SQL：选择与第一顺序匹配特定的ID和特定类型

Question

的行我有如下表：

id name type 
-- ---- ----- 
1 Joe H 
1 Zoe F 
2 Ken F 
2 Ben H 
3 Lime H 
4 Fine F 
5 Lima F 
5 Sima H 

我想首先选择匹配的行：

id: 5 & type: H 

id: 2 & type: F 

id: 3 & type: H 

并经过他们其余的由ASC订购。

所以最终的结果应该是：

id name type 
-- ---- ----- 
5 Sima H 
2 Ken F 
3 Lime H 
1 Joe H 
1 Zoe F 
2 Ben H 
4 Fine F 
5 Lima F 

我只是找不到如何显示超过一个值排序第一特定行。如果有帮助，我使用PHP和MySqli。

谢谢

Answer 1

您可以用case语句命令：

select * from table 
order by 
    case 
     when id = 5 and type = 'H' then 0 
     when id = 2 and type = 'F' then 1 
     when id = 3 and type = 'H' then 2 
     else 3 
    end, id, name 

Answer 2

既然你问MySQL，我会给你一个MySQL的答案。

SELECT * FROM 
(SELECT * FROM `table` WHERE id = 5 AND `type` = "H" ORDER BY id ASC) t 
    UNION 
(SELECT * FROM `table` WHERE id = 2 AND `type` = "F" ORDER BY id ASC) 
    UNION 
(SELECT * FROM `table` WHERE id = 3 AND `type` = "H" ORDER BY id ASC) 
    UNION 
(SELECT * FROM `table` ORDER BY id ASC); 

结果

+------+------+------+ 
| id | name | type | 
+------+------+------+ 
| 5 | Sima | H | 
| 2 | Ken | F | 
| 3 | Lime | H | 
| 1 | Joe | H | 
| 1 | Zoe | F | 
| 2 | Ben | H | 
| 4 | Fine | F | 
| 5 | Lima | F | 
+------+------+------+ 
8 rows in set (0.00 sec)