Question

我有两张桌子，表＃1 =＆gt; tbl_notification_flow具有以下信息：

app_status
Booked
Notified
Missed
Defaulted

app_status是列名，表＃2 =＆gt; tbl_appointment具有以下信息：

id  app_status
1   Booked
3   Missed
4   Missed
5   Missed
6   Missed
7   Booked
8   Booked
9   Booked
10  Booked
11  Booked
12  Booked
13  Booked
14  Booked
15  Booked
17  Booked
18  Booked
19  Booked
20  Defaulted
21  Notified

id和app_status是列，

我希望根据他们在表＃2上出现的次数来计算所有app_status，

我尝试了以下查询：

SELECT COUNT(tbl_appointment.id)AS total_count, tbl_appointment.`app_status`
FROM tbl_appointment
WHERE (tbl_appointment.`app_status`) IN
(
SELECT tbl_notification_flow.`notification_type` AS app_status FROM tbl_notification_flow WHERE tbl_notification_flow.notification_type !='Other'
)

得到以下结果：

total_count app_status
19  Booked

请帮助您获得最佳查询以获得正确的结果。

Answer 1

您应该按app_status汇总数据。

以下是参考资料：

SELECT COUNT(tbl_appointment.id)AS total_count,tbl_appointment.`app_status`
FROM tbl_appointment
WHERE (tbl_appointment.`app_status`) IN
(
SELECT tbl_notification_flow.`notification_type` AS app_status 
FROM tbl_notification_flow 
WHERE tbl_notification_flow.notification_type !='Other'
)
GROUP BY tbl_appointment.`app_status`

Answer 2

使用GROUP BY计算每种通知类型，并利用LEFT JOIN来说明tbl_appointment中没有特定类型事件的情况，在这种情况下，您会获得0 1}}而不是结果集中缺少的行。

SELECT n.notification_type app_status, COUNT(a.id) count
  FROM tbl_notification_flow n LEFT JOIN tbl_appointment a
    ON n.notification_type = a.app_status
 WHERE n.notification_type <> 'Other' 
 GROUP BY n.notification_type

输出：

+------------+-------------+
| app_status | total_count |
+------------+-------------+
| Booked     |          13 |
| Defaulted  |           1 |
| Missed     |           4 |
| Notified   |           1 |
+------------+-------------+

MySQL根据选择的子查询值选择计数

2 个答案: