仅显示用户使用 mySQL 关注的用户的帖子答案

【问题标题】：Displaying posts from only users the user follows with mySQL仅显示用户使用 mySQL 关注的用户的帖子
【发布时间】：2019-10-29 02:21:00
【问题描述】：

必要的表格

   recipe_post    
    +-----------+-----------+-------+
    | recipe_id | posted_by | title |
    +-----------+-----------+-------+

friend_request (followers table)   
    +----------+-----------+
    | follower | following |
    +----------+-----------+

此查询从 recipe_ingredients 表中获取食材并将其与 users_ingredients 表进行匹配，并返回所有帖子的状态为用户是否可以做饭

SELECT u.uid, ri.recipe_id,
       COUNT(ui.i_id) AS available_ingredients, -- Number of ingredients the user has that are required to cook this recipe
       COUNT(ri.i_id) AS required_ingredients,  -- Number of ingredients that are required to cook this recipe
       CASE
         WHEN COUNT(ui.i_id) = COUNT(ri.i_id) THEN 'can_cook'
         WHEN COUNT(ui.i_id) > 0 THEN 'has_some_ingredients'
         ELSE 'has_no_ingredients'
       END AS state,
rp.recipe_id,rp.name,rp.description
FROM users u
CROSS JOIN recipe_ingredients ri
LEFT JOIN userIngredients ui ON(ri.i_id = ui.i_id AND u.uid = ui.uid)
INNER JOIN recipe_post rp ON rp.recipe_id = ri.recipe_id
WHERE u.uid = 1 --matching with user 1
GROUP BY u.uid, ri.recipe_id, rp.recipe_id
ORDER BY u.uid, ri.recipe_id;

我创建了另一个查询，它只返回他们关注的用户的帖子

SELECT * FROM recipe_post p
INNER JOIN friend_requests f ON (f.following = p.posted_by)
WHERE f.follower = 5;

我无法将此查询添加到第一个查询，以便它只能显示他们关注的用户的帖子

【问题讨论】：

标签： mysql sql relational-database

【解决方案1】：

您可以以subquery 的身份加入您的第二个查询。

SELECT u.uid
        , ri.recipe_id
        , COUNT(ui.i_id) AS available_ingredients -- Number of ingredients the user has that are required to cook this recipe
        , COUNT(ri.i_id) AS required_ingredients -- Number of ingredients that are required to cook this recipe
        , CASE WHEN COUNT(ui.i_id) = COUNT(ri.i_id) THEN 'can_cook'
            WHEN COUNT(ui.i_id) > 0 THEN 'has_some_ingredients'
            ELSE 'has_no_ingredients'
            END AS state
        , rp.recipe_id
        , rp.name
        , rp.description
FROM users u
CROSS JOIN recipe_ingredients ri
LEFT JOIN userIngredients ui ON(ri.i_id = ui.i_id AND u.uid = ui.uid)
INNER JOIN recipe_post rp ON rp.recipe_id = ri.recipe_id
INNER JOIN
    (SELECT * FROM recipe_post p
        INNER JOIN friend_requests f ON f.following = p.posted_by) as t1 on t1.recipe_id =  = rp.recipe_id
WHERE u.uid = 1 --matching with user 1
GROUP BY u.uid, ri.recipe_id, rp.recipe_id
ORDER BY u.uid, ri.recipe_id;

【讨论】：

对不起，我忘了说我试过这个并得到一个语法错误

【解决方案2】：

您可以直接添加您对其他表的加入方式：

SELECT u.uid, ri.recipe_id,
       COUNT(ui.i_id) AS available_ingredients, -- Number of ingredients the user has that are required to cook this recipe
       COUNT(ri.i_id) AS required_ingredients,  -- Number of ingredients that are required to cook this recipe
       CASE
         WHEN COUNT(ui.i_id) = COUNT(ri.i_id) THEN 'can_cook'
         WHEN COUNT(ui.i_id) > 0 THEN 'has_some_ingredients'
         ELSE 'has_no_ingredients'
       END AS state,
rp.recipe_id,rp.name,rp.description
FROM users u
CROSS JOIN recipe_ingredients ri
 LEFT JOIN userIngredients ui ON ri.i_id = ui.i_id AND u.uid = ui.uid
INNER JOIN recipe_post rp ON rp.recipe_id = ri.recipe_id
INNER JOIN friend_requests f ON f.following = rp.posted_by
WHERE u.uid = 1 --matching with user 1
  AND f.follower = 5
GROUP BY u.uid, ri.recipe_id, rp.recipe_id
ORDER BY u.uid, ri.recipe_id;

【讨论】：