else 0 end
                              ) as NumLongReviews
                    FROM reviews
                    GROUP BY user_id
                ) b ON b.user_id= b.user_id

这既包括评论也包括“长期”评论。该计数使用嵌套在case函数中的sum()语句完成。

子查询计数MySQL中的字符串

2 个答案: