假设我们在以下结构中有四个表(表名 - 字段):
person - id, name
doctor - id_person, specialty
pacient - id_person, disease
appointment - doctor_id, pacient_id, date
如何构建查询以返回医生的姓名和专业,患者的姓名和疾病以及预约日期?
到目前为止,我已经到了这里:
SELECT person.name, doctor.specialty, pacient.disease, appointment.date
FROM appointment
INNER JOIN person
ON appointment.pacient_id=person.id
INNER JOIN doctor
ON appointment.doctor_id=doctor.id_person
INNER JOIN pacient
ON appointment.pacient_id=pacient.id_person
但这并没有回归正确的领域。我认为问题在于为同一行中的两个不同的id(doctor和pacient)返回相同的字段(person.name
)。
答案 0 :(得分:1)
您需要对person
表执行两个单独的连接,并使用别名来标识各个表,如下所示:
select
dp.name as DoctorName
, doctor.specialty
, pp.name as PacientName
, pacient.disease
, appointment.date
from appointment
inner join doctor
on appointment.doctor_id = doctor.id_person
inner join person dp
on appointment.doctor_id = dp.id
inner join pacient
on appointment.pacient_id = pacient.id_person
inner join person pp
on appointment.pacient_id = pp.id