Question

我有一个包含三列的数据库表visitors：

id | Name | checkin_date               |
1  | Reg  | 2018-04-20T08:28:54.446Z   |
2  | Meg  | 2018-04-21T08:28:54.446Z   |
3  | Ted  | 2018-04-21T08:28:54.446Z   |
4  | Bin  | 2018-04-23T08:28:54.446Z   |

有一些这样的记录。我只想获取过去7天每天的记录数。现在，我能够使用：

来获取所有日期的每天访客数

select count(id) as no_of_users
     , DATE_FORMAT(checkin_date, '%d %b, %Y') as date  
  from visitors  
 GROUP 
    BY DATE(checkin_date)

但是，这将显示所有记录每天的用户数。如何获取仅过去7天的记录。

Answer 1

select count(id) as no_of_users, DATE_FORMAT(checkin_date, '%d %b, %Y') as date  from visitors  
where checkin_date >= DATE(NOW()) - INTERVAL 7 DAY
GROUP BY DATE(checkin_date)

您要在哪里进行日期字段> =最近7天

Answer 2

根据您的问题。

您需要创建一个日历表，然后在日历表上创建LEFT JOIN。

SELECT DATE(t.dt),count(t1.id) cnt
FROM 
(
  SELECT NOW() dt
  UNION ALL 
  SELECT NOW() - INTERVAL 1 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 2 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 3 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 4 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 5 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 6 DAY
  UNION ALL 
  SELECT NOW() - INTERVAL 7 DAY
) t LEFT JOIN T t1 
ON DATE(t.dt) = DATE(t1.checkin_date)
group by t1.name,DATE(t.dt)

sqlfiddle：http://sqlfiddle.com/#!9/59f49b/5

Answer 3

    select id, count(id) as TOTAL, min (checkin_date) as no_of_users
    from visitors  
where checkin_date between '<Start Date>' and '<End Date>'
 GROUP 
    BY  Id,checkin_date

SQL查询以仅获取最近7天以来的每日访问者数量

3 个答案: