Question

我试图找出＆＃34;在一段时间内，特定船上的人发送了多少个独特的消息，这些消息之间的最短天数是多少？并显示它包括计数。

一个人由＆＃39; id＆＃39;代表，船由＆＃39; id2＆＃39;和＃39;文字＆＃39;。

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＆＃34;期间内具有＆＃34;相同id，id2和文本的记录，还要显示在点击次数和＃34;之间的最小差异天数。

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

如何获得所需的输出？

Answer 1

这应该这样做，假设只丢弃一行的序列：

select id, id2, text, seq, count(id) as total, min(diff) as mindiff
from (
      select t1.row, t2.row row2, t1.id, t1.id2, t1.text, t1.id3, 
             TIMESTAMPDIFF(DAY, t1.timestamp, t2.timestamp) as diff,
             IF (TIMESTAMPDIFF(DAY, t1.timestamp, t2.timestamp) > 2, @seq * (1 and @seq := @seq +1), @seq) as seq
      from (select (@row := @row + 1) as row, id, id2, text, id3, timestamp
            from (select   id, id2, text, id3, timestamp
                  from     stacktable
                  order by id, id2, text) sorted, 
                  (select @row := 0) setup) t1
            left join (select (@row2 := @row2 + 1) as row, id, id2, text, id3, timestamp
                       from (select id, id2, text, id3, timestamp
                             from stacktable
                             order by id, id2, text) sorted, 
                             (select @row2 := 0) setup) t2
            on  (t1.id = t2.id and t1.id2 = t2.id2 and t1.text=t2.text and t1.row = t2.row - 1),
            (select @seq := 1) setup_sequence
     ) t3
group by id, id2, text, seq
having total > 1

为便于阅读，查询使用相同的子查询两次，t1和t2，它所做的就是排序并随后对表的行进行编号：

select (@row := @row + 1) as row, id, id2, text, id3, timestamp
from (select   id, id2, text, id3, timestamp
      from     stacktable
      order by id, id2, text) sorted, 
     (select @row := 0) setup

见fiddle。请注意，序列计数器在所有序列之间并不是唯一的。这不是一个错误。它只在具有相同id，id2，text的序列之间唯一。

序列计数器更新有点棘手：@seq *（1和@seq：= @seq +1）。它依赖于在更新之前为乘法设置的第一个@seq。我不确定这是引擎的确定性还是一致性。但是，也可以通过将t1的记录与前一个记录而不是下一个记录（在t2中）相结合来更改查询以避免它。（没试过）

按ID和TIMESTAMPDIFF分组在同一个表中的一列

1 个答案: