Question

我自己找到了答案。但是导致很多人在寻找我想要的山姆的东西，所以分享我的解决方案：

我的示例表：

t_employes                    t_increases
id | name | salary            employe_id | year
------------------            -----------------
1  | Jon  | 3000              1          | 2005
2  | Ben  | 3000              1          | 2008
3  | Tom  | 2499              2          | 2007

我需要什么：

SELECT 
   t_employes.name,
   t_employes.salary,
   COUNT(t_increases.employe_id) AS count_increases
FROM
   t_employes
   LEFT JOIN t_increases ON t_employes.id = t_increases.employe_id 
WHERE
   t_employes.salary < 2500
   -- OR count_increases < 2 -- (error)
   -- OR COUNT(t_increases.employe_id) -- (error)
GROUP BY t_employes.name

我无法使用HAVING，因为我需要在OR语句中使用我的条件

Answer 1

这个正在使用WHERE clausel中的COUNT语句：

SELECT 
    t_employes.name,
    t_employes.salary,
    COUNT(t_increases.employe_id) AS count_increases
FROM
    t_employes
    LEFT JOIN t_increases ON t_employes.id = t_increases.employe_id 
WHERE
    t_employes.salary < 2500 OR
    t_employes.id IN (SELECT employe_id FROM t_increases GROUP BY employe_id HAVING COUNT(year) < 2)
GROUP BY t_employes.id

Answer 2

2个避免子查询的解决方案。

首先在HAVING子句中检查计数和工资。

SELECT 
   t_employes.name,
   t_employes.salary,
   COUNT(t_increases.employe_id) AS count_increases
FROM
   t_employes
   LEFT JOIN t_increases ON t_employes.id = t_increases.employe_id 
GROUP BY t_employes.name
HAVING t_employes.salary < 2500
OR count_increases < 2

第二个解决方案，它执行2个查询，每个条件一个，并使用UNION将结果合并在一起

SELECT 
   t_employes.name,
   t_employes.salary,
   COUNT(t_increases.employe_id) AS count_increases
FROM
   t_employes
   LEFT JOIN t_increases ON t_employes.id = t_increases.employe_id 
WHERE
   t_employes.salary < 2500
GROUP BY t_employes.name
UNION
SELECT 
   t_employes.name,
   t_employes.salary,
   COUNT(t_increases.employe_id) AS count_increases
FROM
   t_employes
   LEFT JOIN t_increases ON t_employes.id = t_increases.employe_id 
GROUP BY t_employes.name
HAVING count_increases < 2

WHERE OR语句中的MySQL COUNT

2 个答案: