Question

因此，我在这里想要做的是获取有关不同用户发送了多少电子邮件（使用MailChimp之类的应用程序）的报告，但是我希望在一个查询中使用两种不同的指标。我想知道每个用户发送了多少封电子邮件。这意味着如果他们分别向100个联系人发送3封电子邮件，则显示300。但是我还想知道发送了多少个唯一电子邮件，即显示3。

我想得到类似以下内容的东西：

-------------------------------------------------------------
| Full Name   | Username        | Total Sent | Unique Mails |
|-------------|-----------------|------------|--------------|
| John Doe    | jdoe@mail.com   | 12000      | 4            |
| James Smith | jsmith@mail.com | 6000       | 12           |
| Jane Jones  | jjones@mail.com | 4000       | 2            |
| ...         | ...             | ...        | ...          |
-------------------------------------------------------------

所以我知道John向很多联系人发送了几封电子邮件，而James向更少联系人发送了更多电子邮件。

这是我的查询的样子。我已经更改了表名和列名，但这是它的确切表示。

SELECT
    CONCAT(Usernames.FirstName, ' ', Usernames.LastName) AS 'Full Name',
    Usernames.Username,
    COUNT(Sent_Mail_Contacts.IDContact) AS `Total Sent`,
    COUNT(Mass_Mail.IDMass_Mail) AS `Individual E-Mails`
FROM Usernames
LEFT JOIN Sent_Mail_Contacts ON Usernames.Username = Sent_Mail_Contacts.Username
LEFT JOIN Mass_Mail ON Usernames.Username = Mass_Mail.Username
GROUP BY Usernames.Username
ORDER BY `Total Sent`

我有一个包含用户名的表，一个具有联系方式的个人表和一个具有唯一性电子邮件的表。

那么我的查询有意义吗？这有可能吗？因为现在当我运行它时，它给了我这样的东西：

-------------------------------------------------------------
| Full Name   | Username        | Total Sent | Unique Mails |
|-------------|-----------------|------------|--------------|
| John Doe    | jdoe@mail.com   | 12000      | 12000        |
| James Smith | jsmith@mail.com | 6000       | 6000         |
| Jane Jones  | jjones@mail.com | 4000       | 4000         |
| ...         | ...             | ...        | ...          |
-------------------------------------------------------------

我在两栏中都给了我相同的数字，需要7分钟来处理。

以下是三个表格的示例（如果有帮助的话）：

Usernames
------------------------------------------------
| Username        | FirstName | LastName | ... |
|-----------------|-----------|----------|-----|
| jdoe@mail.com   | John      | Doe      | ... |
| jsmith@mail.com | James     | Smith    | ... |
| jjones@mail.com | Jane      | Jones    | ... |
| ...             | ...       | ...      | ... |
------------------------------------------------

Mass_Mail
----------------------------------------------------
| ID_Mass_Mail | Username       | Date       | ... |
|--------------|----------------|------------|-----|
|            1 | jdoe@mail.com  | 2019-01-16 | ... |
|            2 | jdoe@mail.com  | 2019-01-29 | ... |
|            3 | jjones@mail.com| 2019-02-14 | ... |
|          ... | ...            | ...        | ... |
----------------------------------------------------

Sent_Mail_Contacts
---------------------------------------------------------------------
| ID_Mass_Mail | Username       | Contact_ID | Contact_Email  | ... |
|--------------|----------------|------------|----------------|------
|            1 | jdoe@mail.com  |          1 | bob@mail.com   | ... |
|            1 | jdoe@mail.com  |          2 | jim@mail.com   | ... |
|            1 | jdoe@mail.com  |          3 | cindy@mail.com | ... |
|          ... | ...            |        ... | ...            | ... |
|            2 | jdoe@mail.com  |          4 | mike@mail.com  | ... |
|            2 | jdoe@mail.com  |          2 | jim@mail.com   | ... |
|            2 | jdoe@mail.com  |          3 | cindy@mail.com | ... |
|          ... | ...            |        ... | ...            | ... |
---------------------------------------------------------------------

Answer 1

使用COUNT(DISTINCT ...)：

SELECT
    CONCAT(Usernames.FirstName, ' ', Usernames.LastName) AS 'Full Name',
    Usernames.Username,
    COUNT(Sent_Mail_Contacts.IDContact) AS `Total Sent`,
    COUNT(DISTINCT Mass_Mail.IDMass_Mail) AS `Individual E-Mails`
FROM Usernames
LEFT JOIN Sent_Mail_Contacts ON Usernames.Username = Sent_Mail_Contacts.Username
LEFT JOIN Mass_Mail ON Usernames.Username = Mass_Mail.Username
GROUP BY Usernames.Username
ORDER BY `Total Sent`

NB：虽然这样不会使查询更快。首先，您至少应确保在JOIN中使用主键/外键关系：Usernames(Username)，Sent_Mail_Contacts(Username)，Mass_Mail(Username)

Answer 2

假设IDMass_Mail中的值表示唯一的电子邮件，则只需编辑最后一个COUNT即可使用DISTINCT关键字。

COUNT(DISTINCT Mass_Mail.IDMass_Mail) AS `Individual E-Mails`

这将返回按Username分组的唯一值的数量。

如果您能够向Username和Sent_Mail_Contacts表中的Mass_Mail列添加索引，则还应该提高性能。

Answer 3

我设法使用一个查询来做到这一点（除了出于隐私考虑而更改了实际的表名和列名）。

SELECT
    Accounts.Account_Name AS `account`,
    Usernames.Username AS `username`,
    COUNT(Mass_Mail_Reached_Contacts.ID_Contact) AS `total_emails`,
    COUNT(Mass_Mail_Reached_Contacts.ID_Mass_Mail) / 
        (
            SELECT COUNT(*)
                FROM 
                    Mass_Mail_Reached_Contacts 
                WHERE
                    Mass_Mail_Reached_Contacts.DATE >= '2019-02-01'
                    AND
                    Mass_Mail_Reached_Contacts.DATE <= '2019-02-28'
        )
    * 100 AS `%`,
    COUNT(DISTINCT Mass_Mail.ID_Mass_Mail) AS `unique_emails`,
    COUNT(Mass_Mail_Reached_Contacts.ID_Mass_Mail) / 
        COUNT(DISTINCT mass_mail.ID_Mass_Mail) 
        AS `avg_contacts_per_email`

FROM
    Usernames

LEFT JOIN Mass_Mail_Reached_Contacts ON Mass_Mail_Reached_Contacts.Username = Usernames.Username
LEFT JOIN Account ON Account.ID_Account = Usernames.ID_Account
LEFT JOIN Mass_Mail ON Mass_Mail.ID_Mass_Mail = Mass_Mail_Reached_Contacts.ID_mass_mail

WHERE
    Mass_Mail_Reached_Contacts.DATE >= '2019-02-01'
    AND
    Mass_Mail_Reached_Contacts.DATE <= '2019-02-28'

GROUP BY
    Usernames.Username

HAVING COUNT(DISTINCT Mass_Mail.IDMass_Mail) > 0

ORDER BY
    `total_emails` DESC

我现在可以得到一个像这样的表

Emails Stats
--------------------------------------------------------------------------------------
| account  | username     | total_emails |     % | unique_emails | avg_contact_email | 
|----------|--------------|--------------|-------|------------------------------------
| Bob inc. | bob@mail.com | 28,550       | 14.52 |            12 |           2379.17 |
| ...      | ...          | ...          |   ... |           ... |               ... |
--------------------------------------------------------------------------------------

Answer 4

开始于：为什么Mass_Mail和Sent_Mail_Contacts都包含Username？这看起来多余。还是Sent_Mail_Contacts.ID_Mass_Mail为空？

至少对于此查询，我想我们可以完全忽略Username中的Sent_Mail_Contacts。真正链接两个表的是ID_Mass_Mail，并且您在查询中忘记了此连接条件。

select
  ws_concat(' ', u.firstname, u.lastname) as full_name,
  u.username,
  count(smc.idmass_mail) as total_sent,
  count(mm.idmass_mail) as individual_e_mails
from usernames u
left join mass_mail mm on mm.username = u.username
left join sent_mail_contacts smc on smc.id_mass_mail = u.id_mass_mail
group by u.username
order by total_sent;

是否可以在3个表的SQL查询中使用两个COUNT和两个JOIN？

4 个答案: