如何在sql中按两列分组？答案

【问题标题】：how to group by two column in sql?如何在sql中按两列分组？
【发布时间】：2019-01-28 13:05:16
【问题描述】：

我有两张表，一张是“events”，第二张是“tickets”：

事件表：

============================
| event_id |   event_name  |
============================
|   101    | Running Event |
|   102    | Cycling Event |
============================

门票表：

==============================================================
| ticket_id | event_id | ticket_number | ticket_generate_date|
==============================================================
|    1      |    101   |    40001      |  2018-08-21 17:05   |
|    2      |    101   |    40002      |  2018-08-21 18:05   |
|    3      |    102   |    40001      |  2018-08-21 19:05   |
|    4      |    101   |    40003      |  2018-08-23 20:05   |
|    5      |    101   |    40004      |  2018-08-24 20:05   |
==============================================================

现在我想运行查询，以便我的输出如下所示：

================================================
| count ticket |   event_name  | day wise data |
================================================
|       2      | Running Event |  2018-08-21   |
|       1      | Cycling Event |  2018-08-21   |
|       1      | Running Event |  2018-08-23   |
|       1      | Running Event |  2018-08-24   |
================================================

我尝试了以下查询：

SELECT COUNT(ticket_id), ticket_generate_date FROM Tickets 
WHERE ticket_generate_date >= DATE_FORMAT(curdate(), '%Y-%m-01') 
GROUP BY DATE_FORMAT(ticket_generate_date, '%d-%b');

【问题讨论】：

到目前为止，@Santosh，您尝试了什么？
你有没有尝试过？有示例代码吗？
@Rafael 我试过这个查询 SELECT COUNT(ticket_id), ticket_generate_date FROM Tickets WHERE ticket_generate_date >= DATE_FORMAT(curdate(), '%Y-%m-01') GROUP BY DATE_FORMAT( ticket_generate_date, '%d-%b')
Santosh，请将此添加到主要任务中，以便每个人都可以阅读。 :)
Using group by on multiple columns的可能重复

标签： php mysql mysqli

【解决方案1】：

您只需将第二组选项添加到您已有的选项中，并用逗号分隔。

我还使用了别名和 INNER JOIN 来获取事件名称。

SELECT 
    COUNT(t.ticket_id),
    e.event_name,
    DATE_FORMAT(t.ticket_generate_date, '%Y-%m-%d') as day_wise
FROM 
    Tickets t
INNER JOIN
    Events e ON e.event_id = t.event_id
WHERE 
    t.ticket_generate_date >= DATE_FORMAT(curdate(), '%Y-%m-01') 
GROUP BY 
    DATE_FORMAT(t.ticket_generate_date, '%Y-%m-%d'), e.event_name

它在这里工作： http://sqlfiddle.com/#!9/b235c/1/0

【讨论】：

【解决方案2】：

如果您在 group by 子句中包含额外的列名 - 在这种情况下，我选择了 event_name，最终输出或多或少符合预期。要输出请求的所有详细信息，尽管您需要连接两个表（或使用嵌套选择）

+------------+-------------+------+-----+---------+-------+
| Field      | Type        | Null | Key | Default | Extra |
+------------+-------------+------+-----+---------+-------+
| event_id   | int(11)     | NO   |     | NULL    |       |
| event_name | varchar(50) | NO   |     | NULL    |       |
+------------+-------------+------+-----+---------+-------+



+----------------------+-----------+------+-----+-------------------+----------------+
| Field                | Type      | Null | Key | Default           | Extra          |
+----------------------+-----------+------+-----+-------------------+----------------+
| id                   | int(11)   | NO   | PRI | NULL              | auto_increment |
| event_id             | int(11)   | NO   |     | 0                 |                |
| ticket_number        | int(11)   | NO   |     | 0                 |                |
| ticket_generate_date | timestamp | NO   |     | CURRENT_TIMESTAMP |                |
+----------------------+-----------+------+-----+-------------------+----------------+



select 
    count( t.id ) as 'sales',
    e.`event_name` as `event`,
    date_format( t.`ticket_generate_date`, '%y-%m-%d' ) as `day wise data`
from `tickets` t
    join `events` e on e.`event_id`=t.`event_id`
where t.`ticket_generate_date` >= date_format( curdate(), '%y-%m-01' )
group by e.event_name, date_format( t.`ticket_generate_date`, '%d-%b' );


+-------+---------------+---------------+
| sales | event         | day wise data |
+-------+---------------+---------------+
|     1 | Cycling event | 18-08-21      |
|     2 | Running Event | 18-08-21      |
|     1 | Running Event | 18-08-23      |
|     1 | Running Event | 18-08-24      |
+-------+---------------+---------------+

或者，为了进一步优化结果，您可以按日期排序

select 
    count( t.id ) as 'sales',
    e.`event_name` as `event`,
    date_format( t.`ticket_generate_date`, '%y-%m-%d' ) as `day wise data`
from `tickets` t
    join `events` e on e.`event_id`=t.`event_id`
where t.`ticket_generate_date` >= date_format( curdate(), '%y-%m-01' )
group by e.event_name, date_format( t.`ticket_generate_date`, '%d-%b' )
order by `day wise data`;

+-------+---------------+---------------+
| sales | event         | day wise data |
+-------+---------------+---------------+
|     2 | Running Event | 18-08-21      |
|     1 | Cycling event | 18-08-21      |
|     1 | Running Event | 18-08-23      |
|     1 | Running Event | 18-08-24      |
+-------+---------------+---------------+

【讨论】：

【解决方案3】：

试试这个：

select sum(ticket)id) ticket_count
       (select event_name from Events where event_id = t.event_id) event_name,
       cast(ticket_generate_date as date)
from Tickets t
group by event_id, cast(ticket_generate_date as date)

【讨论】：