4

我对这个问题感到非常疯狂，想知道是否有一个天才，对他来说这只是一个孩子的游戏......？

此查询应建议随机用户登录用户，而登录用户和建议用户之间的关系不得以下列方式之一被阻止：

  • 按大洲
  • 按国家
  • 按年龄段
  • 按性别
  • 按 id (blocker<->blockee)

表结构

用户

id | username | country_id_home | timestamp_lastonline | gender | birthdate | allow_gender | age_min | age_max | ...

注释：birthdate是 mysql-date 格式 ( YYYY-MM-DD), gender(0=female,1=male), allow_gender(0=both, 1=female, 2=male)

国家

country_id | name | continent_id | ...

城市

id | city_real | ...

评论：loc_id_home在表中user匹配idcities

封锁国家

id | user_id | country_id | ...

被封锁的大陆

id | user_id | continent_id | ...

阻塞用户

id | user1_id | user2_id | ...

评论：user1 是“blocker”，user2 是“blockee”

我到目前为止的查询是：

SELECT u.id AS user_id, u.username, u.fname, u.country_id_home, u.timestamp_reg, u.timestamp_upd, u.timestamp_lastonline, u.online_status, u.gender, u.birthdate, u.prof_pic,

(SELECT name FROM countries co WHERE co.country_id=u.country_id_home) AS country_name,
(SELECT city_real FROM cities ci WHERE ci.id=u.loc_id_home) AS city ,
(SELECT region_name FROM regions r WHERE r.id=u.region_id_home) AS region,
(SELECT region_short FROM regions r WHERE r.id=u.region_id_home) AS region_short

FROM user_d1 u 

LEFT JOIN (
SELECT DISTINCT user2_id AS blockee_id
FROM blocked_user
WHERE user1_id = :user1_id1
)this_user_blocked ON u.id = this_user_blocked.blockee_id

LEFT JOIN (
SELECT DISTINCT user1_id AS blocker_id
FROM blocked_user
WHERE user2_id = :user1_id2
)blocked_this_user ON u.id = blocked_this_user.blocker_id

WHERE 

(allow_gender=0 OR allow_gender=:user1_allow_gender)

AND age_min<=:user1_age1
AND age_max>=:user1_age2


AND prof_status<>2

AND this_user_blocked.blockee_id IS NULL AND blocked_this_user.blocker_id IS NULL 

AND NOT EXISTS (SELECT 1 FROM blocked_countries bc WHERE bc.user_id=u.id AND bc.country_id =:user1_country)
AND NOT EXISTS (SELECT 1 FROM blocked_countries bc WHERE bc.user_id=:user1_id3 AND bc.country_id = u.country_id_home)

AND EXISTS (
    SELECT 1 FROM user_d1 u2 
    WHERE u2.id=:user1_id4
    AND (u2.allow_gender=0 OR u2.allow_gender=(u.gender+1))
    AND age_min<=(DATEDIFF(NOW(),u.birthdate))
    AND age_max>=(DATEDIFF(NOW(),u.birthdate))
    AND prof_status<>2
    )

LIMIT 0,6

没有最后AND EXISTS...一个结果集不是空的，但是像这样它总是不返回任何结果。然而被大陆封锁不包括在内，它还没有被timestamp_lastonline...订购

我最后做错了AND EXISTS...什么？应该用 a 来完成JOIN吗？我也试过了，但它也给出了一个空的结果......

我已将登录用户的所有必要变量存储在 SESSION-variables 中......如果您在查询中发现可以以更好的方式完成的任何其他内容，我也很乐意为您提供帮助！

非常感谢您！

4

1 回答 1

6

我认为这是因为DATEDIFF，它返回两个日期之间的天数而不是年数。因此DATEDIFF可能总是大于age_max。

编辑：你可以这样做：

...
AND DATE_SUB(NOW(), INTERVAL age_max YEAR) <= u.birthdate
AND DATE_SUB(NOW(), INTERVAL age_min YEAR) >= u.birthdate
...
于 2012-07-06T13:06:33.647 回答