* | * |
---|---|
链接 | SQL Basics: Top 10 customers by total payments amount |
难度 | 6kyu |
状态 | √ |
日期 | 2019-4-4 |
题意
题解1
select customer.customer_id, customer.email,
count(payment.customer_id) as payments_count,
cast(sum(payment.amount) as float) as total_amount
from customer
left join payment
on customer.customer_id = payment.customer_id
group by customer.customer_id
order by total_amount desc
limit 10
题解2,LZY
select C.customer_id, C.email,
count(*) as payments_count,
cast(sum(P.amount) as float) as total_amount
from customer as C, payment as P
where C.customer_id = P.customer_id
group by C.customer_id
order by total_amount desc
limit 10
拆解如下:
首先,参与的表只需要customer表(提供客户基本信息)及payment表(提供支付记录)。两张表合并(根据客户ID来join),可以得到payment表中的每笔支付记录对应到哪个客户。
select C.customer_id, C.email
from customer as C, payment as P
where C.customer_id = P.customer_id
接下来,需要汇总,统计每个客户有多少笔支付(count)以及总共花了多少(sum)。payment表中的amount字段表示了一笔交易记录的金额。汇总操作使用group by,字段显然是客户ID。
select C.customer_id, C.email
count(*) as payments_count,
cast(sum(P.amount) as float) as total_amount
from customer as C, payment as P
where C.customer_id = P.customer_id
group by C.customer_id
cast操作只是转换类型,numerical转换为float(题目要求)。
接下来,就简单了,只需要挑选消费最高的10个客户。根据消费金额倒序排下(order by操作),然后限制选择10个就行(limit操作)。
select C.customer_id, C.email,
count(*) as payments_count,
cast(sum(P.amount) as float) as total_amount
from customer as C, payment as P
where C.customer_id = P.customer_id
group by C.customer_id
order by total_amount desc
limit 10
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