Mysql在新查询中使用查询结果

Question

所以我有两个表，一个叫做points_log，另一个叫做排行榜。

mysql> describe points_log; 
+---------+---------+------+-----+---------+-------+ 
| Field | Type | Null | Key | Default | Extra | 
+---------+---------+------+-----+---------+-------+ 
| user_id | int(11) | NO |  | NULL |  | 
| points | int(11) | YES |  | 0  |  | 
| date | date | NO |  | NULL |  | 
+---------+---------+------+-----+---------+-------+ 
3 rows in set (0.00 sec) 

mysql> describe leaderboard; 
+-----------+--------------+------+-----+---------+-------+ 
| Field  | Type   | Null | Key | Default | Extra | 
+-----------+--------------+------+-----+---------+-------+ 
| bucket | varchar(255) | YES |  | NULL |  | 
| user_id | int(11)  | YES |  | NULL |  | 
| school_id | int(11)  | YES |  | NULL |  | 
+-----------+--------------+------+-----+---------+-------+ 
3 rows in set (0.00 sec) 

我有以下查询：

SELECT leaderboard.user_id FROM leaderboard where 
leaderboard.bucket=(SELECT bucket FROM leaderboard WHERE leaderboard.user_id=$user_id) AND 
leaderboard.school_id = (SELECT school_id FROM leaderboard WHERE leaderboard.user_id=$user_id) 

这将返回与USER_ID的是与传入$ user_ID的斗一行或多行我想要做的是采取所有这些USER_ID的，发现运行下面的查询

SELECT sum(points) FROM points_log WHERE user_id=$user_id AND 
date >= (SELECT subdate(curdate(), INTERVAL (weekday(now())) DAY)) 

的问题是本次查询，如果不能保证返回的东西，这样的情况下，它不返回任何我想要的总和（点）为0。我还需要返回每行的user_id，bucket和sum（points）。

现在我有什么是

SELECT leaderboard.user_id,sum(points_log.points) AS points, leaderboard.bucket 
     FROM points_log LEFT JOIN leaderboard ON points_log.user_id = leaderboard.user_id 
     WHERE points_log.DATE >= (SELECT subdate(curdate(), INTERVAL (weekday(now())) DAY)) 
     AND leaderboard.bucket=(SELECT bucket FROM leaderboard WHERE leaderboard.user_id=$user_id) 
     AND leaderboard.school_id = (SELECT school_id FROM leaderboard WHERE leaderboard.user_id=$user_id) 
     GROUP BY USER_ID ORDER BY SUM(points) DESC 

与此问题是，当存在points_log该用户的价值，它才会起作用。如果没有值，我不确定如何将其设置为0。

任何帮助，非常感谢！

Answer 1

SELECT leaderboard.user_id, COALESCE(sum(points_log.points), 0)AS points, leaderboard.bucket 
FROM points_log RIGTH OUTER JOIN leaderboard ON points_log.user_id = leaderboard.user_id 
WHERE points_log.DATE >= (SELECT subdate(curdate(), INTERVAL (weekday(now())) DAY)) 
AND leaderboard.bucket=(SELECT bucket FROM leaderboard WHERE leaderboard.user_id=$user_id) 
AND leaderboard.school_id = (SELECT school_id FROM leaderboard WHERE leaderboard.user_id=$user_id) 
GROUP BY USER_ID ORDER BY SUM(points) DESC 

试试这个......记下Outer Join和COALESCE函数。