Question

我需要一个MySQL实现的精彩代码来回答这个问题：

我有完全相同的问题 - 但（据我所知）MySQL不支持DENSE_RANK()或ROW_NUMBER()

如果有人能指出我正确的方向，那就太好了。我已经尝试使用基于月份的查询版本：forums.mysql.com: Group By Consecutive Dates和here，但查询陷入了300,000个不是非常连续的6,000个“用户”的记录

非常感谢

Answer 1

行。因此，除了“自上一个范围结束后的月份”字段外，此答案应该为您提供所需的一切。也许你可以在它上面构建并自己弄明白;-)。我已将#data表（在日期连续条目中提及）重命名为payment：

CREATE TABLE payment
(
Contact_reference VARCHAR(55),
Date_payment DATETIME,
Payment_value double
);

INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2003-06-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2004-06-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2004-12-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-04-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-05-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-06-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-07-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-08-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-09-08',12.82);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-10-10',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-11-10',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2005-12-10',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-01-10',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-02-10',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-02-28',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-04-12',12.8205);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2006-05-10',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-06-11',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-07-10',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-08-09',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-09-10',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-10-09',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-11-09',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2007-12-10',19.2308);
INSERT INTO payment VALUES ('18EC3CD2-3065-4FF4-BE40-000004228590','2008-01-10',19.2308);

select Contact_reference,
consecStartDate as Range_start,
max(Date_payment) as Range_end,
count(*) as Payments,
sum(Payment_value) as value
from
(
select Contact_reference,
case when year(Date_payment)*12 + month(Date_payment) - @curMonthNum  <= 1 and @curRef = Contact_reference then 'Y' else 'N' end as consec,
case when year(Date_payment)*12 + month(Date_payment) - @curMonthNum  <= 1 and @curRef = Contact_reference then @consecStartDate := @consecStartDate else @consecStartDate := Date_payment  end as consecStartDate,
Date_payment,Payment_value,
@curMonthNum := year(Date_payment)*12 + month(Date_payment),
@curRef := Contact_reference
from payment
inner join (SELECT @consecStartDate := null,@curMonthNum := null,@consecStart := null,@curRef := null) as t
order by Contact_reference desc,Date_payment asc
) n
group by Contact_reference,consecStartDate;

就像我说的那样 - 它不会给你自上一个范围结束以来的间隔，但我认为答案的一部分总比没有好！

MySQL连续日期条件的实现

1 个答案: