Question

我想显示我关注的人的帖子

我拥有的3张桌子是：

用户：

+---------+------+
| id_user | name | last_logout
+---------+------+
|       1 | A    | 22-02-2018 00:00:10
|       2 | B    |
|       3 | C    |
|       4 | D    |
|       5 | E    |
+---------+------+

社区：

+-------------+-------------+
| id_follower | id_followed |
+-------------+-------------+
|           1 |           2 |
|           1 |           3 |
|           1 |           5 |
+-------------+-------------+

文章：

+---------+--------------+---------------+
| id_post | id_user_post |     post      |  date
+---------+--------------+---------------+
|       1 |            2 | hi            |
|       2 |            3 | hello         |
|       3 |            5 | hey you       | 
|       4 |            4 | come on       |
|       5 |            5 | where are you | 22-02-2018 00:01:00
+---------+--------------+---------------+

我正在使用以下代码，但它不会返回任何内容

SELECT u.name AS n
    ,p.post AS t
FROM community AS c
LEFT JOIN users AS u ON u.id_user = c.id_followed
LEFT JOIN posts AS p ON c.id_followed = p.id_user_post
WHERE u.id_follower = 1

Answer 1

Users.id_follower不存在，所以这就是为什么你什么也得不到，实际上你可能会收到错误（无效的列名'id_follower'。）。

请改用Community.id_follower。我还建议使用更具描述性的列名（如'username'和'comment'）

由于回答您特定问题的纯粹原因，我在查询中使用了“n”和“t”。

SELECT u.name as n, p.post as t
FROM Community c
LEFT JOIN Users u ON c.id_followed = u.id_user
LEFT JOIN Posts p ON c.id_followed = p.id_user_post
WHERE c.id_follower = 1

测试：

;WITH USERS AS(
    SELECT *
    FROM (VALUES (1,'A'),
                 (2,'B'),
                 (3,'C'),
                 (4,'D'),
                 (5,'E')) U(id_user, name))
, Community AS(
    SELECT *
    FROM (VALUES (1,2),
                 (1,3),
                 (1,5)) C(id_follower, id_followed))
, posts AS(
    SELECT *
    FROM (VALUES (1,2,'hi'),
                 (2,3,'hello'),
                 (3,5,'hey you'),
                 (4,4,'come on'),
                 (5,5,'where are you')) P(id_post, id_user_post, post))                


SELECT u.name as n, p.post as t
FROM Community c
LEFT JOIN Users u ON c.id_followed = u.id_user
LEFT JOIN Posts p ON c.id_followed = p.id_user_post
WHERE c.id_follower = 1

连接表格以显示我关注的人的帖子

1 个答案: