Php 如何在sql中按两列进行分组？

我有两张桌子，一张是“活动”，第二张是“门票”：

事件表：

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

票务表：

==============================================================
| 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   |
==============================================================

现在我希望运行查询，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');

---

您只需将第二个组选项添加到已有的选项中，并使用逗号

我还使用别名和内部联接来获取事件名称

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：

---

如果在

group by

子句中包含一个额外的列名-在本例中，我选择了

事件\u name

，则最终输出或多或少与预期一致。要输出请求的所有详细信息，尽管您需要连接两个表（或使用嵌套的select）

或者，为了进一步细化结果，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      |
+-------+---------------+---------------+

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)

---

到目前为止你都试了些什么，桑托什？你试过什么吗？你有一个示例代码吗？@Rafael我尝试过这个查询选择计数（票证id），票证生成日期从票证中

ticket\u生成日期

>=日期格式（curdate（），'%Y-%m-01'）按日期分组格式（

ticket\u生成日期

，'%d-%b'）Santosh，请将此添加到主任务中，以便每个人都能阅读它。：）可能重复的

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)