Question

我正在尝试编写一个SQL查询来返回采用猫贝壳的人的姓名＆＃39;

我有一张adopters表，其中包含将要或已经采用动物的人。

pets=# SELECT * FROM adopters;
 id | first_name | last_name |       address        | phone_number 
----+------------+-----------+----------------------+--------------
  1 | jack       | smith     | 1 ave                | 123-123-1234
  2 | james      | dude      | 2 ave                | 221-234-4444
  3 | Forrest    | Stone     | 3 lane               | 234-667-6543
  4 | craig      | list      | 44 bulvard           | 555-444-3243
  5 | kap        | seven     | no team              | 999-000-0909
  6 | Miles      | Garrett   | First energy stadium | 123-555-2424
(6 rows)

cats表，其中包含猫

pets=# SELECT * FROM cats;
 id |   name   | gender | age |     intake_date     |    adoption_date    
----+----------+--------+-----+---------------------+---------------------
  1 | Mushi    | M      |   1 | 2016-01-09 00:00:00 | 2016-03-22 00:00:00
  2 | Seashell | F      |   7 | 2016-01-09 00:00:00 | 
  3 | Azul     | M      |   3 | 2016-01-11 00:00:00 | 2016-04-17 00:00:00
  4 | Victoire | M      |   7 | 2016-01-11 00:00:00 | 2016-09-01 00:00:00
  5 | Nala     | F      |   1 | 2016-01-12 00:00:00 | 
(5 rows)

还有一张adoptions表，其中记录了所有已采用的记录。

pets=# SELECT * FROM adoptions;
 id | adopters | cat |  dog  |  fee  |    date    
----+----------+-----+-------+-------+------------
  1 |        1 |     | 10001 | 10.50 | 2017-11-22
  2 |        2 |     | 10007 | 10.50 | 2017-11-22
  3 |        3 |   5 |       | 10.50 | 2017-11-22
  4 |        5 |   2 |       | 10.50 | 2016-11-22
(4 rows)

我在加入三张桌子时遇到了麻烦，即使你没有直接回答这个问题，我也非常感谢你指出一个有用的资源方向。

Answer 1

这应该有效：

SELECT p.first_name, p.last_name 
FROM adoptions a
JOIN cats c on c.id = a.cat
JOIN adopters p on p.id = a.adopters
WHERE c.name = 'seashell'

Answer 2

试试这个

SELECT adopters.first_name, adopters.last_name
FROM adopters
INNER JOIN adoptions on adopters.id = adoptions.adopters
INNER JOIN cats on cats.id = adoptions.cats
WHERE cats.id = 2

Answer 3

TRY：

SELECT 
  pers.first_name, 
  pers.last_name 
FROM
    cats ct 
JOIN 
    adoptions adop ON (ct.id = adop.id)  
JOIN 
    adopters pers ON (adop.adopters = pers.id)
WHERE 
  ct.name = "Seashell ";

Answer 4

SELECT a.first_name, a.last_name
FROM adopters a, cats c, adoptions ad
WHERE a.id = ad.adopters AND ad.cat = c.id
AND c.name = "Seashell";

请尝试这个

尝试连接三个表以获得特定结果

4 个答案: