mysql - 我可以对父查询中嵌入查询的结果进行数学运算吗？

Question

考虑以下查询：

SELECT n.nid, 
    (select count(*) as count from view_log where id = n.nid) AS the_count,
    (the_count + 1) as the_bumped_count
   FROM node n

当我运行它时，我得到Unknown column 'the_count' in 'field list'. 有没有办法解决？似乎the_count应该在查询中可见并且可以使用，但显然不是。顺便说一句，我也试过SUM(the_count, 1)，但也失败了。谢谢！

Answer 1

您不能使用在ALIAS您想要对其进行计算的同一级别上定义的。

SELECT n.nid, 
    (select count(*) as count from view_log where id = n.nid) AS the_count,
    ((select count(*) as count from view_log where id = n.nid) + 1) as the_bumped_count
   FROM node n

或者更好地使用子查询，

SELECT  nid, 
        the_count, 
        the_count + 1 AS the_bumped_count
FROM
(   
    SELECT n.nid, 
            (select count(*) as count from view_log where id = n.nid) AS the_count
    FROM node n
) s

Answer 2

我认为这也可以，并且只需要您计算一次：

SELECT nid, the_count, the_count+1 as the_bumped_count
FROM (
    SELECT n.nid,
        COUNT(v.*) the_count, 
    FROM node n
       LEFT JOIN view_log v on n.nid = v.id
    GROUP BY n.nid
) t

或者回到你的语法：

SELECT nid, the_count, the_count+1 as the_bumped_count
FROM (
    SELECT n.nid,
        (select count(*) as count from view_log where id = n.nid) AS the_count
    FROM node n
) t

祝你好运。