mysql - 加入 2 个表以返回结果为零的查询

Question

我有两张桌子，一张叫医生，一张叫约会。他们都有共同关系的医生ID。我需要能够生成医生 ID 列表以及为每个有零个或多个预约的医生进行的预约数量。（表内容见图片）

使用下面的代码，我可以获得医生列表) id 以及预约次数，但它不会显示医生列表中的零预约医生。我需要做什么修改？

SELECT doctor_id, COUNT(DISTINCT appt_time) AS No_APP FROM appointment GROUP BY doctor_id;

这就是我从中得到的

Answer 1

尝试：

SELECT doctor.doctor_id, COUNT(appointment.appt_time) AS No_APP 
FROM doctor
LEFT JOIN appointment 
ON appointment.doctor_id = doctor.doctor_id
GROUP BY doctor.doctor_id;

Answer 2

使用左连接：

SELECT doctor.doctor_id, COUNT(appt_time) AS No_APP FROM doctor LEFT JOIN appointment ON doctor.doctor_id = appointment.doctor_id GROUP BY doctor.doctor_id;

它应该可以解决问题

Answer 3

这应该有效：

select 
  d.doctor_id
  ,count(distinct a.appt_time) as No_APP 
from doctor d 
left join appointment a on d.doctor_id=a.doctor_id
group by d.doctor_id

Answer 4

使用左连接：

SELECT doctor.doctor_id, COUNT(DISTINCT appt_date, appt_time) AS No_APP 
FROM doctor 
LEFT JOIN appointment
ON doctor.doctor_id = appointment.doctor_id
GROUP BY doctor_id;