【发布时间】:2015-01-09 12:54:48
【问题描述】:
我试图找出“在一个时间范围内,有多少独特的消息已发送给特定船上的人,这些文本之间的最短天数是多少”并显示它,包括计数。
人用“id”表示,船用“id2”表示,消息用“text”表示。
CREATE TABLE `stacktable` (
`timestamp` DATETIME NOT NULL,
`id` VARCHAR(15) NOT NULL,
`id2` VARCHAR(3) NULL DEFAULT NULL,
`text` VARCHAR(255) NULL DEFAULT NULL,
`id3` INT(10) NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`id3`)
);
insert into stacktable (timestamp,id,id2,text) VALUES
('2015-01-01 00:00:01',1,10,'ABC'),
('2015-01-01 00:00:01',2,11,'ABC'),
('2015-01-01 00:00:01',3,12,'ABC'),
('2015-01-01 00:00:02',3,12,'ABC'),
('2015-01-01 00:00:02',1,10,'ABC'),
('2015-01-04 00:00:01',1,10,'ABC'),
('2015-01-04 00:00:01',1,10,'BCD'),
('2015-01-04 00:00:01',2,11,'ABC'),
('2015-01-04 00:00:01',2,11,'BCD'),
('2015-01-04 00:00:01',3,12,'ABC'),
('2015-01-04 00:00:01',3,12,'BCD'),
('2015-01-04 00:00:01',3,13,'CDE'),
('2015-01-07 00:00:01',2,11,'BCD'),
('2015-01-07 00:00:01',3,12,'BCD'),
('2015-01-07 00:00:01',3,13,'CDE'),
('2015-01-07 00:00:01',3,13,'DEF'),
('2015-01-08 00:00:01',3,12,'ABC'),
('2015-01-08 00:00:01',4,14,'EFG'),
('2015-01-09 00:00:01',4,14,'EFG'),
('2015-01-09 00:00:02',4,15,'FGH'),
('2015-01-10 00:00:01',4,14,'EFG'),
('2015-01-10 00:00:01',4,14,'FGH'),
('2015-01-10 00:00:01',4,15,'FGH'),
('2015-01-11 00:00:01',4,14,'EFG'),
('2015-01-15 00:00:01',4,14,'EFG');
展示我想要实现的目标:
select * from stacktable where id = 1
timestamp id id2 text id3
2015-01-01 00:00:01 1 10 ABC 1 First entry for id+id2+text (ABC)
2015-01-01 00:00:02 1 10 ABC 5 Second entry for same keys id+id2+text 1 second later
2015-01-04 00:00:01 1 10 ABC 6 Third entry for same keys id+id2+text 2 days later
2015-01-04 00:00:01 1 10 BCD 7 First entry for id+id2+text (BCD)
我只想计算“在 2 天内具有相同 id、id2 和文本”的记录,但也显示“命中之间的最小 diffdate”。
我想要的输出是:
id id2 text count(*) mindiffdatebetweenhits
-------------------------------------------
1 10 ABC 3 0 count id3s 1,5 and 6, minimumdaydiff is between id3 1 and 5 = 0 days
3 12 ABC 3 0 count id3s 3,4 and 10, minimumdaydiff is between id3 3 and 4 = 0 days
4 14 EFG 4 1 count id3s 18,19,21 and 24, minimumdaydiff is equal between all hits = 1 day
4 15 FGH 2 0 count id3s 20 and 23, minimumdaydiff is between id3 20 and 23 = 0 days
我怎样才能得到想要的输出?
【问题讨论】:
-
您确定这将是您的输出吗?因为 id3 = 6 和 1 之间的时间差异超过 2 天。
-
我希望它为 id=1 (id3 1,5,6) 计数 3。这是因为 id3 1 和 id3 5
-
记录 3 和 4 的日期为 2015-01-01,记录 10 的日期为 2015-01-04。所以,间隔超过2天。但是你想计算 3 条记录。似乎与“自上次点击后 2 天内”不一致
-
对不起。在 id+id2+text 上的“最后一次”点击后 2 天内,不要从第一次点击开始持续检查 timediff,而是从最近的一次.. 更改它 1 秒,你是正确的,这将是 3 天。
-
id3 = 3 = 2015-01-01 00:00:01 id3 = 4 = 2015-01-01 00:00:02 .. 1 秒后 id3 = 10 = 2015-01-04 00:00:01 .. 2 天后。 SELECT TIMESTAMPDIFF(day, '2015-01-01 00:00:02', '2015-01-04 00:00:01') from dual
标签: mysql sql group-by datediff