MySQL相关'；项目'；查询 请考虑以下项目表：

非常感谢您的帮助

---

皮特

我正在用平板电脑，所以我会简短地说。将它们视为三个独立的查询，并将它们合并。当我有了键盘后，我会带着sql回来。

谢谢你的提示，Marlin，在阅读和一些实验之后，我想我可能已经破解了它

SELECT 
    p.*,
    (
        IFNULL(themes.matches, 0) + IFNULL(platforms.matches, 0) + IFNULL(clients.matches, 0)
    ) as score
FROM `project` p
LEFT JOIN (
    SELECT t2.project_id, COUNT(*) as matches FROM `project_theme` t1, `project_theme` t2
    WHERE 
      t1.theme_id = t2.theme_id AND t1.project_id = 1
      GROUP BY t2.project_id
) themes ON p.id = themes.project_id
LEFT JOIN (
    SELECT f2.project_id, COUNT(*) as matches FROM `project_platform` f1, `project_platform` f2
    WHERE 
      f1.platform_id = f2.platform_id AND f1.project_id = 1
      GROUP BY f2.project_id
) platforms ON p.id = platforms.project_id
LEFT JOIN (
    SELECT p2.id, COUNT(*) as matches FROM `project` p1, `project` p2
    WHERE 
      p1.client_id = p2.client_id AND p1.id = 1
      GROUP BY p2.id
) clients ON p.id = clients.id
GROUP BY p.`id`
HAVING score > 0
ORDER BY score DESC;

Project A: Themes: 18, 19 Platforms: 1, 4 Client: 22 Project D: Themes: 18, 19 Platforms: 1, 4 Client: 22 Score: 5 Project G: Themes: 18, 21 Platforms: 3, 4 Client: 22 Score: 3 Project B: Themes: 8, 21 Platforms: 2, 4 Client: 1 Score: 1

SELECT 
    `project`.*,
    GROUP_CONCAT(`project_theme`.`theme_id`) as themes,
    GROUP_CONCAT(`project_platform`.`platform_id`) as platforms,
    `project`.`client_id` as client
FROM `project`
LEFT JOIN `project_theme` ON `project`.`id` = `project_theme`.`project_id`
LEFT JOIN `project_platform` ON `project`.`id` = `project_platform`.`project_id`
GROUP BY `project`.`id`

SELECT 
    p.*,
    (
        IFNULL(themes.matches, 0) + IFNULL(platforms.matches, 0) + IFNULL(clients.matches, 0)
    ) as score
FROM `project` p
LEFT JOIN (
    SELECT t2.project_id, COUNT(*) as matches FROM `project_theme` t1, `project_theme` t2
    WHERE 
      t1.theme_id = t2.theme_id AND t1.project_id = 1
      GROUP BY t2.project_id
) themes ON p.id = themes.project_id
LEFT JOIN (
    SELECT f2.project_id, COUNT(*) as matches FROM `project_platform` f1, `project_platform` f2
    WHERE 
      f1.platform_id = f2.platform_id AND f1.project_id = 1
      GROUP BY f2.project_id
) platforms ON p.id = platforms.project_id
LEFT JOIN (
    SELECT p2.id, COUNT(*) as matches FROM `project` p1, `project` p2
    WHERE 
      p1.client_id = p2.client_id AND p1.id = 1
      GROUP BY p2.id
) clients ON p.id = clients.id
GROUP BY p.`id`
HAVING score > 0
ORDER BY score DESC;