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MySQL 我如何创建这个子查询?

转载 作者:可可西里 更新时间:2023-11-01 06:33:29 26 4
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我有以下表格

餐 table 农场

+---------+--------+-------------------+-----------+------------+
| FARM_ID |Stock_ID| FARM_TITLE | Size | FARM_VALUE |
+---------+--------+-------------------+-----------+------------+
| 2 | 1 | AgriZone | M | 202 |
| 3 | 1 | Cow Mill | L | 11 |
| 4 | 2 | Beef Farm | H | 540 |
| 5 | 2 | CattleOne | M | 1080 |
| 6 | 2 | FarmOne | L | 455 |
| 7 | 3 | Perdue | H | 333 |
| 8 | 4 | Holstein | M | 825 |
| 10 | 1 | Dotterers | H | 98 |
+---------+--------+-------------------+-----------+------------+

table 门

+---------+---------+------------+
| GATE_ID | FARM_ID | FARM_VALUE |
+---------+---------+------------+
| 1 | 2 | 0 |
| 1 | 3 | 0 |
| 1 | 4 | 540 |
| 2 | 4 | 550 |
| 3 | 4 | 560 |
| 4 | 4 | 570 |
| 5 | 4 | 580 |
| 6 | 4 | 590 |
| 1 | 5 | 1080 |
| 2 | 5 | 1100 |
| 3 | 5 | 1120 |
| 4 | 5 | 1140 |
| 5 | 5 | 1160 |
| 6 | 5 | 1180 |
| 1 | 6 | 455 |
| 2 | 6 | 536 |
| 3 | 6 | 617 |
| 4 | 6 | 698 |
| 5 | 6 | 779 |
| 6 | 6 | 860 |
| 1 | 7 | 0 |
| 1 | 8 | 0 |
| 1 | 10 | 0 |
+---------+---------+------------+

表格来源

+--------+----------+
| ORI_ID | ORI_NAME |
+--------+----------+
| 1 | US |
| 2 | CA |
| 3 | MX |
+--------+----------+

表库存

+--------+--------+-------------------+
|Stock_ID| ORI_ID | Stock_TITLE |
+--------+--------+-------------------+
| 1 | 1 | P1 |
| 2 | 2 | P3 |
| 3 | 3 | Q4 |
| 4 | 3 | B3 |
+--------+--------+-------------------+

表格结果

+-----------+---------+---------+------------+------------+
| RESULT_ID | FARM_ID | GATE_ID | FARM_VALUE | Score% |
+-----------+---------+---------+------------+------------+
| 1 | 7 | 1 | 333 | 100 |
| 2 | 8 | 1 | 825 | 100 |
| 3 | 6 | 1 | 455 | 40 |
| 4 | 6 | 2 | 536 | 0 |
| 5 | 6 | 3 | 617 | 0 |
| 6 | 6 | 4 | 698 | 100 |
| 7 | 6 | 5 | 779 | 0 |
| 8 | 6 | 6 | 860 | 10 |
| 9 | 4 | 1 | 540 | 100 |
| 10 | 4 | 2 | 550 | 90 |
| 11 | 4 | 3 | 560 | 0 |
| 12 | 4 | 4 | 570 | 100 |
| 13 | 4 | 5 | 580 | 10 |
| 14 | 4 | 6 | 590 | 0 |
| 15 | 5 | 1 | 1080 | 0 |
| 16 | 5 | 2 | 1100 | 0 |
| 17 | 5 | 3 | 1120 | 0 |
| 18 | 5 | 4 | 1140 | 50 |
| 19 | 5 | 5 | 1160 | 0 |
| 20 | 5 | 6 | 1180 | 100 |
| 21 | 3 | 1 | 11 | 100 |
| 22 | 10 | 1 | 98 | 90 |
| 23 | 2 | 1 | 202 | 100 |
+-----------+---------+---------+------------+------------+

注释结果表:同上^

+-----------+---------+---------+------------+------------+
| RESULT_ID | FARM_ID | GATE_ID | FARM_VALUE | Score% |
+-----------+---------+---------+------------+------------+

+-----------+---------+---------+------------+------------+
| 1 | 7 | 1 | 333 | 100 | <--|H-Case {H}
+-----------+---------+---------+------------+------------+

+-----------+---------+---------+------------+------------+
| 2 | 8 | 1 | 825 | 100 | <--|M-Case {M}
+-----------+---------+---------+------------+------------+

+-----------+---------+---------+------------+------------+
| 3 | 6 | 1 | 455 | 40 |
| 4 | 6 | 2 | 536 | 0 |
| 5 | 6 | 3 | 617 | 0 |
| 6 | 6 | 4 | 698 | 100 | <--|L
| 7 | 6 | 5 | 779 | 0 | |
| 8 | 6 | 6 | 860 | 10 | |
+-----------+---------+---------+------------+------------+ |
| 9 | 4 | 1 | 540 | 100 | |
| 10 | 4 | 2 | 550 | 90 | |
| 11 | 4 | 3 | 560 | 0 | |
| 12 | 4 | 4 | 570 | 100 | <--+M-case {H,M,L}
| 13 | 4 | 5 | 580 | 10 | |
| 14 | 4 | 6 | 590 | 0 | |
+-----------+---------+---------+------------+------------+ |
| 15 | 5 | 1 | 1080 | 0 | |
| 16 | 5 | 2 | 1100 | 0 | |
| 17 | 5 | 3 | 1120 | 0 | |
| 18 | 5 | 4 | 1140 | 50 | <--|H
| 19 | 5 | 5 | 1160 | 0 |
| 20 | 5 | 6 | 1180 | 100 |
+-----------+---------+---------+------------+------------+

+-----------+---------+---------+------------+------------+
| 21 | 3 | 1 | 11 | 100 | <--|L
| 22 | 10 | 1 | 98 | 90 | <--+H-case {H,M,L}
| 23 | 2 | 1 | 202 | 100 | <--|M
+-----------+---------+---------+------------+------------+

所需的计算:

  • Type最多只能有三个值:{H, M, L};
  • 当所有值都存在时,它们的分级如下:H=70 M=20 L=10
  • 所有独特的情况都是

  • 案例 {H,M}:H=80 M=20

  • 案例{M,L}:M=60 L=40
  • 案例 {H,L} : H=90 L=10
  • 案例{H}:H=100
  • 案例{M}:M=100
  • 案例{L}:L=100
  • 案例 {H,M,L}:H=70 M=20 L=10

进一步说明

  • Onlyatleast one GATE 的股票,完全满意可以获得最高100分
    1. 示例:Q4 有 3 组 6 个 GATES;只需满足一个 GATE 集(有分数)。
    2. 当前的点必须乘以它所属的特定情况示例:Q4 有情况 {H,M,L} 这意味着 H=70; M=20; L=10 这将导致 (70*100%)+(20*50%)+(10*100%)=90 (回顾上面的结果表注释)
    3. 2.
  • 即使没有完全满足要求,仍应考虑和计算分数。当没有完全满足的门时,应保留获得 MAX 积分的门。 (如有不明白会再解释)

如果我们执行查询来理解表和数据,它将如下所示

+---------+-----------+---------------+-----------+---------+-----------+---------+
| Origin | Stock | Farm Title | Farm Value| Gate | Size | Score |
+---------+-----------+---------------+-----------+---------+-----------+---------+
| US | P1 | Perdue | 333 | 1 | H | 100 |
| US | P3 | Holstein | 825 | 1 | M | 100 |
| CA | Q4 | FarmOne | 455 | 1 | L | 40 |
| CA | Q4 | FarmOne | 536 | 2 | L | 0 |
| CA | Q4 | FarmOne | 617 | 3 | L | 0 |
| CA | Q4 | FarmOne | 698 | 4 | L | 100 |
| CA | Q4 | FarmOne | 779 | 5 | L | 0 |
| CA | Q4 | FarmOne | 860 | 6 | L | 10 |
| CA | Q4 | Beef Farm | 540 | 1 | H | 0 |
| CA | Q4 | Beef Farm | 550 | 2 | H | 90 |
| CA | Q4 | Beef Farm | 560 | 3 | H | 0 |
| CA | Q4 | Beef Farm | 570 | 4 | H | 100 |
| CA | Q4 | Beef Farm | 580 | 5 | H | 10 |
| CA | Q4 | Beef Farm | 590 | 6 | H | 0 |
| CA | Q4 | CattleOne | 1080 | 1 | M | 0 |
| CA | Q4 | CattleOne | 1100 | 2 | M | 0 |
| CA | Q4 | CattleOne | 1120 | 3 | M | 0 |
| CA | Q4 | CattleOne | 1140 | 4 | M | 50 |
| CA | Q4 | CattleOne | 1160 | 5 | M | 100 |
| CA | Q4 | CattleOne | 1180 | 6 | M | 0 |
| MX | B3 | Cow Mill | 11 | 1 | L | 100 |
| MX | B3 | Dotterers | 98 | 1 | H | 90 |
| MX | B3 | AgriZone | 202 | 1 | M | 100 |
+---------+-----------+---------------+-----------+---------+-----------+---------+

渴望结果

+---------+-------------------+-------+
| Origin | Stock | score |
+---------+-------------------+-------+
| US | P1 | 100 |
| US | P3 | 100 |
| CA | Q4 | 90 |
| MX | B3 | 93 |
+---------+-------------------+-------+

Explanation

由于 origin 有一个 stock,它由 3 个不同的 farm 组成,而这些 farm 有 6 个 每个门。只要一个 gate 集(数字匹配 gates)得分达到某个值,我们就可以认为整个 STOCK 已找到。这是 股票 可以被认为是 100 的唯一方式。

此外,重申一下,STOCK Q4 有案例:{H,M,L} 并且在某种程度上找到了所有gate (4)gate 4 的分数 (100% * H) + (50% * M) + (100% * L) 等于 (70*100%) + (20*50%) + (10 *100%) = 90

因此:(取自上面)

  | CA      |   Q4              |    90 |

QED

所以我需要帮助 的是创建子查询/子选择来完成此计算。我在下面的 SQL fiddle 链接中设置了上面场景中的所有内容(以及我一直在使用的正在进行的查询)。

非常感谢 stackoverflow 社区。

> The above problem in SqlFiddle can be found here <

最佳答案

这是我一直在处理的查询。但是,结果与您在问题中发布的结果略有不同:

select o.origin_name, s.stock_title, sum(
case f.size
when 'H' then
case
when sizes = 'H,L,M' then 70
when sizes = 'H,M' then 80
when sizes = 'H,L' then 90
when sizes = 'H' then 100
else 0
end
when 'M' then
case
when sizes = 'H,L,M' then 20
when sizes = 'H,M' then 20
when sizes = 'L,M' then 60
when sizes = 'M' then 100
else 0
end
else
case
when sizes = 'H,L,M' then 10
when sizes = 'L,M' then 40
when sizes = 'H,L' then 10
when sizes = 'L' then 100
else 0
end
end * r.score / 100) FinalScore
from farm f
join (
select f.stock_id, group_concat(distinct f.size order by f.size) sizes
from farm f
join results r on f.farm_id = r.farm_id
group by f.stock_id
) stockSizes on f.stock_id = stockSizes.stock_id
join results r on f.farm_id = r.farm_id
join (
select f.stock_id, r.gate_id
from results r
join farm f on r.farm_id = f.farm_id
group by f.stock_id, r.gate_id
having sum(r.score = 0) = 0
) FullGates
on FullGates.stock_id = f.stock_id and FullGates.gate_id = r.gate_id
join stock s on s.stock_id = f.stock_id
join origin o on o.origin_id = s.origin_id
group by o.origin_id, s.stock_id

结果:

+-------------+-------------+------------+| ORIGIN_NAME | STOCK_TITLE | FINALSCORE |+-------------+-------------+------------+| US          | P1          |         93 || CA          | P3          |         90 || MX          | Q4          |        100 || MX          | B3          |        100 |+-------------+-------------+------------+

让我知道这是否有效。

关于MySQL 我如何创建这个子查询?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9830062/

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