如何显示没有特定值的行？

Question

我对 SQL 比较陌生，我需要显示一个没有特定值的行，而不需要在表中添加一个值。 我正在使用的表是：

 Select * from PERSON;
+-------+-------+---------+--------+---------+----------+
| idnum | fname | lname   | rname  | private | linkblue |
+-------+-------+---------+--------+---------+----------+
| 11111 | Bob   | Bobsley | Robert |       0 | bob1111  |
| 93210 | Bowen | Travis  | Luke   |       1 | tlbo232  |
+-------+-------+---------+--------+---------+----------+
Select * from ADVISOR;
+---------+---------+-------+-------+ You can ignore the sdate and edate
| student | advisor | sdate | edate |
+---------+---------+-------+-------+
|   11111 |   93210 | NULL  | NULL  |
|   93210 |   11111 | NULL  | NULL  |
+---------+---------+-------+-------+
Select * from DIGITAL;
+----------+-------+-------------------------+
| smtype   | idnum | smaddr                  |
+----------+-------+-------------------------+
| email    | 11111 | bob1111@uky.edu         |
| email    | 93210 | tlbo232@uky.edu         |
| facebook | 11111 | bob1111@facebook.com    |
| facebook | 93210 | tlbo232@facebook.com    |
| twitter  | 11111 | twitter.com/bob.bobsley |
+----------+-------+-------------------------+

目标是打印学生姓名、他们的电子邮件、顾问的姓名和他们的推特。 所以像这样：

+------------+-----------+-----------------+--------------------+-------------------+-------------------------+
| First Name | Last Name | Email           | Advisor First Name | Advisor Last Name | Advisor Twitter         |
+------------+-----------+-----------------+--------------------+-------------------+-------------------------+
| Bowen      | Travis    | tlbo232@uky.edu | Bob                | Bobsley           | twitter.com/bob.bobsley |
+------------+-----------+-----------------+--------------------+-------------------+-------------------------+

其中来自：

select 
    P.fname as 'First Name', 
    P.lname as 'Last Name', 
    D.smaddr as 'Email', 
    AdivsorList.fname as 'Advisor First Name',
    AdivsorList.lname as 'Advisor Last Name', 
    AdivsorList.smaddr as 'Advisor Twitter' 
from PERSON P 
LEFT OUTER JOIN DIGITAL D ON P.idnum = D.idnum 
LEFT OUTER JOIN (
    SELECT P1.fname, P1.lname, A.advisor, D1.smaddr, D1.smtype
    from ADVISOR A 
    LEFT OUTER JOIN PERSON P1 ON A.student = P1.idnum 
    LEFT OUTER JOIN DIGITAL D1 ON P1.idnum = D1.idnum
) as AdivsorList ON P.idnum = AdivsorList.advisor 
where D.smtype = 'email' AND AdivsorList.smtype = 'twitter';

输出应显示两行，但其中一行没有推特，因此不会显示。 我正在寻找一种方式来展示它。

Answer 1

这可以通过一系列left join来完成：

select
    ps.fname  student_fname,
    ps.lname  student_lname,
    ds.smaddr student_email
    pa.fname  adviser_fname,
    pa.lname  adviser_lname,
    da.smaddr advisor_twitter
from person ps
left join advisor a  on a.student = ps.idnum
left join person  pa on pa.idnum  = a.advisor
left join digital ds on ds.idnum  = ps.idnum and ds.smtype = 'email'
left join digital da on da.idnum  = pa.idnum and da.smtype = 'twitter'