Question

我想在一个查询中计算实现3个结果：

从当前行计算'app_creationdate'=（＆＃39; month＆＃39;）的所有结果
从当前行计算'app_start'=（＆＃39;月＆＃39;）的所有结果
计算'app_creationsdate'所在的所有结果＆lt; 'app_start'和'app_start'=（＆＃39;月＆＃39;）来自当前行

我的表：

app_id | app_creationdate（timestamp）| app_start（日期时间） 00001 | 2014-11-17 19:39:04 | 2014-11-18 09:30:00

SELECT 
    DATE_FORMAT( app_creationsdate, '%m' ) AS 'month', 
    COUNT( app_id ) AS 'new',
    (SELECT COUNT( app_id ) 
     FROM appointments WHERE MONTH(app_start) = MONTH(NOW())) AS 'act',
    (SELECT COUNT( app_id ) 
     FROM appointments WHERE MONTH(app_creationsdate) < MONTH(app_start)) AS 'prev' 

FROM appointments
WHERE app_owner = 2 AND app_creationsdate > DATE_SUB(now(), INTERVAL 12 MONTH)
GROUP BY DATE_FORMAT( app_creationsdate, '%Y%m' )

Answer 1

我不完全确定你想要完成什么，但我确实注意到一件事：

DATE_FORMAT（a2.app_start，＆＃39;％m＆＃39;）= DATE_FORMAT（＆＃39;月＆＃39;，＆＃39;％m＆＃39;）

这：DATE_FORMAT（＆＃39; month＆＃39;，＆＃39;％m＆＃39;）...评估为NULL，因此将其等同于任何东西永远不会起作用（我不认为）无论如何，但我是MySQL的新手...... :)）。

Answer 2

这可能更接近你想要的。我对prev场景仍感到有点困惑，所以我尽了最大努力。我使用EXTRACT(YEAR_MONTH FROM ...)来获取每个日期的年份和月份，以便我们可以进行每月比较。这可能就是你试图对DATE_FORMAT业务做的事情。

SELECT DATE_FORMAT( app_creationsdate, '%m' ) AS 'month', 
   COUNT( app_id ) AS 'new',
   -- get all other appointments that start in this month
   (SELECT COUNT( act.app_id ) 
      FROM appointments AS act 
     WHERE EXTRACT(YEAR_MONTH FROM act.app_start) = EXTRACT(YEAR_MONTH FROM appointments.app_creationsdate)) AS 'act',
   -- get all appointments that were created before they started (???) and that started before this month

   (SELECT COUNT( prev.app_id ) 
      FROM appointments AS prev
     WHERE EXTRACT(YEAR_MONTH FROM prev.app_creationsdate) < EXTRACT(YEAR_MONTH FROM appointments.app_creationsdate)
       AND EXTRACT(YEAR_MONTH FROM prev.app_start) = EXTRACT(YEAR_MONTH FROM appointments.app_creationsdate)) AS 'prev' 
  FROM appointments
 WHERE app_owner = 2
   AND app_creationsdate > DATE_SUB(now(), INTERVAL 12 MONTH)
 GROUP BY EXTRACT(YEAR_MONTH FROM app_creationsdate)

如何按月显示树结果？

2 个答案: