Question

有没有办法扭转整个列？

示例：

ID    ColX    ColY    ColZ
0     001     010     100
1     002     020     200
2     003     030     300

应为：

ID    ColX    ColY    ColZ
0     003     030     300
1     002     020     200
2     001     010     100

所以列ID应该颠倒，具有最后一个ID的记录应该是第一个，第二个是最后一个第一个和到目前为止。

最新的值有ID = 0和最旧的ID = n，这必须完全相反，否则我无法插入新记录。

Answer 1

您可以使用变量：

SELECT t2.ID, ColX, ColY, ColZ
FROM (SELECT ID, ColX, ColY, ColZ,
             @row_number := @row_number + 1 AS rn
      FROM mytable
      CROSS JOIN (SELECT @row_number := 0) AS var
      ORDER BY ID) AS t1
INNER JOIN (  
      SELECT ID, @rn := @rn + 1 AS rn          
      FROM mytable
      CROSS JOIN (SELECT @rn := 0) AS var
      ORDER BY ID DESC) AS t2
ON t1.rn = t2.rn
ORDER BY t2.ID

Demo here

如果您想要UPDATE，那么您可以在UPDATE语句中使用上述查询，如下所示：

UPDATE mytable AS t
INNER JOIN(
   SELECT ID, ColX, ColY, ColZ,
          @row_number := @row_number + 1 AS rn
   FROM mytable
   CROSS JOIN (SELECT @row_number := 0) AS var
   ORDER BY ID) AS t1 ON t.ID = t1.ID
INNER JOIN (  
      SELECT ID, @rn := @rn + 1 AS rn          
      FROM mytable
      CROSS JOIN (SELECT @rn := 0) AS var
      ORDER BY ID DESC) AS t2 ON t1.rn = t2.rn
SET t.ID = t2.ID

Demo here

无论ID列的值是什么，上述操作都会起作用。

Answer 2

如果您的ID中没有空白，则可以使用this select query：

SELECT
  max_id-ID AS ID,
  ColX,
  ColY,
  ColZ
FROM
  mytable CROSS JOIN (SELECT MAX(ID) AS max_id FROM mytable) m
ORDER BY
  ID

或此update query（但仅当ID不是主键时才有效）：

UPDATE
  mytable m1 CROSS JOIN (SELECT MAX(ID) as max_id FROM mytable) m
SET
  m1.ID = m.max_id - m1.ID

如果它是主键，你可以使用它：

UPDATE
  mytable m1 CROSS JOIN (SELECT MAX(ID) as max_id FROM mytable) m
  INNER JOIN mytable m2 ON m1.ID = m.max_id - m2.ID
SET
  m1.ColX = m2.ColX,
  m1.ColY = m2.ColY,
  m1.ColZ = m2.ColZ

（请参阅here）

但如果您的ID列有间隙（例如0,1,2,5,6），则需要采用不同的方法。

Answer 3

SELECT 
  @x - t.id AS new_id,
  t.* 
FROM tab t,
( SELECT @x:= max(id+1) FROM tab ) m;

MariaDB [tmp]> SELECT
    ->   @x - t.id AS new_id,
    ->   t.*
    -> FROM tab t,
    -> ( SELECT @x:= max(id+1) FROM tab ) m;
+--------+----+--------+---------------------+
| new_id | id | action | date                |
+--------+----+--------+---------------------+
|      7 |  1 |      2 | 2015-09-24 15:28:30 |
|      6 |  2 |      4 | 2015-09-24 15:29:26 |
|      5 |  3 |      2 | 2015-09-24 15:30:01 |
|      4 |  4 |      3 | 2015-09-24 15:30:55 |
|      3 |  5 |      1 | 2015-09-24 16:07:25 |
|      2 |  6 |      5 | 2015-09-24 16:10:25 |
|      1 |  7 |      4 | 2015-09-24 16:29:26 |
+--------+----+--------+---------------------+
7 rows in set (0.00 sec)

SQL反转整列

3 个答案: