MySQL返回同名相同表的联接

我正在使用mysql

我有两张桌子

Students:
stud_num 
prof1
prof2

Prof
prof_id
first_name
last_name

Prof Id是学生表中的外键

我想把stud_num和prof1作为名字和姓氏，prof2作为名字和姓氏

以下是我的查询不起作用：

select s.stud_num, CONCAT(p.first_name, " ", p.last_name) as PROF1, CONCAT(p.first_name, " ", p.last_name) as PROF2
from students s
inner join prof p
on s.prof1 = p.prof
and s.prof2 = p.prof

---

这是行不通的。有没有简单的方法可以做到这一点？

请尝试以下方法：

SELECT s.stud_num, 
       CONCAT(p1.first_name, " ", p1.last_name) as PROF1, 
       CONCAT(p2.first_name, " ", p2.last_name) as PROF2
FROM students s, prof p1, prof p2
WHERE s.prof1 = p1.prof
      AND s.prof2 = p2.prof

---

请尝试以下操作：

SELECT s.stud_num, 
       CONCAT(p1.first_name, " ", p1.last_name) as PROF1, 
       CONCAT(p2.first_name, " ", p2.last_name) as PROF2
FROM students s, prof p1, prof p2
WHERE s.prof1 = p1.prof
      AND s.prof2 = p2.prof

---

两次连接到professor表，但如果没有提供professor的ID，我建议使用左连接

select 
      s.stud_num, 
      CONCAT(p1.first_name, " ", p1.last_name) as PROF1, 
      CONCAT(p2.first_name, " ", p2.last_name) as PROF2
   from 
      students s
         LEFT join prof p1
            on s.prof1 = p1.prof
         LEFT join prof p2
            on s.prof2 = p2.prof

---

两次连接到professor表，但如果没有提供professor的ID，我建议使用左连接

select 
      s.stud_num, 
      CONCAT(p1.first_name, " ", p1.last_name) as PROF1, 
      CONCAT(p2.first_name, " ", p2.last_name) as PROF2
   from 
      students s
         LEFT join prof p1
            on s.prof1 = p1.prof
         LEFT join prof p2
            on s.prof2 = p2.prof

---

Prof表需要以两个不同的名称联接两次。 试试这个：

select s.stud_num,
       CONCAT(p1.first_name, " ", p1.last_name) as PROF1,
       CONCAT(p2.first_name, " ", p2.last_name) as PROF2
from   students s,
       prof p1,
       prof p2
where  s.prof1 = p1.prof_id
and    s.prof2 = p2.prof_id

---

Prof表需要以两个不同的名称联接两次。 试试这个：

select s.stud_num,
       CONCAT(p1.first_name, " ", p1.last_name) as PROF1,
       CONCAT(p2.first_name, " ", p2.last_name) as PROF2
from   students s,
       prof p1,
       prof p2
where  s.prof1 = p1.prof_id
and    s.prof2 = p2.prof_id

---

谢谢，完全同意。我只是在发布了它并观看了另一个答案后想了想。谢谢并完全同意。我只是在发布了它并观看了另一个答案后想。谢谢所有的快速回复！！！！感谢所有的快速回复！！！！