Question

例如：

7.找到同一天访问过两位同一专业医生的病人。

   SELECT       p.pid , d.speciality, v.date
   FROM         Patient p
   JOIN         Visits v ON v.pid = p.pid
   JOIN         Doctors d ON d.did = v.did
   GROUP BY     p.pname, d.speciality, v.date
   HAVING   COUNT(DISTINCT d.did) = 2

你如何为此写出合法的RA？

基本上什么是RA中GroupBy和Having Clause的等价？

同样被问到并没有得到回答here

Answer 1

你真的不需要尝试转换SQL，反思问题也提供了RA解决方案：

“找到在同一天访问过两位同一专业的不同医生的病人”

我们首先将患者，访问和医生结合起来

Patients x Visits x Doctors

由于我们有两次访问和两位医生，我们需要另一个表格访问和医生的实例。我们不能只是按原样使用它们，否则我们将无法区分它们。因此，重命名\rho：

Patients x \rho_V1(Visits) x \rho_D1(Doctors) x \rho_V2(Visits) x \rho_D2(Doctors)

接下来我们需要选择“匹配”组合

\sigma_{Patients.pid = V1.pid /\ 
        Patients.pid = V2.pid /\ 
        V1.date = V2.date /\ 
        V1.did = D1.did /\ 
        V2.did = D2.did /\
        D1.did != D2.did /\
        D1.speciality = D2.speciality}
 (Patients x \rho_V1(Visits) x \rho_D1(Doctors)  x 
             \rho_V2(Visits) x \rho_D2(Doctors))

接下来，我们需要找到患者，即项目pid

\pi_{Patients.pid} 
(\sigma_{Patients.pid = V1.pid /\ 
        Patients.pid = V2.pid /\ 
        V1.date = V2.date /\ 
        V1.did = D1.did /\ 
        V2.did = D2.did /\
        D1.did != D2.did /\
        D1.speciality = D2.speciality}
 (Patients x \rho_V1(Visits) x \rho_D1(Doctors)  x 
             \rho_V2(Visits) x \rho_D2(Doctors)))

通过这种方式，您发现患者在同一天至少访问了两名同一专业的医生。如果您需要找到那些完全两位医生的患者，您应该记住正好2 =至少2 - 至少3 ，即

\pi_{Patients.pid} 
(\sigma_{Patients.pid = V1.pid /\ 
        Patients.pid = V2.pid /\ 
        V1.date = V2.date /\ 
        V1.did = D1.did /\ 
        V2.did = D2.did /\
        D1.did != D2.did /\
        D1.speciality = D2.speciality}
 (Patients x \rho_V1(Visits) x \rho_D1(Doctors)  x 
             \rho_V2(Visits) x \rho_D2(Doctors)))
-
\pi_{Patients.pid} 
(\sigma_{Patients.pid = V1.pid /\ 
        Patients.pid = V2.pid /\ 
        Paitents.pid = V3.pid /\
        V1.date = V2.date /\ 
        V2.date = V3.date /\
        V1.did = D1.did /\ 
        V2.did = D2.did /\
        V3.did = D3.did /\
        D1.did != D2.did /\
        D1.did != D3.did /\
        D2.did != D3.did /\
        D1.speciality = D2.speciality /\
        D2.speciality = D3.speciality}
 (Patients x \rho_V1(Visits) x \rho_D1(Doctors)  x 
             \rho_V2(Visits) x \rho_D2(Doctors)  x
             \rho_V3(Visits) x \rho_D3(Doctors) ))

关系代数中的Group By和Having子句查询的等价物是什么？

1 个答案: