我有一个问题:
SELECT users.name,
COUNT(DISTINCT leads.id) as leads,
COUNT(DISTINCT properties.id) as properties,
COUNT(DISTINCT buyers.id) as buyers,
null
FROM users
LEFT JOIN (SELECT * from leads WHERE date(created_at) BETWEEN '2019-04-13' AND'2019-04-13') leads ON users.id = leads.user_id
LEFT JOIN (SELECT * from properties WHERE date(contract_start_date) BETWEEN '2019-04-13' AND'2019-04-13') as properties ON users.id = properties.user_id
LEFT JOIN (SELECT * from buyers WHERE date(created_at) BETWEEN '2019-04-13' AND'2019-04-13') as buyers ON users.id = buyers.user_id
LEFT JOIN buyer_searches ON buyers.id = buyer_searches.buyer_id
GROUP BY users.id
UNION
SELECT users.name,
null, null, null,
COUNT(DISTINCT buyer_searches.id) as buyer_searches
FROM users
LEFT JOIN buyers ON users.id = buyers.user_id
LEFT JOIN (SELECT * from buyer_searches WHERE date(created_at) BETWEEN '2019-04-08' AND'2019-04-13') as buyer_searches ON buyers.id = buyer_searches.buyer_id
GROUP BY users.id当然,我得到了一个包含重复用户的表,但参数正确:
users field1 field2 field3 etc....
user1 data . null . null etc...
user2 data . null . null etc...
user1 null . data . data etc...
user2 null . data .data etc...
etc...我想按用户连接这些行。
发布于 2019-05-04 18:46:39
在SELECT中使用子查询执行以下操作:
SELECT u.name,
(SELECT COUNT(*)
FROM leads l
WHERE l.user_id = u.id AND
l.created_at >= '2019-04-13' AND
l.created_at < '2019-04-14'
) as num_leads,
(SELECT COUNT(*)
FROM properties p
WHERE p.user_id = u.id AND
p.created_at >= '2019-04-13' AND
p.created_at < '2019-04-14'
) as num_buyers,
(SELECT COUNT(*)
FROM buyers b
WHERE b.user_id = u.id AND
b.created_at >= '2019-04-13' AND
b.created_at < '2019-04-14'
) as num_buyers,
(SELECT COUNT(*)
FROM buyers b JOIN
buyer_searches bs
ON b2.buyer_id = b.id AND
WHERE b.user_id = u.id AND
bs.created_at >= '2019-04-13' AND
bs.created_at < '2019-04-14'
) as num_buyer_searches
FROM users u;通过将这些表JOIN在一起,您正在为每个用户创建一个(潜在的)巨大的笛卡尔产品。然后对此进行筛选,以计算不同的ids。直接对它们进行计数效率更高。
每个用户都应该有一行。
请注意,我已经更改了日期比较。列上的函数几乎总是排除了索引的使用。这个版本的查询可以利用表上的索引--比如leads(user_id, created_at) --如果它们可用的话。
https://stackoverflow.com/questions/55981703
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