Mysql SQL脚本未按预期工作：如何在虚拟列中使用聚合对结果排序？

我有这个：

SELECT * FROM `npt_articles` WHERE (date_discovered >= 1464534366 AND
           date_discovered <= 1464534366 AND
                      npt_site_id = 4 AND
                      (@total_social := (comments_html + fb_total  + google_plus_one + pinterest + linked_in ) + 0.0) >
                      865.0 AND
                      http_status_code = 200) ORDER BY @total_social DESC;

我想我想要的是这样的东西：

SELECT *, ((comments_html + fb_total  + google_plus_one + pinterest + linked_in ) + 0.0) as total_social  FROM `npt_articles` WHERE (date_discovered >= 1464534366 AND
                      npt_site_id = 4 AND
                      total_social >
                      865.0 AND
                      http_status_code = 200) ORDER BY total_social DESC; 

WITH inner_table AS (
    SELECT 
        ((comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0) AS total_social
    FROM
        `npt_articles`
)
select * FROM inner_table
WHERE
    (date_discovered >= 1464534366
        AND npt_site_id = 4
        AND total_social > 865.0
        AND http_status_code = 200)
ORDER BY total_social DESC;   

SELECT *, 
       (comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0 AS total_social FROM `npt_articles` WHERE (
  date_discovered >= 1464534366 
  AND 
  npt_site_id = 4 
  AND 
  http_status_code = 200 
) 
HAVING total_social > 865.0 ORDER BY total_social DESC;

唯一的问题是MySQL Workbench说，

total_social

不存在

我试过使用这样的CTE：

SELECT *, ((comments_html + fb_total  + google_plus_one + pinterest + linked_in ) + 0.0) as total_social  FROM `npt_articles` WHERE (date_discovered >= 1464534366 AND
                      npt_site_id = 4 AND
                      total_social >
                      865.0 AND
                      http_status_code = 200) ORDER BY total_social DESC; 

WITH inner_table AS (
    SELECT 
        ((comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0) AS total_social
    FROM
        `npt_articles`
)
select * FROM inner_table
WHERE
    (date_discovered >= 1464534366
        AND npt_site_id = 4
        AND total_social > 865.0
        AND http_status_code = 200)
ORDER BY total_social DESC;   

SELECT *, 
       (comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0 AS total_social FROM `npt_articles` WHERE (
  date_discovered >= 1464534366 
  AND 
  npt_site_id = 4 
  AND 
  http_status_code = 200 
) 
HAVING total_social > 865.0 ORDER BY total_social DESC;

---

就像在“”中一样，但MySQL Workbench不会接受它，因为它不喜欢在该位置带有的

，我知道现在MySQL不支持它。
对，所以解决方案是使用我从“”中找到的
HAVING
子句，我认为这基本上是一个
WHERE
子句，发生在
SELECT
之后

现在，我的SQL如下所示：

SELECT *, ((comments_html + fb_total  + google_plus_one + pinterest + linked_in ) + 0.0) as total_social  FROM `npt_articles` WHERE (date_discovered >= 1464534366 AND
                      npt_site_id = 4 AND
                      total_social >
                      865.0 AND
                      http_status_code = 200) ORDER BY total_social DESC; 

WITH inner_table AS (
    SELECT 
        ((comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0) AS total_social
    FROM
        `npt_articles`
)
select * FROM inner_table
WHERE
    (date_discovered >= 1464534366
        AND npt_site_id = 4
        AND total_social > 865.0
        AND http_status_code = 200)
ORDER BY total_social DESC;   

SELECT *, 
       (comments_html + fb_total + google_plus_one + pinterest + linked_in) + 0.0 AS total_social FROM `npt_articles` WHERE (
  date_discovered >= 1464534366 
  AND 
  npt_site_id = 4 
  AND 
  http_status_code = 200 
) 
HAVING total_social > 865.0 ORDER BY total_social DESC;

现在它执行查询、选择、计算
total\u social
列，然后使用
total\u social
执行第二次查询，然后按
total\u social
降序排序
 毫无疑问，有了样本数据和期望的结果，其他人可能会知道您的实际意图。您不需要变量：
select。。。在哪里（（comments_html+…+linked_in）>865））
那么我如何将其传递给ORDER BY？另外，它不存在于行查询之外，这就是为什么我在查询之外执行它。我想这可能就是我想要的。@MarcB我猜OP正在使用变量，这样他就不必在orderby子句中重新写这行了。就我个人而言，我会在select子句中创建一个变量，就像他在第二个版本中创建的一样。在这一点上，我和Gordon是一致的，我们需要一些样本数据（或者数据来自的网站，我猜他使用的是google scholar之类的），以及他期望的，以及他得到的。