Mysql 重用左连接的问题会导致WHERE和ORDERBY子句

我不能在条件和订购时使用

sa。注意

。为什么？ （我删除了标记的部分，查询工作正常！）

---

我必须在查询中更改什么？如果你能解释一下，那就太好了

您将

外部联接

放在您的

WHERE

标准中，这是在否定您的

外部联接。将其移动到您的
JOIN
，您将获得您的
NULL
记录：

SELECT s.*,
u.username,
u.fullname,
c.title AS ctitle,
c.description AS cdescription,
sa.attention,
sp.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, round(AVG(count),2) AS attention
    FROM points 
    WHERE date > DATE_SUB(CURDATE(),INTERVAL 2 DAY)
    GROUP BY shot_id
) sa ON sa.shot_id = s.id
LEFT OUTER JOIN(
    SELECT shot_id, SUM(count) AS popularity
    FROM points 
    GROUP BY shot_id
) sp ON sp.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.library = 1 AND sa.attention > 40
ORDER BY sa.attention DESC
LIMIT 0,50**
GROUP BY s.id

第二个音符，
groupby
不能放在末尾。我也把它移到了正确的位置。
你把它放在你的
WHERE
标准中，就是在否定你的
外部连接。将其移动到您的
JOIN
，您将获得您的
NULL
记录：

SELECT s.*,
u.username,
u.fullname,
c.title AS ctitle,
c.description AS cdescription,
sa.attention,
sp.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, round(AVG(count),2) AS attention
    FROM points 
    WHERE date > DATE_SUB(CURDATE(),INTERVAL 2 DAY)
    GROUP BY shot_id
) sa ON sa.shot_id = s.id
LEFT OUTER JOIN(
    SELECT shot_id, SUM(count) AS popularity
    FROM points 
    GROUP BY shot_id
) sp ON sp.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.library = 1 AND sa.attention > 40
ORDER BY sa.attention DESC
LIMIT 0,50**
GROUP BY s.id

第二个音符，
groupby
不能放在末尾。我也把它移到了正确的位置。
你把它放在你的
WHERE
标准中，就是在否定你的
外部连接。将其移动到您的
JOIN
，您将获得您的
NULL
记录：

SELECT s.*,
u.username,
u.fullname,
c.title AS ctitle,
c.description AS cdescription,
sa.attention,
sp.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, round(AVG(count),2) AS attention
    FROM points 
    WHERE date > DATE_SUB(CURDATE(),INTERVAL 2 DAY)
    GROUP BY shot_id
) sa ON sa.shot_id = s.id
LEFT OUTER JOIN(
    SELECT shot_id, SUM(count) AS popularity
    FROM points 
    GROUP BY shot_id
) sp ON sp.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.library = 1 AND sa.attention > 40
ORDER BY sa.attention DESC
LIMIT 0,50**
GROUP BY s.id

第二个音符，
groupby
不能放在末尾。我也把它移到了正确的位置。
你把它放在你的
WHERE
标准中，就是在否定你的
外部连接。将其移动到您的
JOIN
，您将获得您的
NULL
记录：

SELECT s.*,
u.username,
u.fullname,
c.title AS ctitle,
c.description AS cdescription,
sa.attention,
sp.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, round(AVG(count),2) AS attention
    FROM points 
    WHERE date > DATE_SUB(CURDATE(),INTERVAL 2 DAY)
    GROUP BY shot_id
) sa ON sa.shot_id = s.id
LEFT OUTER JOIN(
    SELECT shot_id, SUM(count) AS popularity
    FROM points 
    GROUP BY shot_id
) sp ON sp.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.library = 1 AND sa.attention > 40
ORDER BY sa.attention DESC
LIMIT 0,50**
GROUP BY s.id

第二个音符，
groupby
不能放在末尾。我也把它移到了正确的位置