Question

我有2列，'create_time'表示帐户已注册，'last_play'表示帐户上次登录时。我想选择在特定周/月内注册并且在特定周/月的最后2天内有效的帐户。

以下是我如何选择上周的所有条目而不考虑last_play（有效）：

SELECT COUNT(id) FROM account.account 
WHERE WEEKOFYEAR(create_time) = WEEKOFYEAR(NOW()) - 1 
AND YEAR(create_time) = YEAR(NOW());

这是我上周的当前查询，但不起作用：

SELECT COUNT(id) FROM account.account 
WHERE WEEKOFYEAR(create_time) = WEEKOFYEAR(NOW()) - 1 
AND YEAR(create_time) = YEAR(NOW()) 
AND DATE(last_play) BETWEEN 
    ADDDATE(DATE(DATE_SUB(NOW(), INTERVAL 1 WEEK)), 
    INTERVAL 1 - DAYOFWEEK(DATE(NOW())) DAY) 
    AND DATE(NOW());

Answer 1

根据您的第一个工作查询，您可以使用MySQL函数WEEKDAY来识别星期六和星期日：

SELECT COUNT(id) FROM account.account 
WHERE WEEKOFYEAR(create_time) = WEEKOFYEAR(NOW()) - 1 
AND YEAR(create_time) = YEAR(NOW())
AND WEEKOFYEAR(last_play) = WEEKOFYEAR(create_time)  //last_play is in the same week as create_time
AND WEEKDAY(last_play) IN (5,6);  //wekkday is saturday or sunday

这为您提供了在他们注册的同一周的星期六或星期日活动的条目。

编辑：几个月来，您基本上都做同样的事情，但将WEEKOFYEAR替换为MONTH，将WEEKDAY替换为DAYOFMONTH。在给定月份的最后两天，您可以通过为所有可能的案例进行拼图手动查找：

SELECT COUNT(id) FROM account.account 
WHERE MONTH(create_time) = MONTH(NOW()) - 1 
AND YEAR(create_time) = YEAR(NOW())
AND MONTH(last_play) = MONTH(create_time)  //last_play is in the same MONTH as create_time
AND 
   (DAYOFMONTH(last_play) IN (30,31) AND MONTH(last_play) IN (1,3,5,7,8,10,12)
    OR DAYOFMONTH(last_play) IN (29,30) AND MONTH(last_play) IN (4,6,9,11)
    OR DAYOFMONTH(last_play) IN (27,28) AND MONTH(last_play) IN (2))

别介意闰年;-)。或者自己手动再次合并。

MySQL选择特定周/月的最近2天的条目

1 个答案: