sql – 查询滚动日期范围内不同值的计数
我有一组电子邮件地址和日期,这些电子邮件地址已添加到表格中.各种不同日期的电子邮件地址可以有多个条目.例如,如果我有下面的数据集.我希望得到我们在所述日期和3天前之间的不同电子邮件的日期和数量. Date | email -------+---------------- 1/1/12 | test@test.com 1/1/12 | test1@test.com 1/1/12 | test2@test.com 1/2/12 | test1@test.com 1/2/12 | test2@test.com 1/3/12 | test@test.com 1/4/12 | test@test.com 1/5/12 | test@test.com 1/5/12 | test@test.com 1/6/12 | test@test.com 1/6/12 | test@test.com 1/6/12 | test1@test.com 如果我们使用3的日期,结果集看起来像这样 date | count(distinct email) -------+------ 1/1/12 | 3 1/2/12 | 3 1/3/12 | 3 1/4/12 | 3 1/5/12 | 2 1/6/12 | 2 我可以使用下面的查询获得日期范围的明确计数,但是希望按天计算范围,这样我就不必手动更新数百个日期的范围. select test.date,count(distinct test.email) from test_table as test where test.date between '2012-01-01' and '2012-05-08' group by test.date; 感谢帮助. 解决方法测试用例:CREATE TEMP TABLE tbl (day date,email text); INSERT INTO tbl VALUES ('2012-01-01','test@test.com'),('2012-01-01','test1@test.com'),'test2@test.com'),('2012-01-02',('2012-01-03',('2012-01-04',('2012-01-05',('2012-01-06','test1@test.com`'); 查询 – 仅返回tbl中存在条目的天数: SELECT day,(SELECT count(DISTINCT email) FROM tbl WHERE day BETWEEN t.day - 2 AND t.day -- period of 3 days ) AS dist_emails FROM tbl t WHERE day BETWEEN '2012-01-01' AND '2012-01-06' GROUP BY 1 ORDER BY 1; 或者 – 返回指定范围内的所有日期,即使当天没有行: SELECT day,(SELECT count(DISTINCT email) FROM tbl WHERE day BETWEEN g.day - 2 AND g.day ) AS dist_emails FROM (SELECT generate_series('2012-01-01'::date,'2012-01-06'::date,'1d')::date) AS g(day) 结果: day | dist_emails -----------+------------ 2012-01-01 | 3 2012-01-02 | 3 2012-01-03 | 3 2012-01-04 | 3 2012-01-05 | 1 2012-01-06 | 2 这听起来像window functions的工作,但我没有找到一种方法来定义合适的窗框.另外,per documentation:
所以我用相关的子查询解决了它.我猜这是最聪明的方式. 我将您的日期列重命名为day,因为使用类型名称作为标识符是不好的做法. 顺便说一下,“在所述日期和3天前之间”将是4天的时间段.你的定义在那里是矛盾的. 有点短,但只有几天慢: SELECT day,count(DISTINCT email) AS dist_emails FROM (SELECT generate_series('2013-01-01'::date,'2013-01-06'::date,'1d')::date) AS g(day) LEFT JOIN tbl t ON t.day BETWEEN g.day - 2 AND g.day GROUP BY 1 ORDER BY 1; (编辑:开发网_开封站长网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |