Question

表：

*shots*
id user_id cat_id name extension size title description library created updated

此表包含所有图像

favorites
id shot_id user_id date

此表包含所有添加的用户最喜欢的图像

stars
id shot_id user_id date

此表格包含用户图像的所有给定星标

views
id shot_id user_id date

此表包含每个用户的每个视图

points
id shot_id count date

用户的每个动作（addFavvorite，giveStar，viewShot）都会为特定镜头添加点数

是否可以在一个查询中执行以下查询？

SELECT * FROM shots WHERE id = ?

获取单个镜头的信息。

SELECT COUNT(*) FROM favorites WHERE shot_id = ?

获取向其收藏夹添加镜头的用户数

SELECT COUNT(*) FROM star WHERE shot_id = ?

获取提供星标的用户数

SELECT COUNT(*) FROM views WHERE shot_id = ?

获取访问此镜头的人数

SELECT AVG(count) AS attention FROM points WHERE shot_id = ? AND date > DATE_SUB(DATE(),INTERVAL 2 DAY) ORDER BY shot_id

目前注意射击（过去两天平均获得的分数）

SELECT SUM(count) AS attention FROM points WHERE shot_id = ? ORDER BY shot_id

镜头的受欢迎程度（上传镜头后的点数）

如何在注意力和受欢迎程度标准之后选择镜头（例如，显示每个注意力超过40，或者人气超过100的镜头）

Answer 1

您可以像这样完成基本需求： -

SELECT s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    c.title AS ctitle,
    c.description AS cdescription,
    COUNT(DISTINCT f.id) AS favorites,
    COUNT(DISTINCT st.id) AS stars,
    COUNT(DISTINCT v.id) AS views
FROM shots s
INNER JOIN users u ON u.id = s.user_id
INNER JOIN categories c ON c.id = s.cat_id
LEFT OUTER JOIN favorites f ON f.shot_id = s.id
LEFT OUTER JOIN stars st ON st.shot_id = s.id
LEFT OUTER JOIN views v ON v.shot_id = s.id
WHERE s.id = 18
GROUP BY s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    ctitle,
    cdescription
ORDER BY s.id

LEFT OUTER JOIN用于拍摄看似可能没有收藏/星/视图，并且使用内部连接中的一个丢失将导致没有返回记录。

它还使用了 COUNT（DISTINCT ...），因为一个镜头可能有多个收藏夹，所以喜欢的ID可能会出现多次（这是因为查询得到了每个可能的收藏夹组合，明星和意见）。

要添加问题的第二部分，可以使用几个子查询来完成。

如果您要检索许多镜头的详细信息，那么通过加入子查询可能最快： -

SELECT s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    c.title AS ctitle,
    c.description AS cdescription,
    sub_att.current_attention,
    sub_pop.popularity,
    COUNT(DISTINCT f.id) AS favorites,
    COUNT(DISTINCT st.id) AS stars,
    COUNT(DISTINCT v.id) AS views
FROM shots s
INNER JOIN users u ON u.id = s.user_id
INNER JOIN categories c ON c.id = s.cat_id
LEFT OUTER JOIN
(
    SELECT shot_id, AVG(count) AS current_attention 
    FROM points 
    WHERE date > DATE_SUB(DATE(),INTERVAL 2 DAY)
    GROUP BY shot_id
) sub_att ON sub_att.shot_id = s.id
LEFT OUTER JOIN
(
    SELECT shot_id, SUM(count) AS popularity
    FROM points 
    GROUP BY shot_id
) sub_pop ON sub_pop.shot_id = s.id
LEFT OUTER JOIN favorites f ON f.shot_id = s.id
LEFT OUTER JOIN stars st ON st.shot_id = s.id
LEFT OUTER JOIN views v ON v.shot_id = s.id
WHERE s.id = 18
GROUP BY s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    ctitle,
    cdescription,
    sub_att.current_attention,
    sub_pop.popularity
ORDER BY s.id

对于单个镜头，那么最好只将子查询放在语句的SELECT部分中（这样它会强制每个返回的行执行一次子查询，其中返回的大量记录将会效率低下： -

SELECT s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    c.title AS ctitle,
    c.description AS cdescription,
    (
        SELECT AVG(count) AS current_attention 
        FROM points 
        WHERE shot_id = s.id AND date > DATE_SUB(DATE(),INTERVAL 2 DAY)
    ) AS current_attention,
    (
        SELECT SUM(count) AS popularity
        FROM points 
        WHERE shot_id = s.id 
    ) AS popularity,
    COUNT(DISTINCT f.id) AS favorites,
    COUNT(DISTINCT st.id) AS stars,
    COUNT(DISTINCT v.id) AS views
FROM shots s
INNER JOIN users u ON u.id = s.user_id
INNER JOIN categories c ON c.id = s.cat_id
LEFT OUTER JOIN favorites f ON f.shot_id = s.id
LEFT OUTER JOIN stars st ON st.shot_id = s.id
LEFT OUTER JOIN views v ON v.shot_id = s.id
WHERE s.id = 18
GROUP BY s.id,
    s.user_id,
    s.cat_id,
    s.name,
    s.extension,
    s.size,
    s.title,
    s.description,
    s.library,
    s.created,
    s.updated,
    u.username,
    u.fullname,
    ctitle,
    cdescription,
    sub_att.current_attention,
    sub_pop.popularity
ORDER BY s.id

将所有SELECT语句放在一起

1 个答案: