Mysql 获取每只哺乳动物携带的水果总量

我正在努力解决特定的基于MySQL查询的问题

我有这样的结果集：

Table: Report
 mammal_id    mammal_name   fruit_name   gets  |  total  |
    3         rabbit        apple         4    |    5    |
    3         rabbit        carrot        4    |    4    |
    3         rabbit        cabbage       1    |    3    |
    2         squirrel      nuts          1    |    3    |
    2         squirrel      cabbage       2    |    2    |
    1         chipmunk      nuts          2    |    2    |
    1         chipmunk      apple         1    |    1    |

And I want to filter like this:

Table: Filtered
 mammal_id    mammal_name   fruit_name   has  
    3         rabbit        apple         4
    3         rabbit        carrot        4
    3         rabbit        cabbage       1
    2         squirrel      nuts          1
    2         squirrel      cabbage       2
    1         chipmunk      nuts          2
    1         chipmunk      apple         1

洞的要点是得到每只哺乳动物携带的水果总量

现在我有：

SELECT a.mammal_id, b.mammal_id, a.mammal_name, b.mammal_name, a.fruit_name, b.fruit_name, (b.total - a.total) as has
FROM (SELECT * FROM Report (result set)) as a 
INNER JOIN (SELECT * FROM Report (result set)) as b 
ON a.fruit_name=b.fruit_name WHERE  a.mammal_id = b.mammal_id-1

在这个查询之后，我得到如下结果：

Table: Result
a.mammal_id  b.mammal_id  a.mammal_name  b.mammal_name  a.fruit_name  b.fruit_name  has
   2             3            squirrel      rabbit         cabbage        cabbge     1
   1             2            chipmunk      squirrel       nuts           nuts       1

---

感谢您对这个问题的任何指导。

下面的查询将为您提供所需的准确输出

SELECT mammal_id, mammal_name, fruit_name, SUM(gets) as has
FROM Report GROUP BY mammal_name, fruit_name

---

下面的查询应该准确地提供您所需的输出

SELECT mammal_id, mammal_name, fruit_name, SUM(gets) as has
FROM Report GROUP BY mammal_name, fruit_name

---

要获取每个动物在“获取”列中的项目总数，您需要

SELECT mammal_id, mammal_name, SUM(gets)
FROM Report 
GROUP BY mammal_id, mammal_name

---

如果这不是你要问的，请澄清问题。

要获取每只动物在“获取”列中的项目总数，你需要

SELECT mammal_id, mammal_name, SUM(gets)
FROM Report 
GROUP BY mammal_id, mammal_name

---

如果这不是您要问的问题，请澄清问题。

谢谢！比我想象的容易多了。谢谢！比我想象的容易多了。