只有当值存在时，MySQL才会与一列联接

在我的查询中，表上有一组连接，如下所示：

SELECT a.id, a.firstnames, a.surname,b.subject_grade_id
FROM students a
LEFT JOIN student_subject b ON a.id=b.studentid
LEFT JOIN subject_grade c on b.subject_grade_id=c.id
WHERE b.subject_grade_id=?
ORDER by a.surname ASC

id   firstnames   surname   subject_grade_id
--------------------------------------------- 
8      John        Doe           17
9      Bob         Mitchell      17
10     Mary        Smith         17

id   term     studentid  subjectid   mark
--------------------------------------------- 
 1     2         8           17        25

id   firstnames   surname   subject_grade_id    mark
----------------------------------------------------- 
8      John        Doe           17              25
9      Bob         Mitchell      17              null
10     Mary        Smith         17              null

id   firstnames   surname   subject_grade_id    mark
--------------------------------------------------------
8      John        Doe           17              25
9      Bob         Mitchell      17              25
10      Mary        Smith         17              25

这给了我这样一个结果：

SELECT a.id, a.firstnames, a.surname,b.subject_grade_id
FROM students a
LEFT JOIN student_subject b ON a.id=b.studentid
LEFT JOIN subject_grade c on b.subject_grade_id=c.id
WHERE b.subject_grade_id=?
ORDER by a.surname ASC

id   firstnames   surname   subject_grade_id
--------------------------------------------- 
8      John        Doe           17
9      Bob         Mitchell      17
10     Mary        Smith         17

id   term     studentid  subjectid   mark
--------------------------------------------- 
 1     2         8           17        25

id   firstnames   surname   subject_grade_id    mark
----------------------------------------------------- 
8      John        Doe           17              25
9      Bob         Mitchell      17              null
10     Mary        Smith         17              null

id   firstnames   surname   subject_grade_id    mark
--------------------------------------------------------
8      John        Doe           17              25
9      Bob         Mitchell      17              25
10      Mary        Smith         17              25

现在，我还有一个

marks

表，我想将其加入到我的查询中。假设该表只有一个值，如下所示：

SELECT a.id, a.firstnames, a.surname,b.subject_grade_id
FROM students a
LEFT JOIN student_subject b ON a.id=b.studentid
LEFT JOIN subject_grade c on b.subject_grade_id=c.id
WHERE b.subject_grade_id=?
ORDER by a.surname ASC

id   firstnames   surname   subject_grade_id
--------------------------------------------- 
8      John        Doe           17
9      Bob         Mitchell      17
10     Mary        Smith         17

id   term     studentid  subjectid   mark
--------------------------------------------- 
 1     2         8           17        25

id   firstnames   surname   subject_grade_id    mark
----------------------------------------------------- 
8      John        Doe           17              25
9      Bob         Mitchell      17              null
10     Mary        Smith         17              null

id   firstnames   surname   subject_grade_id    mark
--------------------------------------------------------
8      John        Doe           17              25
9      Bob         Mitchell      17              25
10      Mary        Smith         17              25

我的预期输出如下：

SELECT a.id, a.firstnames, a.surname,b.subject_grade_id
FROM students a
LEFT JOIN student_subject b ON a.id=b.studentid
LEFT JOIN subject_grade c on b.subject_grade_id=c.id
WHERE b.subject_grade_id=?
ORDER by a.surname ASC

id   firstnames   surname   subject_grade_id
--------------------------------------------- 
8      John        Doe           17
9      Bob         Mitchell      17
10     Mary        Smith         17

id   term     studentid  subjectid   mark
--------------------------------------------- 
 1     2         8           17        25

id   firstnames   surname   subject_grade_id    mark
----------------------------------------------------- 
8      John        Doe           17              25
9      Bob         Mitchell      17              null
10     Mary        Smith         17              null

id   firstnames   surname   subject_grade_id    mark
--------------------------------------------------------
8      John        Doe           17              25
9      Bob         Mitchell      17              25
10      Mary        Smith         17              25

我怎样才能做到这一点

我试过这个：

select a.id, a.firstnames, a.surname,b.subject_grade_id,d.mark
from students a
left join student_subject b ON a.id=b.studentid
left join subject_grade c on b.subject_grade_id=c.id
outer join marks d on d.subjectid=b.subject_grade_id
WHERE b.subject_grade_id=17
ORDER by a.surname ASC

但它给了我这样一个结果：

SELECT a.id, a.firstnames, a.surname,b.subject_grade_id
FROM students a
LEFT JOIN student_subject b ON a.id=b.studentid
LEFT JOIN subject_grade c on b.subject_grade_id=c.id
WHERE b.subject_grade_id=?
ORDER by a.surname ASC

id   firstnames   surname   subject_grade_id
--------------------------------------------- 
8      John        Doe           17
9      Bob         Mitchell      17
10     Mary        Smith         17

id   term     studentid  subjectid   mark
--------------------------------------------- 
 1     2         8           17        25

id   firstnames   surname   subject_grade_id    mark
----------------------------------------------------- 
8      John        Doe           17              25
9      Bob         Mitchell      17              null
10     Mary        Smith         17              null

id   firstnames   surname   subject_grade_id    mark
--------------------------------------------------------
8      John        Doe           17              25
9      Bob         Mitchell      17              25
10      Mary        Smith         17              25

---

我做错了什么？

不要

外部连接

您的

标记

表，

左连接

它。

LEFT JOIN

的目的是在不满足

ON

条件时保留JOIN右侧的行，并用

NULL

值替换JOIN左侧的值

select a.id, a.firstnames, a.surname,b.subject_grade_id,d.mark
from students a
left join student_subject b ON a.id=b.studentid
left join subject_grade c on b.subject_grade_id=c.id
left join marks d on d.id=a.id  // You don't need to make join on subject Id just need to join on student id
WHERE b.subject_grade_id=17
ORDER by a.surname ASC

---

您可以尝试上面的查询。

向

连接标记d

添加一个条件，将其连接到表

a

，例如

和a.id=d.studentid

，我明白了！这起作用了。我现在明白我的错误了。非常感谢！非常感谢。我现在更明白了。