php - 按日期获取打开和关闭的帖子计数(间隔 1 周)？

Question

我需要获取已关闭和打开的帖子的帖子计数，按日期分组，间隔为 1 周。我现在得到的是。

我想要的是这样的。

抱歉，解释不好。 谢谢!

Answer 1

请尝试一下:

SELECT 
date_posts - INTERVAL WEEKDAY(date_posts) DAY AS date_start,
date_posts + INTERVAL (6 - WEEKDAY(date_posts)) DAY AS date_start,
COUNT(*) totalPosts,
COUNT(CASE WHEN Close_count = 1 THEN 1 END) closeCount,
COUNT(CASE WHEN Open_count = 1 THEN 1 END) openCount
FROM your_table
GROUP BY YEARWEEK(date_posts,1);

注意:

YEARWEEK(date) 函数假定开始日期是星期日。

YEARWEEK(date,1) 函数假定开始日期为星期一。

WEEKDAY(date) 函数返回相应周的日期索引，假设 Monday 作为一周的开始日。它返回0表示星期一，1表示星期二......以及6 > 周日。

编辑:

对于日期范围搜索:

SELECT 
date_posts - INTERVAL WEEKDAY(date_posts) DAY AS date_start,
date_posts + INTERVAL (6 - WEEKDAY(date_posts)) DAY AS date_start,
COUNT(*) totalPosts,
COUNT(CASE WHEN Close_count = 1 THEN 1 END) closeCount,
COUNT(CASE WHEN Open_count = 1 THEN 1 END) openCount
FROM your_table
WHERE date_posts >= YOUR_START_DATE - INTERVAL WEEKDAY(YOUR_START_DATE) DAY
AND date_posts <= YOUR_END_DATE + INTERVAL (6 - WEEKDAY(YOUR_END_DATE)) DAY 
GROUP BY YEARWEEK(date_posts,1);

因此，基于此，如果您想获取过去 5 周(包括本周)的结果，您需要运行以下查询:

SELECT 
date_posts - INTERVAL WEEKDAY(date_posts) DAY AS date_start,
date_posts + INTERVAL (6 - WEEKDAY(date_posts)) DAY AS date_start,
COUNT(*) totalPosts,
COUNT(CASE WHEN Close_count = 1 THEN 1 END) closeCount,
COUNT(CASE WHEN Open_count = 1 THEN 1 END) openCount
FROM your_table
WHERE date_posts >= (CURDATE() - INTERVAL 5 WEEK) - INTERVAL WEEKDAY(CURDATE() - INTERVAL 5 WEEK) DAY
AND date_posts <= CURDATE() + INTERVAL (6 - WEEKDAY(CURDATE())) DAY 
GROUP BY YEARWEEK(date_posts,1);