指望 mysql GROUP BY

Question

我想使用以下两个（简化的）表格计算每个用户访问我网站的唯一天数：

Table: Users
+--------------+------------------+
| Field        | Type             |
+--------------+------------------+
| userid       | int(10)          |
| firstname    | varchar(50)      |
| lastname     | varchar(50)      |
+--------------+------------------+

Table: Visitors
+--------------+------------------+
| Field        | Type             |
+--------------+------------------+
| visitorid    | int(10)          |
| userid       | int(10)          |
| visitdate    | timestamp        |
+--------------+------------------+

首先，我按 DATE(visitdate) 分组以按日期对记录进行分组。然后我按userid分组，统计每个用户访问的天数。

我对此的查询是：

SELECT count(userid) as count, userid, firstname, lastname FROM 
    (SELECT users.userid, firstname, lastname, DATE(`visitdate`) AS dateonly FROM 
    visitors JOIN users ON visitors.userid = users.userid 
    GROUP BY DATE(`visitdate`)) b 
GROUP BY userid ORDER BY count DESC

然而，这给出了一个不正确的计数，当我简单地添加一个 WHERE 子句来限制用户标识时，这一点得到了验证。以下查询给出了单个用户的正确计数。

SELECT count(userid) as count, userid, firstname, lastname FROM 
    (SELECT users.userid, firstname, lastname, DATE(`visitdate`) AS dateonly FROM 
    visitors JOIN users ON visitors.userid = users.userid 
    WHERE users.userid = 167
    GROUP BY DATE(`visitdate`)) b 
GROUP BY userid ORDER BY count DESC

显然我弄错了 - 任何帮助将不胜感激。谢谢。

Answer 1

如果我理解你的问题，我想这就是你想要的：

SELECT u.userid, u.firstname, u.lastname, COUNT(DISTINCT CAST(v.visitdate AS DATE)) AS visits
FROM users u
LEFT JOIN visitors v ON v.userid = u.userid 
GROUP BY u.userid, u.firstname, u.lastname

这将向您显示每个用户的用户 ID、名字、姓氏和不同访问日期的计数（如果没有访问，则为 0，因为它使用左连接将缺少项目的用户包括在访问者中 table).

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