1

假设我有一个返回以下结果的查询：

| val   |  type  | i |
----------------------
| 59    |   1    | 1 |
| 40    |   2    | 2 |
| 12    |   1    | 3 |
| 3     |   2    | 4 |
| 24    |   1    | 5 |
| 30    |   1    | 6 |
| 98    |   2    | 7 |
| 45    |   2    | 8 |
| 46    |   1    | 9 |

val = an arbitrary number
type = 1 or 2
i = auto increment column, just for reference

我想执行以下减法：

A - B

A:  Value of row of type '2'
B:  Value of row above A of type '1' if it exists, but without crossing a row of type '2'

我真的希望这是有道理的..

一些例子：

Row i=2 is of type 2.  So the first row above it of type 1 is i=1.  
So the subtraction will be:  40 - 59 = -19 

Row i=7 is of type 2.  So the first row above it of type 1 is i=6
So the subtraction will be: 98 - 30 = 68

Row i=8 is of type 2.  But if we go above we cross a value of type 2.
So the subtraction should return 0, or be ignored, whichever is simplest.

最后，结果应该返回一列 type=1 和减去的 value 的值。

例如：

| val | diff |
--------------
| 59  | -19  |
| 12  | -9   |
| 24  | 0    |  ***
| 30  | 68   |
| 46  | 0    |  ***

*** Return diff=0 if the subtraction was never made for this row.

我有获取我的初始表的查询。我知道如何做减法（在实际情况下 val 是一个日期，但我在这个例子中保持简单）。

我唯一不知道该怎么做是根据 MySQL 中的不同行执行逻辑。例如，如何根据某些条件找到最近的行，然后引用该行以对其执行某些操作？是否有一种方法可以引用当前的“i”并对（i 处的行）减（i-1 处的行）或类似的东西进行减法。

我已经在这上面花了很多时间，我正要放弃，改用 php 来做，并降低性能。作为最后的手段，我在这里问是否有办法直接在 mysql 中完成所有这些操作？

我不需要完整的解决方案，因为这是非常复杂的要求，但我会接受你能给我的任何建议或指示。我在 MySQL 方面不是最好的，所以也许我错过了一些简单的东西。

谢谢。

4

3 回答 3

2 
 SELECT t1.val AS val, IFNULL( t2.val, t1.val ) - t1.val AS diff FROM my_table AS t1
 LEFT JOIN my_table AS t2 ON( t2.type = 2 AND t2.i = t1.i + 1 )
 WHERE t1.type = 1 ORDER BY t1.i;

对您的数据进行测试，结果：

| val | diff |
--------------
| 59  | -19  |
| 12  | -9   |
| 24  | 0    |
| 30  | 68   |
| 46  | 0    |
于 2011-08-30T18:38:52.080 回答
0

您需要的是一个带有光标的存储过程。这应该是一个很好的起点。

CREATE PROCEDURE curdemo()
BEGIN
  DECLARE done INT DEFAULT 0;
  DECLARE val, type, prev_val, prev_type INT;
  DECLARE cur1 CURSOR FOR SELECT val, type, i from your_table;
  DECLARE CONTINUE HANDLER FOR NOT FOUND SET done = 1;

  OPEN cur1;

  read_loop: LOOP
    FETCH cur1 INTO val, tpye, i;
    IF done THEN
      LEAVE read_loop;
    END IF;

    -- your logic goes here, sorry, not sure I understood correctly, but roughly here it goes
    IF type = 2 THEN
      IF prev_type is not null and prev_type = 1 THEN
        INSERT INTO result_table VALUES (prev_val, val - prev_val);
      END IF;
    END IF;
    SET prev_val = val;
    SET prev_type = type;
  END LOOP;

  CLOSE cur1;
END;
于 2011-08-30T18:03:24.487 回答
0

也许你可以做这样的事情：

select t1.val, (t1.val - t2.val) as diff
     from table as t1, table as t2
     where t1.i = t2.i + 1 
       and t1.type = 1
       and t2.type = 2

用零选择你可以做一个这样的黑客：

select t1.val, (t1.val - t2.val)*(t2.type - t1.type) as diff
     from table as t1, table as t2
     where t1.i = t2.i + 1 
       and t1.type = 1

要在有一个时获得最后一个零，请加入：

select val, 0 as diff from
   select val
      from table
      order by i desc
      limit 1
于 2011-08-30T18:10:24.223 回答