Question

我在mysql数据库的webinar_timing表中有以下数据

start_time和end_time的类型为datetime


   id  | webinar_id | start_time            | end_time
-------------------------------------------------------------------
    1  |     5      | 3/18/2015  6:00:00 PM | 3/18/2015  7:00:00 PM
    2  |     5      | 3/19/2015  6:00:00 PM | 3/19/2015  7:00:00 PM
    3  |     5      | 3/20/2015  6:00:00 PM | 3/20/2015  7:00:00 PM
    4  |     5      | 3/21/2015  6:00:00 PM | 3/21/2015  7:00:00 PM
    5  |     5      | 3/22/2015  6:00:00 PM | 3/22/2015  7:00:00 PM
    6  |    11      | 3/20/2015  8:00:00 PM | 3/20/2015  9:00:00 PM
    7  |    11      | 3/21/2015  8:00:00 PM | 3/21/2015  9:00:00 PM
    8  |    11      | 3/22/2015  8:00:00 PM | 3/22/2015  9:00:00 PM
    9  |    22      | 3/25/2015  8:00:00 PM | 3/25/2015  9:00:00 PM
   10  |    22      | 3/27/2015  8:00:00 PM | 3/27/2015  9:00:00 PM
   11  |    22      | 3/29/2015  8:00:00 PM | 3/27/2015  9:00:00 PM

基本上，对于每个网络研讨会，我都希望完成或剩余的课程总数和课程数量以及NEXT即将开课的课程

Egs：当我运行此查询时，请在2015年3月21日下午4:00说 - 这是我期待的结果

 webinar_id | total     | Classes Completed | Next Class
----------------------------------------------------------
     5      |   5       | 3                 | 3/21/2015  6:00:00 PM
    11      |   3       | 1                 | 3/21/2015  8:00:00 PM
    22      |   3       | 0                 | 3/25/2015  8:00:00 PM

OR

 webinar_id | total     | Classes Remaining | Next Class
----------------------------------------------------------
     5      |   5       | 2                 | 3/21/2015  6:00:00 PM
    11      |   3       | 2                 | 3/21/2015  8:00:00 PM
    22      |   3       | 3                 | 3/25/2015  8:00:00 PM

任何帮助将不胜感激提前致谢

Answer 1

只需使用SUM(IF(...))

即可

select webinar_id, count(*) AS total, 
SUM(IF(end_time<NOW(), 1, 0)) AS completed, 
SUM(IF(start_time>=NOW(), 1, 0)) AS remaining
from webinar_times
group by webinar_id;

你必须弄清楚你是在看start_time还是end_time，以及如何处理equals和那些小细节......

Answer 2

SELECT COUNT(*), W.webinar_id, X.COMPLETED, Y.[NOT COMPLETE]
FROM #Webinars W
JOIN (SELECT COUNT(*) AS 'COMPLETED', webinar_id
       FROM #Webinars
       WHERE end_tim <= '3/18/2015  7:00:00'
       GROUP BY webinar_id) X
ON X.webinar_id = W.webinar_id  
JOIN (SELECT COUNT(*) AS 'NOT COMPLETE', webinar_id
       FROM #Webinars
       WHERE end_tim > '3/18/2015  7:00:00'
       GROUP BY webinar_id) Y
ON Y.webinar_id = W.webinar_id     
GROUP BY W.webinar_id, X.COMPLETED, Y.[NOT COMPLETE]

MySQL查询以确定剩余或已完成的类

2 个答案: