需要一些帮助来构建查询,这是我目前的方案:
users:
+----+------------+
| id | username |
+----+------------+
| 1 | rob |
| 2 | john |
| 3 | jane | <--- jane never has donated
| 4 | mike |
+----+------------+
donations:
+--------------------+------------+
| uid | amount | date |
+---------+----------+------------+
| 1 | 20 | 2013-10-10 |
| 2 | 5 | 2013-10-03 |
| 2 | 50 | 2013-09-25 |
| 2 | 5 | 2013-10-01 |
| 4 | 100 | 2012-10-01 | <-- past year
+---------+----------+------------+
Result I want:
+---------+-------------+---------+-------------+---------------+----------+
| id | username | amount | monthly | totalamount | total |
+---------+-------------+---------+-------------+ --------------+----------+
| 1 | rob | 20 | 1 | 20 | 1 |
| 2 | john | 60 | 3 | 60 | 3 |
| 3 | jane | 0 | 0 | 0 | 0 |
| 4 | mike | 0 | 0 | 100 | 1 |
+---------+-------------+-----------------------+---------------+----------+
这是我的疑问:
SELECT
u.*,
COALESCE(sum(d.amount), 0) amount,
COUNT(d.uid) monthly,
COUNT(d.amount) as Total, <-- need to get sum all time donations and number of times donated
FROM users u
LEFT JOIN donations d
ON u.id = d.uid
AND (month(d.date), year(d.date)) = (month(CURDATE()), year(CURDATE()))
GROUP BY u.id ORDER BY u.id ASC
所以我需要从相同的数据中添加2个不同的总和。
编辑:http://sqlfiddle.com/#!2/20a974/9架构和数据
我怎么能这样做?
答案 0 :(得分:2)
为此,我们需要过滤选择而不是连接上的数据。
删除此条件:
AND (month(d.date), year(d.date)) = (month(CURDATE()), year(CURDATE()))
并将其添加到选择:
SUM (CASE WHEN (month(d.date), year(d.date)) = (month(CURDATE()), year(CURDATE())) THEN 1 ELSE 0 END) as monthly
编辑:
整个查询:
SELECT users.id, users.username,
COALESCE(sum(CASE WHEN (month(donations.date), year(donations.date)) = (month(CURDATE()), year(CURDATE())) THEN donations.amount ELSE 0 END), 0) monthly_sum,
COALESCE(sum(CASE WHEN (month(donations.date), year(donations.date)) = (month(CURDATE()), year(CURDATE())) THEN 1 ELSE 0 END), 0) monthly_amount,
COALESCE(sum(donations.amount), 0) total_sum,
count(*) total_amount
from users
left join donations
on donations.uid = users.id
group by users.id, users.username
答案 1 :(得分:1)
对我来说,考虑单独分组信息的最简单方法是将其放入单独的查询中,然后将结果重新加入。这可能不是最有效的,但它有助于让事情有效。
select auo.id, auo.username,
coalesce(monthly_count, 0), coalesce(monthly_total, 0),
coalesce(total, 0), coalesce(total_amount, 0)
from aaa_users auo
left join (
select au.id as id, count(adm.amount) as monthly_count, SUM(adm.amount) as monthly_total
from aaa_users au join aaa_donations adm on au.id = adm.uid and adm.donate_date > GETDATE()-30
group by au.id
) as monthly on monthly.id = auo.id
left join (
select au.id as id, count(ady.amount) total, SUM(ady.amount) as total_amount
from aaa_users au join aaa_donations ady on au.id = ady.uid and ady.donate_date > getDate()-450
group by au.id
) as yearly on yearly.id = auo.id
正如@CompuChip所说,加入捐赠表两次更加清晰,但是我的连接逻辑出了问题,因为john的值正在重复。我认为需要有一个捐赠专栏,以防止每月和全部捐款被合并。无论如何,这是一个例子,即使它不能正常工作
select au.id, au.username,
count(adm.amount), SUM(adm.amount) as monthly_total,
count(ady.amount), SUM(ady.amount) as total_amount
from aaa_users au
left outer join aaa_donations adm on au.id = adm.uid and adm.donate_date > GETDATE()-60
left outer join aaa_donations ady on au.id = ady.uid and ady.donate_date > getDate()-450
group by au.id, au.username
order by au.id, au.username
答案 2 :(得分:0)
你可以再做一次捐赠,给它一个不同的别名:在d2.uid = u.id上LEFT JOIN捐款d2。然后对最后两个字段的d2.amount求和,例如
SELECT u.*,
COALESCE(sum(d.amount), 0) amount,
COUNT(d.uid) monthly,
COUNT(d.amount) as Total,
COALESCE(sum(d2.amount), 0) amountAll,
COUNT(d2.uid) monthlyAll,
COUNT(d2.amount) as TotalAll
FROM users u
LEFT JOIN donations d ON u.id = d.uid AND (month(d.date), year(d.date)) = (month(CURDATE()), year(CURDATE()))
LEFT JOIN donations d2 ON u.id = d2.uid
GROUP BY u.id ORDER BY u.id ASC