Question

我的查询中有一个复杂的排序问题。

原始无序数据：

+------+--------+-----------+
| id   | job_id | action_id |
+------+--------+-----------+
|    1 |      2 |         1 |  
|    2 |      2 |         2 |  
|    3 |      1 |         1 |  
|    4 |      2 |         3 |  
|    5 |      4 |         1 |  
|    6 |      1 |         2 |  
|    7 |      3 |         1 |  
|    8 |      3 |         2 |  
|    9 |      4 |         2 |  
+------+--------+-----------+

必填排序：

+------+--------+-----------+
| id   | job_id | action_id |
+------+--------+-----------+
|    7 |      3 |         1 |  
|    8 |      3 |         2 |  
|      |        |           |    * blank lines added for clarity,
|    5 |      4 |         1 |      not desired in actual data
|    9 |      4 |         2 |  
|      |        |           | 
|    3 |      1 |         1 |  
|    6 |      1 |         2 |  
|      |        |           | 
|    1 |      2 |         1 |  
|    2 |      2 |         2 |  
|    4 |      2 |         3 |  
+------+--------+-----------+

这种排序背后的理论：

最大的ID是最近添加的条目
action_id为1
后跟具有相同job_id
然后是下一个最近的action_id 1
ad infinitum

编辑：我无法在表格中添加列以帮助排序，正如我在其他解决方案中看到的那样排序问题。

非常感谢任何帮助！

Answer 1

我最好的拍摄是：

SELECT * FROM tbl 
    ORDER BY FIND_IN_SET(job_id,
        (SELECT GROUP_CONCAT(job_id ORDER BY ID DESC) 
         FROM tbl WHERE action_id = 1));

Answer 2

我没有找到方法轻松完成这项工作，您如何看待以下代码：

select c.id, c.job_id, c.action_id
  from (select a.id, a.job_id, a.action_id, min(b.id) as related_id
          from myTable a 
          inner join myTable b
             on a.job_id=b.job_id 
         group by a.job_id) c
 group by c.id
 order by c.related_id desc, c.action_id

MySQL复杂的ORDER BY问题

2 个答案: