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我使用了 mysql 5.7 版，我为每种产品、数量和使用许多这样的探险都有一个表格制作

+---------+-----------------+------+--------+---------+
| Product | Type_Expedition | Pack | Amount |  Weight |
+---------+-----------------+------+--------+---------+
| Chicken | A               |    1 |      2 |       2 |
| Beef    | A               |    1 |      2 |       2 |
| Lamb    | B               |    1 |      2 |       2 |
| Beef    | B               |    2 |      2 |       4 |
| Chicken | A               |    3 |      2 |       6 |
| Lamb    | A               |    1 |      1 |       1 |
| Lamb    | A               |    1 |      1 |       1 |
+---------+-----------------+------+--------+---------+

如何计算 type_expedition B 和非 B 的重量和金额的总和（除 B 之外的所有类型的探险）？

我假设使用这种语法（对不起，我想使用 dbfiddle.uk 但这是错误的）

select product, type_expedition, pack, amount, weight, (sum(amount) where type_expedition = B), (sum(weight) where type_expedition = B) from my_table

预期成绩

+---------------------------------------------------+---+----+
|   Total amount and weight for type_expedition B   | 4 | 6  |
+---------------------------------------------------+---+----+
| Total amount and weight for type_expedition NON B | 8 | 12 |
+---------------------------------------------------+---+----+
4

1 回答 1

1

您可以对最后 2 行使用 UNION ALL：

select t.Product, t.Type_Expedition, t.Pack, t.Amount, t.Weight
from (
  select *, 0 sort from my_table
  union all
  select 'Total Amount and Weight for expedition B', null, null,
    sum(amount),
    sum(weight), 1
  from my_table  
  where Type_Expedition = 'B'
  union all
  select 'Total Amount and Weight for expedition not B', null, null,
    sum(amount),
    sum(weight), 2
  from my_table 
  where Type_Expedition <> 'B'
) t
order by t.sort

请参阅演示。
结果：

| Product                                      | Type_Expedition | Pack | Amount | Weight |
| -------------------------------------------- | --------------- | ---- | ------ | ------ |
| Beef                                         | A               | 1    | 2      | 2      |
| Chicken                                      | A               | 3    | 2      | 6      |
| Lamb                                         | B               | 1    | 2      | 2      |
| Lamb                                         | A               | 1    | 1      | 1      |
| Chicken                                      | A               | 1    | 2      | 2      |
| Beef                                         | B               | 2    | 2      | 4      |
| Lamb                                         | A               | 1    | 1      | 1      |
| Total Amount and Weight for expedition B     |                 |      | 4      | 6      |
| Total Amount and Weight for expedition not B |                 |      | 8      | 12     |

如果您只想要最后 2 行的总数：

select 
  case Type_Expedition 
    when 'B' then 'Total Amount and Weight for expedition B'
    else 'Total Amount and Weight for expedition not B'
  end type,
  sum(amount),
  sum(weight)
from my_table
group by type

请参阅演示。
结果：

| Total Amount and Weight for expedition B     | 4           | 6           |
| Total Amount and Weight for expedition not B | 8           | 12          |
于 2020-04-15T08:48:52.693 回答