mysql查询结果未显示某些行

我对如何查询下表感到困惑。我只想显示在特定日期订购和接收的数量

日历表

id  |  date
1   |  2013-07-01 
2   |  2013-07-02 
3   |  2013-07-03 
4   |  2013-07-04 
5   |  2013-07-05
6   |  2013-07-06
7   |  2013-07-07

电话号码详情表

id   |  dateOrdered    | dateReceived | upg
1    |  2013-07-01     | 2013-07-05   | exp
2    |  2013-07-02     | 2013-07-05   | post
3    |  2013-07-02     | 2013-07-06   | upgrade
4    |  2013-07-07     | 2013-07-07   | upggrade
5    |  2013-07-03     | 2013-07-04   | exp
6    |  2013-07-01     | 2013-07-02   | exp

我希望结果是什么

calendar.date  | noOrdered   |  noReceived  | # of post  | # of exp  | # of upgrade
2013-07-01     | 2           |              |            | 2         |
2013-07-02     | 2           |  1           | 1          |           | 1
2013-07-03     | 1           |              |            | 1         |
2013-07-04     |             |  1           |            |           |
2013-07-05     |             |  2           |            |           |
2013-07-06     |             |  1           |            |           |
2013-07-07     | 1           |  1           |            | 1         | 

我的问题是：

select calendar.date,DAYNAME(calendar.date) as `day`,
sum(if((`phone_details`.`upg` = 'Post'),1,0)) AS `Post Paid`,
sum(if((`phone_details`.`upg` = 'Upgrade'),1,0)) AS `Upgrade`,
sum(if(((`phone_details`.`upg` = 'Exp') or (`phone_details`.`upg` = 'Future Exp')),1,0)) AS `Exp`,
(select count(phone_ID) FROM phone_details 
        WHERE dateReceived = calendar.date 
        )AS `received`

from `phone_details` JOIN calendar
where calendar.date = phone_details.dateOrdered

group by calendar.date DESC

此查询的问题是，如果某个日期没有任何订单，它不会在结果中显示，因此即使在该日期有接收，也不会显示。我的结果只是看起来像下表，而不是上表。如果我对每一列进行子查询，我就能够生成我想要的结果，但是处理时间似乎要慢很多

calendar.date  | noOrdered   |  noReceived  | # of post  | # of exp  | # of upgrade
2013-07-01     | 2           |              |            | 2         |
2013-07-02     | 2           |  1           | 1          |           | 1
2013-07-03     | 1           |              |            | 1         |
2013-07-07     | 1           |  1           |            | 1         | 

---

请提供一些指导。非常感谢。

我认为您的加入不正确，因为您应该使用on而不是where：

from `phone_details` JOIN calendar
where calendar.date = phone_details.dateOrdered

尝试使用：

from `phone_details` JOIN calendar
on calendar.date = phone_details.dateOrdered

---

您需要一个

左联接

来为

日历

中的每一行生成结果，即使它们与您的

电话\u详细信息

表中的任何内容都不匹配

SELECT c.date "calendar_date", 
  SUM(c.date=pd.dateOrdered) noOrdered,
  SUM(c.date=pd.dateReceived) noReceived,
  SUM(c.date=pd.dateOrdered AND upg='post')    "# of post",
  SUM(c.date=pd.dateOrdered AND upg='exp')     "# of exp",
  SUM(c.date=pd.dateOrdered AND upg='upgrade') "# of upgrade"
FROM calendar c
LEFT JOIN phone_details pd
  ON c.date = pd.dateOrdered
  OR c.date = pd.dateReceived
GROUP BY c.date;

这个很好用。

---

您可以在此处检查加入，以及添加OP想要但未包含在查询中的附加列。非常感谢，这非常有效，我在发布此处之前尝试了左加入，我想我得到了日期条件颠倒，这就是为什么它第一次不起作用的原因。再次感谢

select calendar.date,DAYNAME(calendar.date) as `day`,
sum(if((`phone_details`.`upg` = 'Post'),1,0)) AS `Post Paid`,
sum(if((`phone_details`.`upg` = 'Upgrade'),1,0)) AS `Upgrade`,
sum(if(((`phone_details`.`upg` = 'Exp') or (`phone_details`.`upg` = 'Future Exp')),1,0)) AS `Exp`,
(select count(id) FROM phone_details 
        WHERE dateReceived = calendar.date 
        )AS `received`

from `phone_details` JOIN calendar
ON calendar.date = phone_details.dateOrdered or calendar.date = phone_details.dateReceived

group by calendar.date DESC