如何在下表中只选择一些行，以便它们总和到某个值？
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
让我们说，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，直到该57值，并丢弃其他行。在这个例子中，我会得到以下结果：
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，所以我丢弃第3行和第3行4 ...
现在我希望最高值是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
由于这些值总计为50，而来自row2的7，因此省略了qty4 ... （顺便说一句，这些行是以这种特殊的方式排序的，因为这是我希望考虑qtys总和的顺序......例如，总结第一行1，然后是3，然后是2和4，这是无效的......它们应该始终按照1,2,3,4 ......的顺序进行控制。
如果我想补充这个怎么办？我的意思是，在最后一个结果中我没有得到另外两行。
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，那么如何获得：
id | qty1 | qty2 | qty3 | qty4
------------------------------
1 | 0.0 | 0.0 | 10 | 20
因为将第2行的原始qtys加起来会超过50的值，我将其丢弃...... 这种情况下的补充应该是：
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
）
答案 0 :(得分：13)
括号中的简化选项也不错：
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
;
（你没有提到表的名字，所以我选择了foo。）
补充将是：
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
)
;
不明白的选项更加棘手;这是可行的，但你最好使用stored procedure。
答案 1 :(得分：7)
让我们这样说：如果SQL是一种宗教信仰，那么我会为提供这种解决方案而下地狱。 SQL并不是要解决这类问题，所以任何解决方案都会很糟糕。我也不例外:)）
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，结果如下：
+----+---------+---------+---------+---------+
| ID | NEWQTY1 | NEWQTY2 | NEWQTY3 | NEWQTY4 |
+----+---------+---------+---------+---------+
| 1 | 0 | 0 | 10 | 20 |
| 2 | 1.5 | 0 | 7.5 | 11 |
+----+---------+---------+---------+---------+
补充：
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，结果如下：
+----+---------+---------+---------+---------+
| ID | NEWQTY1 | NEWQTY2 | NEWQTY3 | NEWQTY4 |
+----+---------+---------+---------+---------+
| 2 | 0 | 0 | 0 | 7 |
| 3 | 1 | 2 | 7.5 | 18 |
| 4 | 0 | 0.5 | 5 | 13 |
+----+---------+---------+---------+---------+
享受！
答案 2 :(得分：4)
让我们从问题中加载您的样本数据
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>
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);
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设置两次。
请告诉我有这个......
答案 3 :(得分：4)
您应该仅调整init子查询中的@limit变量初始化。第一个查询输出数据到限制，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:=@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)
;
补充：
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)
;