Question

我有以下名为tbl_pet_owners的MySQL表：

+----+--------+----------+--------+--------------+
| id | name   | pet      | city   | date_adopted |
+----+--------+----------+--------+--------------+
|  1 | jane   | cat      | Boston | 2017-07-11   |
|  2 | jane   | dog      | Boston | 2017-07-11   |
|  3 | jane   | cat      | Boston | 2017-06-11   |
|  4 | jack   | cat      | Boston | 2016-07-11   |
|  5 | jim    | snake    | Boston | 2017-07-11   |
|  6 | jim    | goldfish | Boston | 2017-07-11   |
|  7 | joseph | cat      | NYC    | 2016-07-11   |
|  8 | sam    | cat      | NYC    | 2017-07-11   |
|  9 | drew   | dog      | NYC    | 2016-07-11   |
+----+--------+----------+--------+--------------+

在之前的Stack Overflow帖子中，我使用COUNT寻求帮助以获得每个城市的宠物数量，但如果一个人拥有两个或更多相同类型的宠物，那么这些宠物将被计为一个。宠物类型列在另一个名为tbl_pet_types的MySQL表中：

+----------+-------------+
| pet      | type        |
+----------+-------------+
| cat      | mammal      |
| dog      | mammal      |
| goldfish | fish        |
| goldfish | seacreature |
| snake    | reptile     |
+----------+-------------+

以下是此代码的工作代码：

select count(*), result.city from (
    select owners.city, types.type, owners.name
    from tbl_pet_owners owners
    left join tbl_pet_types types on owners.pet = types.pet group by owners.city, owners.name, types.type
) as result
group by result.city;

我试图修改代码，以便只在2017-01-01＆＃39;之间采用宠物。和＆＃39; 2017-08-01＆＃39;被计算在内。因此，在这个例子中，杰克的猫，约瑟夫的猫和画狗不会被计算在内。

我尝试在我的查询中添加where语句，但是我收到了很多语法错误：

select count(*), result.city from (
    select owners.city, types.type, owners.name
    from tbl_pet_owners owners
    left join tbl_pet_types types on owners.pet = types.pet group by owners.city, owners.name, types.type
) as result where result.date_adopted > '2017-01-01' 
and result.date_adopted < '2017-08-01'
group by result.city;

关于如何实现这一目标的任何提示？

Answer 1

尝试这样的事情：

select count(*), result.city from (
    select owners.city, types.type, owners.name
    from tbl_pet_owners owners
    left join tbl_pet_types types on owners.pet = types.pet 
    where owners.date_adopted BETWEEN '2012-12-25 00:00:00' AND '2012-12-25 23:59:59'
    group by owners.city, owners.name, types.type
) as result 
group by result.city;

过滤MySQL COUNT查询结果

1 个答案: