如何在 MySQL 2 中连接多个表？答案

【问题标题】：How to join multiple tables in MySQL 2?如何在 MySQL 2 中连接多个表？
【发布时间】：2015-06-19 11:16:58
【问题描述】：

我想看一个公民的全名

reportBasket,reportNoBasket 表中的 notifyEmployee ='1'

并且 reportNoBasket.groupEmail 不为空

并计算来自reportBasket 的电子邮件，citizenEmail reportNOBasket 在一个计数器中

表格：

公民（电子邮件，名字，姓氏）

reportBasket(电子邮件,notifyEmployee)

reportNoBasket(citizenEmail, notifyEmployee , groupEmail)

之前我代表我的查询以这种方式显示他们的电子邮件，效果很好：

SELECT email,count(email) as email_count 
FROM (
    SELECT email
    FROM reportBasket
    WHERE notifyEmployee='1'
    UNION ALL 
    SELECT citizenEmail
    FROM reportNoBasket
    WHERE notifyEmployee='1'
) as T
GROUP BY email
ORDER BY email ASC

但现在我想显示公民的全名而不是电子邮件地址，例如concat(citizen.firstname,' ' citizen.lastname)

如何加入？

【问题讨论】：

标签： mysql database join

【解决方案1】：

试试这样的：

SELECT concat(firstname, ' ', lastname), count(*) as citizen_count 
FROM (
    SELECT c.firstname, c.lastname
    FROM reportBasket rb
    JOIN citizen c
        ON rb.email = c.email
    WHERE notifyEmployee='1'
    UNION ALL 
    SELECT c.firstname, c.lastname
    FROM reportNoBasket rnb
    JOIN citizen c
        ON rnb.citizenEmail = c.email
    WHERE notifyEmployee='1'
) as T
GROUP BY firstname, lastname
ORDER BY firstname, lastname ASC

【讨论】：

只需将其添加到所有 3 个 SELECT 语句和 GROUP BY

【解决方案2】：

我认为您仍想通过电子邮件对项目进行分组（因为它们是唯一的），然后只需加入公民表以添加名称

SELECT email,count(email) as email_count, concat(citizen.firstname,' ' citizen.lastname)
FROM ( SELECT email
       FROM reportBasket
       WHERE notifyEmployee='1'
       UNION ALL 
       SELECT citizenEmail
       FROM reportNoBasket
       WHERE notifyEmployee='1')as T
LEFT JOIN citizen USING (email)
GROUP BY email
ORDER BY email ASC

【讨论】：

【解决方案3】：

试试这个：

SELECT count(email) as email_count, concat(citizen.firstname,' ' citizen.lastname) as name
FROM ( SELECT email
       FROM reportBasket
       WHERE notifyEmployee='1'
       UNION ALL 
       SELECT citizenEmail
       FROM reportNoBasket
       WHERE notifyEmployee='1')as T
LEFT JOIN citizen USING (email)
GROUP BY name
ORDER BY name ASC

【讨论】：