每个区的用户数包括0计数

Question

我需要找到每个地区的用户数量。

我的数据库结构：

Table Name : countries
country_id  country_name
1           India
2           Singapore
3           Malaysia
4           United States

Table Name : states
state_id    state_name  country_id
1           Tamil Nadu  1
2           Andhra      1
.
.
.
20          Arizona     4


Table Name : districts
district_id     district_name   state_id
1               Trichy          1
2               Chennai         1
3               Hyderabad       2
.
.
.
100             Phoenix         20

Table Name : user_addresses
user_address_id user_id doorno  street  ... districtName    stateName   countryName     active 
1               1       10      ....    ... Trichy          Tamil Nadu  India           1
2               2       A12     ....    ... Chennai         Tamil Nadu  India           0
3               3       A2      ....    ... Chennai         Tamil Nadu  India           1
4               4       B2      ....    ... Chennai         Tamil Nadu  India           1
5               41      89      ....    ... Phoenix         Arizona     United States   1

user_addresses表包含用户的多个地址。但只有一个是活跃的。我需要找到一个地区和国家/地区的用户数量。

我使用了这个查询：

SELECT country_name, district_name, COUNT(*) 
FROM districts as d 
LEFT JOIN states as s USING (state_id) 
LEFT JOIN countries as c USING (country_id) 
LEFT JOIN user_addresses as ua on ua.districtName = d.district_name 
WHERE ua.active=1 
GROUP BY ua.district

我得到user_addresses中的所有地区及其数量。但我仍然在districts表中有一些区域，我需要将其显示为0。

我得到了什么：

country_name    district_name   COUNT(*)
India           Trichy          1
India           Chennai         2
United States   Phoenix         1

我需要什么：

country_name    district_name   COUNT(*)
India           Trichy          1
India           Chennai         2
India           Hyderabad       0
Singapore       ...             0
Malaysia        ...             0
...             ...             0
...             ...             0
United States   Phoenix         1

Answer 1

time_A = np.array(A['time'])
time_B = np.array(B['time'])
time_inter = np.intersect1d(time_A,time_B)
index_A = np.where(time_A == time_inter)
A1 = A[index_A]

一般SELECT country_name, district_name, COUNT(ua.active) FROM districts as d LEFT JOIN states as s USING (state_id) LEFT JOIN countries as c USING (country_id) LEFT JOIN user_addresses as ua on ua.districtName = d.district_name AND ua.active=1 GROUP BY country_name, district_name 规则说：

如果指定了GROUP BY子句，则SELECT列表中的每个列引用都必须标识分组列或者是set函数的参数。

当GROUP BY时，您通常不会在LEFT JOIN子句中放置右侧表格条件，如果您获得内部联接结果。转到WHERE子句以获得真实的ON结果！