我有一个查询：
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
我得到的表中当然有重复的用户，但是参数正确：
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...
我想按用户连接行。
答案 0 :(得分：0)
为此在<img src="assets/img/app-nec/baby_and_kids-compressor.jpg">
中使用子查询：
SELECT通过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;
将这些表放在一起，您将为每个用户创建一个（可能）巨大的笛卡尔积。然后，您正在筛选以计数不同的ID。直接计数它们会更有效。
每个用户应该有一行。
请注意，我已经更改了日期比较。列上的函数几乎总是禁止使用索引。此版本的查询可以利用表上的索引（例如JOIN）（如果可用）。