php - SQL 计算营业时间为早上 10 点到早上 6 点的客户的不同访问次数

Question

我正在为每天早上 10 点到早上 6 点营业的俱乐部编写客户忠诚度软件。数据存储在 MYSQL 中，我想统计客户当月的总访问量。

我使用的是 count(distinct(date)) 但如果玩家在下午 5 点到达并一直待到凌晨 3 点，并在晚上 10 点和凌晨 2 点进行了 2 笔交易。它将被计为 2 次访问，而不是 1 次。

我有一个事务表，其中列出了以下列:

ps: anything in the brackets () is not real data. I get about 2000 transactions a day. I am also able to change the table structure

 Transaction_ID | Date(not Date/Time) | Customer_ID | Item | price | timestamp
   1            | 11-06-2015    (6pm) | Jane        | drink| 2.00  | 156165166
   2            | 09-06-2015    (2pm) | Jane        | drink| 2.00  | 1433858493
   3            | 10-06-2015    (3am) | Jane        | drink| 2.00  | 1433906073
   4            | 06-06-2015    (6pm) | Jane        | drink| 2.00  | 156165166

当前代码返回 {4, Jane}。我正在寻找的答案是 {3,Jane}。事务 {2,3} 应被视为一次访问

SELECT count(distinct(Date)) as visit, Customer_ID 
FROM transaction  
GROUP BY Customer_ID 
WHERE timestamp BETWEEN $timestamp1 AND $timestamp2

$timestamp1 = strtotime("first day of february +10am");
$timestamp2 = strtotime("first day of march +6am");

您建议如何准确统计下面的总访问量？我能够将表结构从日期更改为日期/时间。

The easiest answer with least changes to my codes.

SELECT count(DISTINCT(DATE(DATE_SUB(from_unixtime(timestamp),INTERVAL 6 HOUR))) as visit, Customer_ID 
FROM transaction  
GROUP BY Customer_ID 
WHERE timestamp BETWEEN $timestamp1 AND $timestamp2

Answer 1

最简单的方法是在 SQL 语句中将日期时间 (date,timestamp?) 字段向后移动 6 小时，然后您将得到一天的时间间隔从凌晨 4 点到中午 12 点:

DISTINCT(DATE(DATE_SUB(dt,INTERVAL 6 HOUR)))

SQLFiddle demo