使用 count inside count 时将结果限制为一个的 SQL 查询

Question

我正在尝试选择特定项目的点赞数。我想出的主意是

CAST(count(uploads.ID in (SELECT uploadID from votes)) as decimal) as numberoflikes

这可行，但查询只返回一件事。

整个查询

SELECT DISTINCT users.NAME AS username
    ,users.ID AS userID
    ,subjects.NAME AS subjectname
    ,uploads.TIME
    ,uploads.description
    ,uploads.NAME
    ,uploads.ID
    ,CASE 
        WHEN uploads.ID IN (
                SELECT uploadID
                FROM votes
                WHERE userID = 2
                )
            THEN CAST(1 AS DECIMAL)
        ELSE CAST(0 AS DECIMAL)
        END AS liked
    ,CASE 
        WHEN uploads.ID IN (
                SELECT uploadID
                FROM bookmarks
                WHERE userID = 2
                )
            THEN CAST(1 AS DECIMAL)
        ELSE CAST(0 AS DECIMAL)
        END AS bookmarked
    ,CAST(count(uploads.ID IN (
                SELECT uploadID
                FROM votes
                )) AS DECIMAL) AS numberoflikes
FROM uploads
INNER JOIN subjects ON (subjects.ID = uploads.subjectID)
INNER JOIN users ON (users.ID = uploads.userID)
INNER JOIN uploadGrades ON (uploads.ID = uploadGrades.uploadID)
INNER JOIN grades ON (grades.ID = uploadGrades.gradeID)
WHERE uploads.active = 1
    AND subjects.ID IN (
        SELECT subjectID
        FROM userSubjects
        INNER JOIN users ON (users.ID = userSubjects.userID)
        WHERE userSubjects.userID = 2
        )
    AND grades.ID IN (
        SELECT userGrades.gradeID
        FROM uploadGrades
        INNER JOIN userGrades ON (uploadGrades.gradeID = userGrades.gradeID)
        WHERE userGrades.userID = 2
        )
ORDER BY uploads.trueRating DESC;

Answer 1

count(uploads.ID in (SELECT uploadID from votes)) as numberoflikes

按uploads.Id ORDER BY uploads.trueRating DESC 分组

我设法做到了这一点。如果我在那时添加了该组，它会将点赞数分成几行并返回多于一行。感谢您的帮助！

Answer 2

试试这样的...

SELECT users.name as username, users.ID as userID, subjects.name as subjectname,
       uploads.time, uploads.description, uploads.name, uploads.ID, 
       count(userVotes.userId), count(bookmarksMade.userId),
FROM uploads 
     join subjects     on(subjects.ID = uploads.subjectID) 
     join users        on(users.ID = uploads.userID)
     join uploadGrades on(uploads.ID = uploadGrades.uploadID) 
     join grades       on(grades.ID = uploadGrades.gradeID)
     left join (select userId, uploadId from votes where userId = 2) as userVotes on uploads.id = userVotes.uploadId
     left join (select userId, uploadId from bookmarks where userId = 2) as bookmarksMade on uploads.id = bookmarksMade.uploadId
     join userSubjects on subjects.id = userSubjects.subjectID
WHERE uploads.active = 1 AND 
      userSubjects.userID = 2
ORDER BY uploads.trueRating DESC;

但是，我忽略了 userGrades 的事情，因为你在那里做了一个我不太理解的时髦连接（在看起来不是两个表上的整个主键的东西上连接两个表）。

无论如何，你真的需要去更多像这样或 Oropeza 在他的回答中建议的东西。更直接地了解您想要什么。这个查询看起来像一个怪物，它一直在增长，并根据您的需要添加了“IN”子句。是时候回到绘图板上思考你想要什么以及如何直接实现它了。

Answer 3

让我们试试你的查询的简化版本，这是获得更好答案的基础

1. 我将初始查询减少到 user 和 upload 开始。同时删除您已经知道如何计算的字段。

.

SELECT DISTINCT users.NAME AS username
    ,users.ID AS userID
    ,uploads.NAME
    ,uploads.ID
    ,CAST(count(uploads.ID IN (
                SELECT uploadID
                FROM votes
                )) AS DECIMAL) AS numberoflikes
FROM uploads
INNER JOIN users ON (users.ID = uploads.userID)
WHERE uploads.active = 1
ORDER BY uploads.trueRating DESC;

1. 然后用LEFT JOIN添加投票以替换COUNT中的SELECT，这样如果不匹配，您将获得NULL，正如我在评论中所说COUNT不计入NULL's

.

SELECT DISTINCT users.NAME AS username
    ,users.ID AS userID
    ,uploads.NAME
    ,uploads.ID
    ,CAST(count(votes.uploadID)) AS DECIMAL) AS numberoflikes
FROM uploads
INNER JOIN users ON (users.ID = uploads.userID)
LEFT JOIN votes ON (uploads.ID = votes.uploadID)
WHERE uploads.active = 1
ORDER BY uploads.trueRating DESC;