我的项目中有两个表。
posts表

p_id     p_user_id      p_title     p_description
.................................................
1        1              Post 1      lorem ipsum * 1
2        1              Post 2      lorem ipsum * 2
3        2              Post 3      lorem ipsum * 3
4        3              Post 4      lorem ipsum * 4

followers表

f_id     f_following_users_id    f_followed_users_id     f_date
...................................................................
1        2                       1                       2018-01-25
2        2                       3                       2018-01-25
3        3                       2                       2018-01-25

现在我想根据登录用户id获取所有用户的列表。登录用户id取自$\u session['user\u id']。
我想要的结果如下。
第一种情况：如果logged-in-users user\u id为1，即$\u session['user\u id']=1，或第二个表中的以下用户\u id为1，则结果应为：

Array
(
[0] => Array
    (
        [p_id] => 1
        [p_user_id] => 1
        [p_title] => Post 1
        [p_description] => lorem ipsum * 1
        [f_id] => 
        [f_following_users_id] => 
        [f_followed_users_id] => 
        [f_date] =>
    )

[1] => Array
    (
        [p_id] => 2
        [p_user_id] => 1
        [p_title] => Post 2
        [p_description] => lorem ipsum * 2
        [f_id] => 
        [f_following_users_id] => 
        [f_followed_users_id] => 
        [f_date] => 
    )

[2] => Array
    (
        [p_id] => 3
        [p_user_id] => 2
        [p_title] => Post 3
        [p_description] => lorem ipsum * 3
        [f_id] => 
        [f_following_users_id] => 
        [f_followed_users_id] => 
        [f_date] => 
    )

)

数组的最后四个字段应为空，因为用户id为的用户没有跟踪任何人
第二种情况：如果logged-in-users user\u id为2，即$\u session['user\u id']=2，或第二个表中的以下用户\u id为2，则结果应为：

Array
(
[0] => Array
    (
        [p_id] => 1
        [p_user_id] => 1
        [p_title] => Post 1
        [p_description] => lorem ipsum * 1
        [f_id] => 1
        [f_following_users_id] => 2 
        [f_followed_users_id] => 1
        [f_date] => 2018-01-25
    )

[1] => Array
    (
        [p_id] => 2
        [p_user_id] => 1
        [p_title] => Post 2
        [p_description] => lorem ipsum * 2
        [f_id] => 1
        [f_following_users_id] => 2 
        [f_followed_users_id] => 1
        [f_date] => 2018-01-25
    )

[2] => Array
    (
        [p_id] => 3
        [p_user_id] => 2
        [p_title] => Post 3
        [p_description] => lorem ipsum * 3
        [f_id] => 
        [f_following_users_id] => 
        [f_followed_users_id] => 
        [f_date] => 
    )

)

• 前2个数组的前四个字段应填充followers表的第一行，因为用户id为的用户跟随了用户id为1的用户。第3个帖子的4行应该为空，因为用户id为2的用户没有跟随用户id为2的用户***

第三种情况：如果logged-in-users user\u id为3，即$\u session['user\u id']=3，或第二个表中的以下用户\u id为3，则结果应为：

Array
(
[0] => Array
    (
        [p_id] => 1
        [p_user_id] => 1
        [p_title] => Post 1
        [p_description] => lorem ipsum * 1
        [f_id] => 
        [f_following_users_id] => 
        [f_followed_users_id] => 
        [f_date] => 
    )

[1] => Array
    (
        [p_id] => 2
        [p_user_id] => 1
        [p_title] => Post 2
        [p_description] => lorem ipsum * 2
        [f_id] => 
        [f_following_users_id] =>  
        [f_followed_users_id] => 
        [f_date] => 
    )

[2] => Array
    (
        [p_id] => 3
        [p_user_id] => 2
        [p_title] => Post 3
        [p_description] => lorem ipsum * 3
        [f_id] => 3
        [f_following_users_id] => 3
        [f_followed_users_id] => 2
        [f_date] => 2018-01-25
    )

)

前2个数组4最后一列应该为空，因为用户id 3没有跟踪这些用户的帖子。但最后四列的第三个数组应该被填充，因为用户标识3紧跟着用户标识2。
我寻找了所有的解决办法，但都白费了。也许我找不到正确的帖子怎么做，即使在stackoverflow。
我尝试的mysql查询是：

SELECT * FROM posts AS p LEFT JOIN followers AS f ON p.p_user_id=f.f_followed_users_id WHERE f.f_followed_users_id = p_user_id AND f.f_following_users_id = 1 ORDER BY p.p_id

但我没有得到我想要的结果。我尝试了左连接右连接外连接和内连接。但没有成功。我能把这两张table合起来吗？如果是，请帮助我，否则让我知道如何解决这个问题。抱歉，问题这么大。我是新来的，所以不知道怎么问问题。任何帮助都将不胜感激。