Question

我的目标是找到过去7天内访问过我网站的访问者数量，今天也访问过了。这意味着如果有人访问了05-02，今天就会在05-02上计算。

我的查询存在的问题是，如果有人访问了05-01和05-02，而且今天也只会访问05-01。但我希望查询能够在05-02上计算他。

如何编辑下面的查询来执行此操作？

SELECT COUNT( DISTINCT v.`hash` ) hashCount, DATE( v.`timestamp` ) AS views
FROM audience v
INNER JOIN behaviour b ON v.`hash` = b.`hash` 
WHERE v.timestamp
BETWEEN SUBDATE( CURDATE( ) , 7 ) 
AND CURDATE( ) 
AND DATE( b.timestamp ) = CURDATE( ) 
GROUP BY views

Answer 1

为了方便起见，请带走所有访问过的人并寻找在过去7天内访问，如果找到一个返回的最后日期访问。我认为这种方法更容易理解：

select b.`hash` ,  
       audience.last_visit 
from behaviour b
 inner join (select v.`hash`, max(v.timestamp) as last_visit from audience v
  where DATE(v.timestamp) between date_sub(current_date, interval 7 day) and 
  date_sub(current_date, interval 1 day)  
  group by v.`hash`) as audience 
 on(b.`hash` = audience.`hash`)
where DATE(b.timestamp) = CURDATE();

我们加入了一个已经包含我们需要的信息的选择（过去7天内每个哈希的最新访问次数）。



返回的行数是访问过您的访问者数量     今天和上周的页面。

您也可以从您的选择中计算，以获得您要查找的数字作为查询结果：

select count(*) from    
(select b.`hash` ,  
       audience.last_visit 
from behaviour b
 inner join (select v.`hash`, max(v.timestamp) as last_visit from audience v
  where DATE(v.timestamp) between date_sub(current_date, interval 7 day) and 
  date_sub(current_date, interval 1 day)  
  group by v.`hash`) as audience 
 on(b.`hash` = audience.`hash`)
where DATE(b.timestamp) = CURDATE() ) as my_visitors;

- Testdata

    drop table if exists your_schema.behaviour;
    create table your_schema.behaviour(`hash` varchar(255), `timestamp` timestamp) ;

    insert into your_schema.behaviour
    values ('ab','2016-05-23'),('ac','2016-05-23');

    drop table if exists your_schema.audience;
    create table your_schema.audience (`hash` varchar(255), `timestamp` timestamp) ;
insert into your_schema.audience
values ('ab','2016-05-01'),('ab','2016-05-02'),('ab','2016-05-03'),('ab','2016-05-04'),('ab','2016-05-21'),('ab','2016-05-23'),
       ('ac','2016-05-01'),('ac','2016-05-02'),('ac','2016-05-03'),('ac','2016-05-04'),('ac','2016-05-21'),('ac','2016-05-23'),
       ('ad','2016-05-01'), ('ad','2016-05-02'), ('ad','2016-05-03'),('ad','2016-05-04'),('ad','2016-05-21'),('ad','2016-05-23');

Answer 2

您也可以使用 DATE_SUB ，检查以下查询

SELECT COUNT(DISTINCT v.`hash`) AS hashCount, DATE(v.`timestamp`) AS views
FROM audience v
INNER JOIN behaviour b ON v.`hash` = b.`hash` 
WHERE views >= DATE_SUB(Now(),INTERVAL 7 DAY)
GROUP BY views

有关更多信息，请访问以下网址

https://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html

Answer 3

您只需使用DATE - INTERVAL ? DAY功能即可。检查下面的示例：

SELECT COUNT(DISTINCT v.`hash`) AS hashCount, DATE(v.`timestamp`) AS views
FROM audience v
INNER JOIN behaviour b ON v.`hash` = b.`hash` 
WHERE views >= DATE(NOW()) - INTERVAL 7 DAY
GROUP BY views

转到此处了解更多Date and Time Functions。

如何编辑我的查询以获得更准确的用户保留结果？

3 个答案: