Question

我有一个像这样的结构表：

ID    NAME     SURNAME      EXTRA       TYPE
______________________________________________

1     MARIO     ROSSI       RED          10
2     MARCO     VERDI       YELLOW       10
3     GIANNI    BLU         TEACHER      20
4     LUCA      BLU         STUDENT      20 
5     LUCA      ROSSI       GREEN        10 
6     MARIA     GIALLA      10/08/05     30  
7     MARTA     ROSA        11/01/79     30
8     FRANCO    NERO        BARMAN       20
9     MARY      NERI        05/09/88     30
10    MAX       BLU         06/08/98     30

现在我需要按名称，姓氏和（额外[DESC]，其中type = 30）对查询进行排序，所以对于第三个条件，我的意思是我会按降序排列类型为30的所有额外数据：

....
6     MARIA     GIALLA      10/08/05     30  
10    MAX       BLU         06/08/98     30
9     MARY      NERI        05/09/88     30
7     MARTA     ROSA        11/01/79     30
....

所以我试过这个：

select * from my_table order by name, surname, if(my_table.type=30, extra desc, extra asc)

但它给了我一个Sql错误。

感谢您的支持。

感谢其他程序员，根据我的数据，我用这种方式解决了：ORDER BY name，surname，if（TYPE = 30，STR_TO_DATE（EXTRA，'％d /％m /％Y％H.％我'），null）DESC

Answer 1

尝试此查询： -

Architecture

Answer 2

更有用的是，请考虑以下事项：

    DROP TABLE IF EXISTS my_table;

    CREATE TABLE my_table
    (ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY
    ,NAME VARCHAR(12) NOT NULL     
    ,SURNAME VARCHAR(12) NOT NULL      
    ,EXTRA VARCHAR(12) NOT NULL      
    ,TYPE INT NOT NULL
    );

    INSERT INTO my_table VALUES
    ( 1,'MARIO','ROSSI','RED','10'),
    ( 2,'MARCO','VERDI','YELLOW','10'),
    ( 3,'GIANNI','BLU','TEACHER','20'),
    ( 4,'LUCA','BLU','STUDENT','20'),
    ( 5,'LUCA','ROSSI','GREEN','10'),
    ( 6,'MARIA','GIALLA','2005/08/10','30'),
    ( 7,'MARTA','ROSA','1979/01/11','30'),
    ( 8,'FRANCO','NERO','BARMAN','20'),
    ( 9,'MARY','NERI','1988/09/05','30'),
    (10,'MAX','BLU','1998/08/06','30'),
    (11,'MARIO','ROSSI','PLUMBER','20'),
    (12,'MARCO','VERDI','TAILOR','20'),
    (13,'GIANNI','BLU','YELLOW','10'),
    (14,'LUCA','BLU','BLUE','10'),
    (15,'LUCA','ROSSI','BAKER','20'),
    (16,'MARIO','ROSSI','2004/08/10','30'),
    (17,'MARCO','VERDI','1978/01/11','30'),
    (18,'FRANCO','NERO','RED','10'),
    (19,'FRANCO','NERO','1987/09/05','30'),
    (20,'MARIA','GIALLA','1995/08/06','30');

    SELECT name
         , surname
         , MAX(CASE WHEN type = 10 THEN extra END) colour
         , MAX(CASE WHEN type = 20 THEN extra END) occupation
         , MAX(CASE WHEN type = 30 THEN extra END) date
      FROM my_table
     GROUP 
        BY name
         , surname
     ORDER 
        BY name
         , surname
         , extra
         , type
         , date DESC;
+--------+---------+--------+------------+------------+
| name   | surname | colour | occupation | date       |
+--------+---------+--------+------------+------------+
| FRANCO | NERO    | RED    | BARMAN     | 1987/09/05 |
| GIANNI | BLU     | YELLOW | TEACHER    | NULL       |
| LUCA   | BLU     | BLUE   | STUDENT    | NULL       |
| LUCA   | ROSSI   | GREEN  | BAKER      | NULL       |
| MARCO  | VERDI   | YELLOW | TAILOR     | 1978/01/11 |
| MARIA  | GIALLA  | NULL   | NULL       | 2005/08/10 |
| MARIO  | ROSSI   | RED    | PLUMBER    | 2004/08/10 |
| MARTA  | ROSA    | NULL   | NULL       | 1979/01/11 |
| MARY   | NERI    | NULL   | NULL       | 1988/09/05 |
| MAX    | BLU     | NULL   | NULL       | 1998/08/06 |
+--------+---------+--------+------------+------------+

您可以加入这样的结果（或实际上更简单的替代方案 - 但我现在仍然坚持这一点）回到原始表格，以便按照您的喜好订购结果。我在STR_TO_DATE函数中添加了仅供参考。显然，这里没有必要，与你的版本需要的方式略有不同......

 SELECT x.*
      , y.date
   FROM my_table x
   LEFT
   JOIN
      ( SELECT name
             , surname
             , MAX(CASE WHEN type = 30 THEN extra END) date
          FROM my_table
         GROUP
            BY name
             , surname
      ) y
     ON y.name = x.name
    AND y.surname = x.surname
  ORDER
     BY x.name
      , x.surname
      , STR_TO_DATE(y.date,'%Y/%m/%d') DESC;

+----+--------+---------+------------+------+------------+
| ID | NAME   | SURNAME | EXTRA      | TYPE | date       |
+----+--------+---------+------------+------+------------+
| 18 | FRANCO | NERO    | RED        |   10 | 1987/09/05 |
| 19 | FRANCO | NERO    | 1987/09/05 |   30 | 1987/09/05 |
|  8 | FRANCO | NERO    | BARMAN     |   20 | 1987/09/05 |
| 13 | GIANNI | BLU     | YELLOW     |   10 | NULL       |
|  3 | GIANNI | BLU     | TEACHER    |   20 | NULL       |
| 14 | LUCA   | BLU     | BLUE       |   10 | NULL       |
|  4 | LUCA   | BLU     | STUDENT    |   20 | NULL       |
| 15 | LUCA   | ROSSI   | BAKER      |   20 | NULL       |
|  5 | LUCA   | ROSSI   | GREEN      |   10 | NULL       |
| 17 | MARCO  | VERDI   | 1978/01/11 |   30 | 1978/01/11 |
|  2 | MARCO  | VERDI   | YELLOW     |   10 | 1978/01/11 |
| 12 | MARCO  | VERDI   | TAILOR     |   20 | 1978/01/11 |
|  6 | MARIA  | GIALLA  | 2005/08/10 |   30 | 2005/08/10 |
| 20 | MARIA  | GIALLA  | 1995/08/06 |   30 | 2005/08/10 |
| 11 | MARIO  | ROSSI   | PLUMBER    |   20 | 2004/08/10 |
| 16 | MARIO  | ROSSI   | 2004/08/10 |   30 | 2004/08/10 |
|  1 | MARIO  | ROSSI   | RED        |   10 | 2004/08/10 |
|  7 | MARTA  | ROSA    | 1979/01/11 |   30 | 1979/01/11 |
|  9 | MARY   | NERI    | 1988/09/05 |   30 | 1988/09/05 |
| 10 | MAX    | BLU     | 1998/08/06 |   30 | 1998/08/06 |
+----+--------+---------+------------+------+------------+

Answer 3

应该是这样的：

SELECT *
FROM Tabletest
order by 
  CASE WHEN type <> 30 THEN 1 ELSE 2 END,
  CASE WHEN type <> 30 THEN Name END,
  CASE WHEN type <> 30 THEN Surname END,
  CASE WHEN type = 30 THEN STR_TO_DATE(extra,'%m/%d/%Y') ELSE 0 END DESC

这会将结果命令为：

如何在我的sql中解决这个ORDER BY？

3 个答案: