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[英]Write a query that displays the names of cities and their countries when the capital city is the largest of all cities listed for that country
[英]Find countries by a user to city relationship where a user is related to all cities of a country
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。
用户可以与0到许多城市相关(希望访问)。
我该如何编写MySQL查询(或DQL)来查找与用户完全相关的地区和国家(即希望访问其所有城市)?
对于地区而言,比较城市数与最大城市数很容易
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'
您可以对国家/地区做同样的事情。 让我知道您是否无法解决。
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
)
从User
级别开始。 然后加入更大的实体,并使用distinct减少重复项。
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国家的国家
发布问题更新后,让我们仅进入用户希望访问其中所有城市的国家/地区。 在这种情况下,我们从国家/地区开始,一直到用户。 然后,我们合计查看一个国家的城市所占的百分比。
关键是仅加入我们关注的用户(尽管不对其进行过滤,这将消除NULL
,即用户与城市不匹配)。 毕竟,简单的HAVING
将有助于我们仅针对完全匹配的国家/地区进行过滤。
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)
使用联接
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
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