在 SQLite 和 Count 中连接两个表答案

【问题标题】：Join two tables in SQLite and Count在 SQLite 和 Count 中连接两个表
【发布时间】：2014-12-08 01:17:02
【问题描述】：

我有两个名为“likes”和“cmets”的表，我想要一个表，其中包含每个特定用户的喜欢和 cmets 计数，我在 SQLite 中编写了以下查询，但结果并非对所有用户都是正确的，计数两个表中用户的值都是喜欢数和 cmet 数的倍数。

SELECT 
  likes.liker_name, likes.liker_id, likes.profile_picture , 
  COUNT(comments.commenter_name) AS comment_count, COUNT( likes.liker_id) AS like_count
FROM likes  
LEFT JOIN comments 
  ON likes.liker_name = comments.commenter_name 
GROUP BY 
  likes.liker_name 
ORDER BY 
  COUNT( likes.liker_id) DESC

如何获得两个表中都存在的用户的正确计数值？

【问题讨论】：

标签： sqlite join count

【解决方案1】：

问题是：有些用户有 cmets 但没有喜欢，有些用户有喜欢但没有 cmets，有些人两者都有，有些人没有。因此，我建议使用联合查询并再次对其求和

SELECT
    u.name, u.id, u.profile_picture,
    SUM(u.like_count) AS like_count, SUM(u.comment_count) AS comment_count
FROM (
    SELECT
        liker_name AS name, liker_id AS id, profile_picture,
        COUNT(*) AS like_count, 0 AS comment_count
    FROM
        likes
    GROUP BY
        liker_name, liker_id, profile_picture
    UNION ALL
    SELECT
        commenter_name AS name, commenter_id AS id, profile_picture,
        0 AS like_count, COUNT(*) AS comment_count
    FROM
        comments
    GROUP BY
        commenter_name, commenter_id, profile_picture
) AS u
GROUP BY
    u.name, u.id, u.profile_picture

如果您有一个单独的用户表，您还可以将喜欢计数和 cmets 计数子查询加入用户表

SELECT
    u.name, u.id, u.profile_picture, l.cnt AS like_count, c.cnt AS comment_count
FROM
    users u
    LEFT JOIN
        (SELECT liker_id, COUNT(*) AS cnt
        FROM likes
        GROUP BY liker_id
        ) AS l
        ON u.user_id = l.liker_id
    LEFT JOIN
        (SELECT commenter_id, COUNT(*) AS cnt
        FROM comments
        GROUP BY commenter_id
        ) AS c
        ON u.user_id = c.commenter_id
WHERE l.cnt > 0 OR c.cnt > 0

不管你怎么做，你必须在单独的子查询中计算 cmets 和 likes。如果您在加入后进行计数，您正在对可能重复记录的结果（左侧的记录）求和，并且您得到错误的计数。

【讨论】：