Question

我有三张桌子：

members (id, name, surname, usr_img)

user_uploads (imgID, user_id, filename, description, up_time)

img_likes (likeID, img_id, user_id)

我需要从user_uploads获取所有内容，并使用img_id，从成员获取上传者信息并在img_likes表中进行2次计数（检查user_id（来自会话变量）和img_id是否存在并获得img总喜欢）。< / p>

SELECT user_uploads.* AS uu, members.*, COUNT(img_id, user_id) AS usr_liked, COUNT(img_id) AS total_likes
FROM user_uploads
INNER JOIN members AS m ON m.id = uu.user_id -- owner info
INNER JOIN img_likes AS il ON il.img_id = uu.imgID AND il.user_id = ? -- check if logged in user already liked
INNER JOIN img_likes AS ilt ON ilt.img_id = uu.imgID -- total likes
GROUP BY img_id, user_id
ORDER BY up_time DESC

我不需要从img_id中获取任何内容，只需计算行数。我不知道是否需要内连接，可能指定user_id？

并获得计数：

$usr_liked = row['usr_liked'];
$total_likes = rwo['total_likes'];

这会起作用吗？

编辑：添加新查询：

SELECT
    user_uploads.*,
    (
        SELECT *
        FROM members m
        WHERE m.id = user_uploads.user_id
    ),
    (
        SELECT COUNT(*)
        FROM img_likes t
        WHERE t.img_id = user_uploads.imgID AND t.user_id = ?
    ) AS user_likes,
    (
        SELECT COUNT(*)
        FROM img_likes t
        WHERE t.img_id = user_uploads.imgID
    ) AS total_likes
FROM user_uploads
ORDER BY up_time DESC

Answer 1

您可以使用子查询：

SELECT 
    uu.*, 
    m.*,
    (
        SELECT COUNT(t.*)
        FROM img_likes t
        WHERE t.img_id = uu.imgID AND t.user_id = ?
    ) AS user_likes,
    (
        SELECT COUNT(t.*)
        FROM img_likes t
        WHERE t.img_id = uu.imgID 
    ) AS total_likes
FROM 
    user_uploads AS uu
    INNER JOIN members AS m ON 
        m.id = uu.user_id
ORDER BY
    uu.up_time DESC

如何内连接三个表并获得计数

1 个答案: