SQL选择表中的某些行，以便它们总计为某个值

问题描述：

如何才能在下表中选择一些行，以便它们总计为某个值？SQL选择表中的某些行，以便它们总计为某个值

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，前值我要的是57 ...

所以我需要从上表中选择行这样qty1 + qty2 + qty3 +每一行的qty4，直到获得那57值，并放弃其他行。在这个例子中，我会得到以下：

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，所以丢弃行3 & 4 ...

现在我希望顶值是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

因为这些值总和为50，并从ROW2 7，qty4被冷落...... （BTW行以这种特别的方式订购，因为这是我希望说明qtys的总和的顺序......总结第一个r是无效的ow1，然后3，然后2然后4，例如...他们应该总是按顺序1,2,3,4 ...）

如果我想补充这个？我的意思是，其他两行我没有得到最后的结果。

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

（如果第二种情况过于复杂，如何获取如何：

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

因为添加了原来qtys第2行将超过50的值，我放弃它... 在这种情况下的补充应该仅仅是：

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

）

我已经写了很多复杂的查询，我喜欢承担一个挑战，但这是那些只是乞求你写程序代码少见的情况用你选择的语言。 – 2012-03-01 23:40:18

即使是简化的第二种情况？在帖子末尾括号中的最后一个......？ – 2012-03-01 23:41:42

简化的第二种情况可作为查询使用。如果你修改你的问题来问这个问题（或者创建一个新的问题，那就要求），我可以提供帮助。 – 2012-03-01 23:54:06

答

让我们这样说：如果SQL是一种宗教，我会去提供这种解决方案。 SQL并不是为了解决这类问题，所以任何解决方案都是可怕的。矿也不例外:)

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 | 
+----+---------+---------+---------+---------+

和补体：

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！

只是尝试了所提供的解决方案，但对于补充案例，使用相同的样本数据，我没有得到你显示的结果，但只是从数据表中的原始4行... – 2012-03-06 15:49:55

现在它的工作原理！我不知道我上次做错了什么:)让我测试它，并将其应用于我的真实案例，我会给你一个关于赏金的答案;） – 2012-03-07 15:59:47

当然，慢慢来:) – 2012-03-07 19:54:32

答

括号中的简化方法是不是太糟糕了：

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 
;

（你没有提到表的名字，所以我foo去）。

补是：

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 
     ) 
;

非熟练的选项更棘手;这是可行的，但你会更好使用stored procedure。

答

让我们从问题

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>

最终查询，充分WORKS

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);

最终查询补，充分WORKS

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;

这里加载样本数据是57

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>

输出下面是50

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>

输出请记住设置数量在(SELECT 50 mylimit)子查询mylimit各两次。

请告诉我这一个...

谢谢:)但是这只是解决了我提出的简单情况，而不是复杂的情况;） – 2012-03-07 16:04:23

@Javier什么是复杂的？如果你的意思是使用其他一些数字，例如50，只需在两个查询（FINAL和FINAL COMPLEMENT）中用'（SELECT 50 mylimit）'替换'（SELECT 57 mylimit）'，它们将完美工作。否则，请解释复杂的情况。 – RolandoMySQLDBA 2012-03-07 16:09:31

我会尝试用示例来解释我使用的样本数据。在mylimit = 50的情况下，查询结果应为id = 2：qty1 = 1.5，qty2 = 0.0，qty3 = 7.5，qty4 = 11。但是，您的查询给我没有id = 2的行。如您所见，qty4 = 11不是数据表中的数据，而是每行添加qty1 + qty2 + qty3 + qty4的值直到达到mylimit值的结果。也就是说，当你添加50时，就会达到：row1.qty1 + row1.qty2 + row1.qty3 + row1.qty4 then + row2.qty1 + row2.qty2 + row2.qty3 then +7从原始row2.qty4中减去= 18，这给出了你在期望的结果上看到的7个 – 2012-03-07 18:58:02

答

你应该调整只在init子查询的@limit变量初始化。第一个查询输出数据达到极限，secnd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:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty2 >=0, 
      @limit:=(@limit-(@qty2:=qty2)), 
      @qty2:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty3 >=0, 
      @limit:=(@limit-(@qty3:=qty3)), 
      @qty3:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty4 >=0, 
      @limit:=(@limit-(@qty4:=qty4)), 
      @qty4:[email protected] + LEAST(@limit, @limit:=0))),0) 
;

补语：

SELECT 
    id, 
    IF([email protected], qty1, [email protected]) as qty1, 
    IF([email protected], qty2, [email protected]) as qty2, 
    IF([email protected], qty3, [email protected]) as qty3, 
    IF([email protected], qty4, [email protected]) 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:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty2 >=0, 
     @limit:=(@limit-(@qty2:=qty2)), 
     @qty2:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty3 >=0, 
     @limit:=(@limit-(@qty3:=qty3)), 
     @qty3:[email protected] + LEAST(@limit, @limit:=0)), 
     IF(@limit-qty4 >=0, 
     @limit:=(@limit-(@qty4:=qty4)), 
     @qty4:[email protected] + LEAST(@limit, @limit:=0)), 
     @limit), 0, 1) 
;

谢谢！它可以处理样本数据。让我用我的真实数据进行测试，然后我会回答你关于赏金的事情...... – 2012-03-07 16:07:15

SQL选择表中的某些行，以便它们总计为某个值

最终查询，充分WORKS

最终查询补，充分WORKS

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