Question

我正在尝试获取属于某个用户的所有帖子，然后对每个帖子都计算。

在这里，我获取了属于某个用户的所有帖子：

SELECT Posts.id,
Posts.uuid,
    Posts.caption,
    Posts.path,
    Posts.date,
    USERS.id,
    USERS.username,
    USERS.fullname,
    USERS.profileImage
    FROM A.Posts JOIN A.USERS ON
    Posts.id = 145 AND USERS.id = 145 ORDER by date DESC
    LIMIT 0, 5

这就是我如何查询每个帖子的喜欢的内容：

SELECT COUNT(uuidPost)
 FROM Activity
 WHERE type =  "like" AND uuidPost = "FA4C8196-CEA3-4373-94B2-59F387BB1906"

不确定如何组合它们？

如果有人可以提供帮助或向我提供有关查询的提示，我将非常感谢您的帮助！

提前致谢！

SELECT Posts.id,
    Posts.uuid,
    Posts.caption,
    Posts.path,
    Posts.date,
    USERS.id,
    USERS.username,
    USERS.fullname,
    USERS.profileImage,
    coalesce(A.LikeCNT,0),
IF( A.uuidPost IS NOT NULL , 1, 0 ) AS CurrentUser
FROM Posts 
INNER JOIN USERS 
ON Posts.id = 145 
AND USERS.id = 145
LEFT JOIN (SELECT COUNT(A.uuidPost) LikeCNT, A.UUIDPost, A.id
FROM Activity A
WHERE type = 'like' 
GROUP BY A.UUIDPOST) A
on A.UUIDPost=Posts.uuid
AND A.id = Posts.id
WHERE Posts.id = 145
ORDER BY date DESC
LIMIT 0, 5

Answer 1

执行此操作的一种方法是使用内联选择...使用相关查询。

SELECT Posts.id,
    Posts.uuid,
    Posts.caption,
    Posts.path,
    Posts.date,
    USERS.id,
    USERS.username,
    USERS.fullname,
    USERS.profileImage,
   (SELECT COUNT(A.uuidPost)
    FROM Activity A
    WHERE type =  'like' 
      AND A.uuidPost = Posts.uuid) as LikeCNT
FROM Posts 
INNER JOIN USERS 
  ON Posts.id = 145 
 AND USERS.id = 145
ORDER BY date DESC
LIMIT 0, 5

虽然我不是大型数据集的忠实粉丝...我通常更喜欢......

   SELECT Posts.id,
        Posts.uuid,
        Posts.caption,
        Posts.path,
        Posts.date,
        USERS.id,
        USERS.username,
        USERS.fullname,
        USERS.profileImage,
        coalesce(A.LikeCNT,0)
    FROM Posts 
    INNER JOIN USERS 
      ON Posts.id = 145 
     AND USERS.id = 145
    LEFT JOIN (SELECT COUNT(A.uuidPost) LikeCNT, A.UUIDPost
        FROM Activity A
        WHERE type =  'like' 
        GROUP BY A.UUIDPOST) A
     on A.UUIDPost=Posts.uuid
    ORDER BY date DESC
    LIMIT 0, 5

由于引擎生成计数数据集一次，而不必为每个UUID执行。我们必须使用coalesce作为UUID，因为帖子可能没有喜欢，因此不存在记录，因此左边连接上的空值。所以为了显示0，我们需要取第一个非空值，或者是数字，或者使用0。

--- UPDATE：

您是否意识到您添加的A.ID将是活动表中uuidpost权利的随机ID？

我修改了这个以包括当前用户（145？）通过向子查询中添加一个名为CurrentUserLiked的新列来“喜欢”该帖子。

我不得不假设活动表上userID的列名是UserID;必要时改变它。我还假设当前用户被定义为145，这最终将通过php传递。以及列出的其他两个145。

我不确定您为当前用户尝试使用A.uuidPost，所以我暂时不管它。

SELECT Posts.id,
    Posts.uuid,
    Posts.caption,
    Posts.path,
    Posts.date,
    USERS.id,
    USERS.username,
    USERS.fullname,
    USERS.profileImage,
    coalesce(A.LikeCNT,0),
IF( A.uuidPost IS NOT NULL , 1, 0 ) AS CurrentUser,
    A.CurrentUserLiked
FROM Posts 
INNER JOIN USERS 
ON Posts.id = 145 
AND USERS.id = 145
LEFT JOIN (SELECT COUNT(A.uuidPost) LikeCNT, A.UUIDPost, A.id, sum(CASE WHEN A.USERID = 145 then 1 else 0 end) as CurrentUserLiked
           FROM Activity A
           WHERE type = 'like' 
           GROUP BY A.UUIDPOST) A
on A.UUIDPost=Posts.uuid
AND A.id = Posts.id
WHERE Posts.id = 145
ORDER BY date DESC
LIMIT 0, 5

MYSQL查询 - 为每个帖子选择帖子和类似计数

1 个答案: