组连续包含相应的数据

Question

这是我的表结构：

orders
--------------------------
id        | customer_name
--------------------------
23        | John Doe
24        | Jane Doe

order_comments
--------------------------------------------------------------------
id | order_id  | username | created_at  | comment
--------------------------------------------------------------------
1  | 23        | Bob      | 2019-04-01  | my first comment
2  | 23        | Jim      | 2019-04-03  | another comment
3  | 24        | Jim      | 2019-04-05  | testing
4  | 24        | Jim      | 2019-04-06  | testing again

我想选择换行连接的注释，但也包括用户名和created_at。这是我到目前为止：

select *
from (SELECT order_id, GROUP_CONCAT(`comment` order by id desc SEPARATOR '\n') as comments
      FROM `order_comments`
      group by order_id) comments

结果：

order_id | comments
--------------------------------
23       | my first comment
         | another comment
--------------------------------
24       | testing
         | testing again

这是我想要的结果，包括用户名，并为每个评论concat创建：

order_id | comments
--------------------------------
23       | Bob on 2019-04-01:
         | my first comment
         |
         | Jim on 2019-04-03:
         | another comment
---------------------------------
24       | Jim on 2019-04-05:
         | testing
         |
         | Jim on 2019-04-06:
         | testing again

我怎样才能得到理想的结果？

Answer 1

GROUP_CONCAT或只是CONCAT：

SELECT order_id,CONCAT(username,' on ',created_at,' ',comments) 
FROM order_comments 
ORDER BY order_id;

另一种方式是qazxsw poi>与分隔符连接：

CONCAT_WS

Answer 2

试试这个吧

SELECT order_id,CONCAT_WS(' ',username,'on',created_at,comments) 
FROM order_comments 
ORDER BY order_id;