Question

以下是我的表架构： - 约会

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| schID | appointment_date | amount | location |
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我想点击一个查询，我可以获得金额总额，今年总预约数，即2016年和今年剩余的预约日期，即2016年。

所以我编写了以下查询来计算上述字段： -

SELECT sum(a.amount) as total, 
       count(distinct a.appointment_date) as total_appointment, 
       count(distinct a2.appointment_date) as remaining appointments 
from Appointments a 
LEFT JOIN Appointments a2 ON a.schID = a2.schID 
WHERE a2.appointment_date > GETDATE() AND year(a.appointment_date) = 2016
group by a.location

上述查询根据要求不返回值：（

数据库属于SQL Server。

Answer 1

您可以使用条件聚合：

SELECT sum(amount) as total, 
       count(appointment_date) as total_appointment, 
       count(DISTINCT CASE 
                WHEN appointment_date > GETDATE() AND YEAR(appointment_date) = 2016 
                  THEN DATE(appointment_date)
             END) as remaining appointments 
from Appointments a 
group by a.location

Answer 2

对于此类查询，您不需要join：

SELECT sum(a.amount) as total, count(a.appointment_date) as total_appointment, 
       sum(case when a.appointment_date > getdate() then 1 else 0
           end) as remaining appointments
from Appointments a 
where year(a.appointment_date) = year(GETDATE() );

如果您需要按位置细分，请在location和select条款中加入group by。

SQL：根据两种不同的条件计算值

2 个答案: