mysql - 使用 count inside count 时将结果限制为 1 的 SQL 查询

Question

我正在尝试选择特定项目的点赞数。我想出的想法是

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

让我们尝试减少查询的版本，这是获得更好答案的基础

1. 我减少了对用户的初始查询，并开始上传。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;

然后使用LEFT JOIN添加投票来替换COUNT中的SELECT，这样如果不匹配，您将得到NULL 正如我在评论中所说的 COUNT 不算 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;