用户与用户与用户与国家所有城市相关的城市关系找到国家

Question

我有四个表：用户，国家，地区和城市

国家有许多城市。

用户可以与0到多个城市有关（希望访问）。

我如何编写一个MySQL查询（或DQL），该查询（或DQL）找到了用户是完全的国家/地区，即希望访问所有城市？

Answer 1

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
)

这只会找到用户希望访问所有城市的区域，但是，如果您想要各个国家，那只是一个微小的修改

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
)

Answer 2

对于区域，很容易将城市数与最大城市数量进行比较

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'

您可以对国家做同样的事情。让我知道您是否无法弄清楚。

Answer 3

使用加入

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 4

从User级别开始。然后加入越来越大的实体，并使用独特的方式来减少重复。

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所有

的国家

发布问题更新，让我们只能在用户想要访问其中所有城市的国家/地区拉动。在这种情况下，我们从国家开始，然后向用户努力。然后，我们汇总以查看一个国家的城市被捕获的百分比。

关键是仅加入我们关心的用户（不过要过滤他们，这将消除用户不匹配城市的NULL）。毕竟，一个简单的HAVING将帮助我们过滤完全匹配的国家。

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)