通过用户与城市关系查找国家/地区，其中用户与一个国家/地区的所有城市相关

Question

我有四个表：用户、国家、地区和城市

国家有很多地区，有很多城市。

一个用户可以关联到 0 对多的城市（希望访问）。

我如何编写 MySQL 查询（或 DQL）来找出用户与他们完全相关的地区和国家，即希望访问 他们所有个城市？

Answer 1

从 User 级开始。然后加入更大的实体，并使用 distinct 来减少重复。

SELECT distinct
  u.user_name
  ,r.region_name
  ,co.country_name
FROM users u
LEFT JOIN cities c ON u.wish_to_visit = c.city_name
LEFT JOIN regions r ON c.region = r.region_name
LEFT JOIN countries co ON r.country = co.country_name
WHERE u.user_name = 'John Doe'

编辑：仅查找用户想要访问的所有国家/地区

Post 问题更新，让我们只拉入用户想要访问其中所有城市的国家。在这种情况下，我们从国家开始，一直到用户。然后我们汇总以查看一个国家/地区的城市被捕获的百分比。

关键是只加入我们关心的用户（不过不要按他们过滤，这将消除 NULL，其中用户与城市不匹配）。毕竟，一个简单的 HAVING 将帮助我们过滤出完全匹配的国家。

SELECT
  c.country_name
  ,count(ci.city_name) as count_all_cities
  ,count(u.wish_to_visit) as count_user_cities
FROM countries c
LEFT JOIN regions r ON c.country_name = r.country
LEFT JOIN cities ci ON r.region_name = ci.region
LEFT JOIN users u ON ci.city_name = u.wish_to_visit AND u.user_name = 'John Doe'
GROUP BY c.country_name
HAVING count(ci.city_name) = count(u.wish_to_visit)

Answer 2

使用连接

select cu.country_name, r.region_name from User u join City c
on u.cityID=c.cityID join Region r on
c.regionID=r.regionID join Country cu
on r.countryID=cu.countryID
where u.userID=SOMEID

Answer 3

对于区域，比较城市数量和最大城市数量会很容易

SELECT user_id,
 Region_count.region_id
FROM User
JOIN 
(SELECT user_id, region_id, count(*) as reg_user_count
 FROM City
 WHERE user_id='ThisUser'
 GROUP BY region_id) Region_Count
  ON Region_Count.user_id=User.user_id 
JOIN 
(SELECT region_id, count(*) as reg_max
 FROM City
 GROUP BY region_id) as Region_Max
 ON Region_Max.region_id=Region_Count.region_id 
   AND Region_Max.reg_max=Region_Count.reg_user_count
 WHERE user_id='ThisUser'

你可以对国家做同样的事情。如果你想不通，请告诉我。

Answer 4

使用NOT EXISTS和LEFT JOIN

SELECT *
FROM Region r
WHERE NOT EXISTS(
   SELECT 1 
   FROM City c
   LEFT JOIN User u ON c.id_city = u.id_city and u.id_user = 'user_id'
   WHERE c.id_reg = r.id_reg and u.id_user IS NULL
)

这将仅查找用户希望访问所有城市的区域，但是，如果您想要访问国家/地区，则只需稍作修改

SELECT *
FROM Country ctr
WHERE NOT EXISTS(
   SELECT 1 
   FROM Region r
   LEFT JOIN City c ON r.id_reg = c.id_reg
   LEFT JOIN User u ON c.id_city = u.id_city and u.id_user = 'user_id'
   WHERE ctr.id_country = r.id_country and u.id_user IS NULL
)

通过用户与城市关系查找国家/地区，其中用户与一个国家/地区的所有城市相关

Find countries by a user to city relationship where a user is related to all cities of a country

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