【问题标题】：SQL select some rows in a table so that they sum up to certain valueSQL 选择表中的某些行，以便它们总和为某个值
【发布时间】：2012-03-20 12:11:43
【问题描述】：

我怎样才能只选择下表中的一些行，以便它们总和为某个值？

Table
-----
id | qty1 | qty2 | qty3 | qty4
------------------------------
1  | 0.0  | 0.0  | 10   | 20
2  | 1.5  | 0.0  | 7.5  | 18
3  | 1.0  | 2.0  | 7.5  | 18
4  | 0.0  | 0.5  | 5    | 13

比方说，我想要的最高值是 57...

所以我需要从上一个表中选择行，使得每行的 qty1+qty2+qty3+qty4，直到 57 值，然后丢弃其他行。在这个例子中，我会得到以下信息：

id | qty1 | qty2 | qty3 | qty4
------------------------------
1  | 0.0  | 0.0  | 10   | 20
2  | 1.5  | 0.0  | 7.5  | 18

因为 10+20+1.5+7.5+18 = 57，所以我舍弃了第 3 行和第 4 行...

现在我希望最高值为 50，那么我应该得到：

id | qty1 | qty2 | qty3 | qty4
------------------------------
1  | 0.0  | 0.0  | 10   | 20
2  | 1.5  | 0.0  | 7.5  | 11

由于这些值的总和为 50，并且 row2,qty4 中的 7 被忽略... （顺便说一句，行以这种特殊方式排序，因为这是我希望计算 qtys 总和的顺序......例如，先总结第一行，然后是 3，然后是 2，然后是 4，这是无效的......它们应始终按 1、2、3、4 的顺序相加...）

如果我想要这个的补充怎么办？我的意思是，我在最后一个结果中没有得到的另外两行。

第一种情况：

id | qty1 | qty2 | qty3 | qty4
------------------------------
3  | 1.0  | 2.0  | 7.5  | 18
4  | 0.0  | 0.5  | 5    | 13

第二种情况：

id | qty1 | qty2 | qty3 | qty4
------------------------------
2  | 0.0  | 0.0  | 0.0  | 7
3  | 1.0  | 2.0  | 7.5  | 18
4  | 0.0  | 0.5  | 5    | 13

（如果第二种情况太复杂，获取如何：

id | qty1 | qty2 | qty3 | qty4
------------------------------
1  | 0.0  | 0.0  | 10   | 20

因为将第 2 行的原始数量加起来会超过 50 的值，所以我将其丢弃... 这种情况下的补码应该是：

id | qty1 | qty2 | qty3 | qty4
------------------------------
2  | 1.5  | 0.0  | 7.5  | 18
3  | 1.0  | 2.0  | 7.5  | 18
4  | 0.0  | 0.5  | 5    | 13

)

【问题讨论】：

我写了很多复杂的查询，我喜欢接受挑战，但这是极少数情况之一，只需要你用你选择的语言编写程序代码。
即使是简化的第二种情况？文章末尾括号中的最后一个...？
简化的第二种情况可用作查询。如果您修改您的问题以仅要求该问题（或创建一个仅要求该问题的新问题），我可以提供帮助。
fo 300 我将不得不向您解释我所面临的实际问题的每一个细节，而不是在这里仅展示抽象的特定细节作为示例，因此您提供的解决方案不仅仅是概括我必须翻译或改编，但我需要解决我的问题的实际解决方案。我会让你签署一份保密协议，因为这些细节是我工作中的事情，不应该为任何人所知...... ;)

标签： mysql select sum

【解决方案1】：

括号里的简化选项还不错：

SELECT foo1.*
  FROM foo AS foo1
  JOIN foo AS foo2
    ON foo2.id <= foo1.id
 GROUP
    BY foo1.id
HAVING SUM(foo2.qty1 + foo2.qty2 + foo2.qty3 + foo2.qty4) <= 57
;

（你没有提到表的名称，所以我选择了foo。）

补语是：

SELECT *
  FROM foo
 WHERE id NOT IN
        ( SELECT foo1.id
            FROM foo AS foo1
            JOIN foo AS foo2
              ON foo2.id <= foo1.id
           GROUP
              BY foo1.id
          HAVING SUM(foo2.qty1 + foo2.qty2 + foo2.qty3 + foo2.qty4) <= 57
        )
;

不带括号的选项要复杂得多；这是可行的，但你最好使用stored procedure。

【讨论】：

【解决方案2】：

让我们这样说吧：如果 SQL 是一种宗教，我会因为提供这个解决方案而下地狱。 SQL 并不是为了解决这类问题，所以任何解决方案都会很糟糕。我的也不例外:)

set @limitValue := 50;
select id, newQty1, newQty2, newQty3, newQty4 from (
  select id,
  if(@limitValue - qty1 > 0, qty1, greatest(@limitValue, 0)) newQty1,
  @limitValue := @limitValue - qty1 Total1,
  if(@limitValue - qty2 > 0, qty2, greatest(@limitValue, 0)) newQty2,
  @limitValue := @limitValue - qty2 Total2,
  if(@limitValue - qty3 > 0, qty3, greatest(@limitValue, 0)) newQty3,
  @limitValue := @limitValue - qty3 Total3,
  if(@limitValue - qty4 > 0, qty4, greatest(@limitValue, 0)) newQty4,
  @limitValue := @limitValue - qty4 Total4
  from (
    select id, qty1, qty2, qty3, qty4,
      @rowTotal < @limitValue Useful,
      @previousRowTotal := @rowTotal PreviousRowTotal,
      @rowTotal := @rowTotal + qty1 + qty2 + qty3 + qty4 AllRowsTotal,
      @rowTotal - @previousRowTotal CurrentRowTotal 
    from t,
    (select @rowTotal := 0, @previousRowTotal := 0) S1
  ) MarkedUseful
  where useful = 1
) Final

对于提供的数据，这会导致：

+----+---------+---------+---------+---------+
| ID | NEWQTY1 | NEWQTY2 | NEWQTY3 | NEWQTY4 |
+----+---------+---------+---------+---------+
|  1 | 0       |       0 | 10      |      20 |
|  2 | 1.5     |       0 | 7.5     |      11 |
+----+---------+---------+---------+---------+

还有补语：

set @limitValue := 50;
select t1.id,
  coalesce(t1.qty1 - newQty1, t1.qty1) newQty1,
  coalesce(t1.qty2 - newQty2, t1.qty2) newQty2,
  coalesce(t1.qty3 - newQty3, t1.qty3) newQty3,
  coalesce(t1.qty4 - newQty4, t1.qty4) newQty4
from t t1 left join (
    select id,
    if(@limitValue - qty1 > 0, qty1, greatest(@limitValue, 0)) newQty1,
    @limitValue := @limitValue - qty1 Total1,
    if(@limitValue - qty2 > 0, qty2, greatest(@limitValue, 0)) newQty2,
    @limitValue := @limitValue - qty2 Total2,
    if(@limitValue - qty3 > 0, qty3, greatest(@limitValue, 0)) newQty3,
    @limitValue := @limitValue - qty3 Total3,
    if(@limitValue - qty4 > 0, qty4, greatest(@limitValue, 0)) newQty4,
    @limitValue := @limitValue - qty4 Total4
    from (
      select id, qty1, qty2, qty3, qty4,
        @rowTotal < @limitValue Useful,
        @previousRowTotal := @rowTotal PreviousRowTotal,
        @rowTotal := @rowTotal + qty1 + qty2 + qty3 + qty4 AllRowsTotal,
        @rowTotal - @previousRowTotal CurrentRowTotal 
      from t,
      (select @rowTotal := 0, @previousRowTotal := 0) S1
    ) MarkedUseful
    where useful = 1
) Final
on t1.id = final.id
where Total1 < 0 or Total2 < 0 or Total3 < 0 or Total4 < 0 or final.id is null

对于提供的数据，这会导致：

+----+---------+---------+---------+---------+
| ID | NEWQTY1 | NEWQTY2 | NEWQTY3 | NEWQTY4 |
+----+---------+---------+---------+---------+
|  2 |       0 | 0       | 0       |       7 |
|  3 |       1 | 2       | 7.5     |      18 |
|  4 |       0 | 0.5     | 5       |      13 |
+----+---------+---------+---------+---------+

享受吧！

【讨论】：

刚刚尝试了提供的解决方案，但是对于补充情况，使用相同的示例数据，我没有得到您显示的结果，而是数据表中的原始 4 行...
现在可以了！我不知道我上次做错了什么：）让我测试一下并将其应用到我的真实案例中，我会给你一个关于赏金的答案；）

【解决方案3】：

您应该只调整init 子查询中的@limit 变量初始化。第一个查询输出数据达到限制，第二个查询输出其补码。

SELECT
  id,
  @qty1 as qty1,
  @qty2 as qty2,
  @qty3 as qty3,
  @qty4 as qty4
FROM quantities q,
  (SELECT @qty1:=0.0, @qty2:=0.0,
          @qty3:=0.0, @qty4:=0.0,
          @limit:=50.0) init
WHERE
  IF(@limit > 0,
     GREATEST(1,
       IF(@limit-qty1 >=0,
          @limit:=(@limit-(@qty1:=qty1)),
          @qty1:=@limit + LEAST(@limit, @limit:=0)),
       IF(@limit-qty2 >=0,
          @limit:=(@limit-(@qty2:=qty2)),
          @qty2:=@limit + LEAST(@limit, @limit:=0)),
       IF(@limit-qty3 >=0,
          @limit:=(@limit-(@qty3:=qty3)),
          @qty3:=@limit + LEAST(@limit, @limit:=0)),
       IF(@limit-qty4 >=0,
          @limit:=(@limit-(@qty4:=qty4)),
          @qty4:=@limit + LEAST(@limit, @limit:=0))),0)
;

补语：

SELECT
  id,
  IF(qty1=@qty1, qty1, qty1-@qty1) as qty1,
  IF(qty2=@qty2, qty2, qty2-@qty2) as qty2,
  IF(qty3=@qty3, qty3, qty3-@qty3) as qty3,
  IF(qty4=@qty4, qty4, qty4-@qty4) as qty4
FROM quantities q,
  (SELECT @qty1:=0.0, @qty2:=0.0,
          @qty3:=0.0, @qty4:=0.0,
          @limit:=50.0) init
WHERE
  IF(
    LEAST(
      IF(@limit-qty1 >=0,
         @limit:=(@limit-(@qty1:=qty1)),
         @qty1:=@limit + LEAST(@limit, @limit:=0)),
      IF(@limit-qty2 >=0,
         @limit:=(@limit-(@qty2:=qty2)),
         @qty2:=@limit + LEAST(@limit, @limit:=0)),
      IF(@limit-qty3 >=0,
         @limit:=(@limit-(@qty3:=qty3)),
         @qty3:=@limit + LEAST(@limit, @limit:=0)),
      IF(@limit-qty4 >=0,
         @limit:=(@limit-(@qty4:=qty4)),
         @qty4:=@limit + LEAST(@limit, @limit:=0)),
      @limit), 0, 1)
;

【讨论】：

谢谢！它确实适用于样本数据。让我用我的真实数据测试一下，我会回答你关于赏金的问题......

【解决方案4】：

让我们从问题中加载您的示例数据

mysql> drop database if exists javier;
Query OK, 1 row affected (0.02 sec)

mysql> create database javier;
Query OK, 1 row affected (0.01 sec)

mysql> use javier
Database changed
mysql> create table mytable
    -> (
    ->     id int not null auto_increment,
    ->     qty1 float,qty2 float,qty3 float,qty4 float,
    ->     primary key (id)
    -> );
Query OK, 0 rows affected (0.08 sec)

mysql> insert into mytable (qty1,qty2,qty3,qty4) values
    -> ( 0.0 , 0.0 , 10  , 20 ),( 1.5 , 0.0 , 7.5 , 18 ),
    -> ( 1.0 , 2.0 , 7.5 , 18 ),( 0.0 , 0.5 , 5   , 13 );
Query OK, 4 rows affected (0.05 sec)
Records: 4  Duplicates: 0  Warnings: 0

mysql> select * from mytable;
+----+------+------+------+------+
| id | qty1 | qty2 | qty3 | qty4 |
+----+------+------+------+------+
|  1 |    0 |    0 |   10 |   20 |
|  2 |  1.5 |    0 |  7.5 |   18 |
|  3 |    1 |    2 |  7.5 |   18 |
|  4 |    0 |  0.5 |    5 |   13 |
+----+------+------+------+------+
4 rows in set (0.00 sec)

mysql>

完全有效的最终查询

select BBBB.* from (select id,sums FROM (select A.id,A.sums from
(select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB 
where BB.id<=AA.id) sums from mytable AA order by id) A 
INNER JOIN (SELECT 50 mylimit) B ON A.sums <= B.mylimit) AAA
UNION
(select A.id,A.sums from (select id,(select sum(qty1+qty2+qty3+qty4)  
from mytable BB where BB.id<=AA.id) sums from mytable AA order by id) A
where A.sums=(select min(A.sums) sums from (select id, 
(select sum(qty1+qty2+qty3+qty4) from mytable BB where BB.id<=AA.id) sums 
from mytable AA order by id) A INNER JOIN (SELECT 50 mylimit) B 
ON A.sums >= B.mylimit))) AAAA JOIN mytable BBBB USING (id);

完全有效的最终查询补充

select BBBB.* from  mytable BBBB LEFT JOIN
(select id,sums FROM (select A.id,A.sums from ( 
select id,(select sum(qty1+qty2+qty3+qty4)  
from mytable BB where BB.id<=AA.id) sums 
from mytable AA order by id) A INNER JOIN
(SELECT 50 mylimit) B ON A.sums <= B.mylimit) AAA
UNION
(select A.id,A.sums from (select id, 
(select sum(qty1+qty2+qty3+qty4) from mytable BB 
where BB.id<=AA.id) sums from mytable AA order by id) A
where A.sums=(select min(A.sums) sums from ( 
select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB 
where BB.id<=AA.id) sums from mytable AA order by id) A 
INNER JOIN (SELECT 50 mylimit) B ON A.sums >= B.mylimit))) AAAA
USING (id) WHERE AAAA.id IS NULL;

这是 57 的输出

mysql>     select BBBB.* from (select id,sums FROM (select A.id,A.sums from
    ->     (select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     INNER JOIN (SELECT 57 mylimit) B ON A.sums <= B.mylimit) AAA
    ->     UNION
    ->     (select A.id,A.sums from (select id,(select sum(qty1+qty2+qty3+qty4)
    ->     from mytable BB where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     where A.sums=(select min(A.sums) sums from (select id,
    ->     (select sum(qty1+qty2+qty3+qty4) from mytable BB where BB.id<=AA.id) sums
    ->     from mytable AA order by id) A INNER JOIN (SELECT 57 mylimit) B
    ->     ON A.sums >= B.mylimit))) AAAA JOIN mytable BBBB USING (id);
+----+------+------+------+------+
| id | qty1 | qty2 | qty3 | qty4 |
+----+------+------+------+------+
|  1 |    0 |    0 |   10 |   20 |
|  2 |  1.5 |    0 |  7.5 |   18 |
+----+------+------+------+------+
2 rows in set (0.00 sec)

mysql>     select BBBB.* from  mytable BBBB LEFT JOIN
    ->     (select id,sums FROM (select A.id,A.sums from (
    ->     select id,(select sum(qty1+qty2+qty3+qty4)
    ->     from mytable BB where BB.id<=AA.id) sums
    ->     from mytable AA order by id) A INNER JOIN
    ->     (SELECT 57 mylimit) B ON A.sums <= B.mylimit) AAA
    ->     UNION
    ->     (select A.id,A.sums from (select id,
    ->     (select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     where A.sums=(select min(A.sums) sums from (
    ->     select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     INNER JOIN (SELECT 57 mylimit) B ON A.sums >= B.mylimit))) AAAA
    ->     USING (id) WHERE AAAA.id IS NULL;
+----+------+------+------+------+
| id | qty1 | qty2 | qty3 | qty4 |
+----+------+------+------+------+
|  3 |    1 |    2 |  7.5 |   18 |
|  4 |    0 |  0.5 |    5 |   13 |
+----+------+------+------+------+
2 rows in set (0.00 sec)

mysql>

这是 50 的输出

mysql>     select BBBB.* from (select id,sums FROM (select A.id,A.sums from
    ->     (select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     INNER JOIN (SELECT 50 mylimit) B ON A.sums <= B.mylimit) AAA
    ->     UNION
    ->     (select A.id,A.sums from (select id,(select sum(qty1+qty2+qty3+qty4)
    ->     from mytable BB where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     where A.sums=(select min(A.sums) sums from (select id,
    ->     (select sum(qty1+qty2+qty3+qty4) from mytable BB where BB.id<=AA.id) sums
    ->     from mytable AA order by id) A INNER JOIN (SELECT 50 mylimit) B
    ->     ON A.sums >= B.mylimit))) AAAA JOIN mytable BBBB USING (id);
+----+------+------+------+------+
| id | qty1 | qty2 | qty3 | qty4 |
+----+------+------+------+------+
|  1 |    0 |    0 |   10 |   20 |
|  2 |  1.5 |    0 |  7.5 |   18 |
+----+------+------+------+------+
2 rows in set (0.00 sec)

mysql>     select BBBB.* from  mytable BBBB LEFT JOIN
    ->     (select id,sums FROM (select A.id,A.sums from (
    ->     select id,(select sum(qty1+qty2+qty3+qty4)
    ->     from mytable BB where BB.id<=AA.id) sums
    ->     from mytable AA order by id) A INNER JOIN
    ->     (SELECT 50 mylimit) B ON A.sums <= B.mylimit) AAA
    ->     UNION
    ->     (select A.id,A.sums from (select id,
    ->     (select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     where A.sums=(select min(A.sums) sums from (
    ->     select id,(select sum(qty1+qty2+qty3+qty4) from mytable BB
    ->     where BB.id<=AA.id) sums from mytable AA order by id) A
    ->     INNER JOIN (SELECT 50 mylimit) B ON A.sums >= B.mylimit))) AAAA
    ->     USING (id) WHERE AAAA.id IS NULL;
+----+------+------+------+------+
| id | qty1 | qty2 | qty3 | qty4 |
+----+------+------+------+------+
|  3 |    1 |    2 |  7.5 |   18 |
|  4 |    0 |  0.5 |    5 |   13 |
+----+------+------+------+------+
2 rows in set (0.01 sec)

mysql>

请记住在 (SELECT 50 mylimit) 子查询中为 mylimit 设置数字，每次两次。

请告诉我有这个...

【讨论】：

谢谢 :) 但这只是解决了我提出的简单案例，而不是复杂案例；）
@Javier 复杂的是什么？如果您的意思是使用其他数字，例如 50，只需在两个查询（最终和最终补码）中将 (SELECT 57 mylimit) 替换为 (SELECT 50 mylimit)，它们将完美运行。否则，请解释复杂情况。
我将尝试通过示例来解释，使用我提供的示例数据。在 mylimit=50 的情况下，id=2 的查询结果应为：qty1=1.5, qty2=0.0, qty3=7.5, qty4=11。但是，您的查询没有给我 id=2 的行。可以看到，qty4=11 不是表的数据，而是每行的qty1+qty2+qty3+qty4的值相加，直到达到mylimit值的结果。也就是说，当您添加时达到 50： row1.qty1+row1.qty2+row1.qty3+row1.qty4 然后 +row2.qty1+row2.qty2+row2.qty3 然后 +7 从原始 row2.qty4 中减去=18，这给出了您在所需结果上看到的 7
嗨哈维尔，刚刚签到。请检查最终查询和最终查询补充。它们完全适用于 57 和 50。
我清理了我的答案，以便将工作查询放在首位。我包括了 57 和 50 被执行的完整示例。