Question

我试图按字符串上的某个字符排序，例如，

之前：

+----+---------+
| id |  name   |
+----+---------+
| 1  | red     |
| 2  | red-a   |
| 3  | red-xy  |
| 4  | blue    |
| 5  | blue-a  |
| 6  | blue-xy |
+----+---------+

后：

+----+---------+
| id |  name   |
+----+---------+
| 4  | blue    |
| 1  | red     |
| 5  | blue-a  |
| 2  | red-a   |
| 6  | blue-xy |
| 3  | red-xy  |
+----+---------+

是否有任何方法可以使用ORDER BY

基于 -a 或 -xy 进行分类

提前谢谢。

Answer 1

SELECT 
  CASE 
    WHEN RIGHT(`name`,LENGTH(`name`)-INSTR(`name`,'-')) = `name` THEN ''
    ELSE RIGHT(`name`,LENGTH(`name`)-INSTR(`name`,'-')) 
  END AS `suffix`,
  `name`
FROM 
  `table1`
ORDER BY
  `suffix`, `name`

如果未找到后缀，则会将记录放在第一行中。警告：第一个破折号用于将单词与前缀分开。

Answer 2

这将满足您的需求。如果你有很多行，不愿意承诺表现出色：

select id, name from 
(
  select id, 
         name, 
         if (substring_index(name,'-', -1) = name, '', substring_index(name,'-', -1)) as grouping
  from Table1
  order by grouping, name
) as subTable

SQLFiddle here

[编辑]实际上，可以通过以下方式将其简化为单个选择：

select id, 
       name
from Table1
order by  if (substring_index(name,'-', -1) = name, '', substring_index(name,'-', -1)), name

Answer 3

这些查询更具可读性，这可能是对后缀进行排序的最简单方法

SELECT 
 *
 , IF (LOCATE('-', name) = 0
     , 0
     , LENGTH(SUBSTRING_INDEX(name, '-', -1))   
   )
     suffix_length

FROM 
 Table1

ORDER BY 
 suffix_length
;



SELECT 
 *

FROM 
 Table1

ORDER BY 
  IF (LOCATE('-', name) = 0
     , 0
     , LENGTH(SUBSTRING_INDEX(name, '-', -1))   
   )
;

请参阅演示http://sqlfiddle.com/#!9/92b63/41

MySQL - 按特定的最后一个字符串排序

3 个答案: