Question

我有一个包含一些值的表，如下所示，我想得出一些值，主题的总和等于另一个值，让我们说像13：

id values

1  4
2  7
3  5
4  6

我想得出那些等于13的值。朋友给了我一个查询：

SELECT t1.* FROM tables AS t1
INNER JOIN tables AS t2
WHERE t1.id != t2.id AND t1.`values` + t2.`values` = 13

但它返回的只有两个值，但是假设我想要两个以上的值，甚至更多 4 + 7 + 5 = 16或4 + 7 + 5 + 6 = 22 我的意思在这里我不想只有两个数字，有时我需要更多，这取决于我想要的价值。请给我一个解决方案。

Answer 1

临时表：

group
  sum     int
  lastid  int
  ids loooooongtext
  phase int

初始化临时表：

SET @SUM = 13;

INSERT INTO group
SELECT
  tables.val AS sum,
  tables.id  AS lastid,
  tables.id  AS ids
  0          AS phase
FROM tables
WHERE tables.val <= @SUM;

还有一些循环（你可以用php或mysql来做循环），半个SQL，半个伪代码：

SET @PHASE = 0;

do
  SET @PHASE = @PHASE + 1;

  -- try to append new values to each group
  INSERT INTO group
  SELECT  ( group.sum   +   tables.val ) AS sum,
                            tables.id    AS lastid,
    CONCAT( group.ids, ",", tables.id  ) AS ids,
            @PHASE                       AS phase
  FROM       group
  INNER JOIN tables  ON group.lastid < tables.id          -- monotonity to prevent (some) duplicates
                    AND group.sum    + tables.val <= @SUM;

  -- delete dead ends
  DELETE
  FROM  group
  WHERE group.phase < @PHASE
    AND group.sum  != @SUM;

while ROW_COUNT() > 0;

如果循环准备就绪，请从group读取每一行，group.ids将包含table.ids。

根据php mysql中的值从多个值返回多个值

1 个答案: