Mysql 将联接和计数添加到现有查询会扩大总和和计数函数的输出

我有一个正常工作的查询：

select problems.problem_id , problem_title , sum( vote ) as totalVotes 
from problems 
left join problem_votes on 
problems.problem_id = problem_votes.problem_id 
left join problem_categories on 
problems.problem_id = problem_categories.problem_id  
where problem_categories.category_id = 1 GROUP BY problem_title;  

当我在

summed\u solutions

表和sum（建议的\u solutions\u表中的列）上添加另一个连接以查看有多少建议的解决方案存在问题时，查询如下所示：

select problems.problem_id , problem_title , sum( vote ) as totalVotes , count(solution_name) from problems 
left join problem_votes on 
problems.problem_id = problem_votes.problem_id 
left join problem_categories on 
problems.problem_id = problem_categories.problem_id  
left join suggested_solutions on
problems.problem_id = suggested_solutions.problem_id
where problem_categories.category_id = 1 GROUP BY problem_title;

第二个查询的问题是，虽然它仍然返回相同数量的行，但原始的sum（）函数返回的计数数量是膨胀的，而建议的\u解决方案的count函数也返回膨胀的数字

知道我做错了什么吗

---

谢谢

对于特定的

问题id

有多个

建议的\u解决方案

行，这意味着原始查询中求和的行会膨胀

您可以在派生表中执行

groupby

，然后加入到该表中

SELECT problems.problem_id,
       problem_title,
       SUM(vote) AS totalVotes,
       Solution_Count
FROM   problems
       LEFT JOIN problem_votes
         ON problems.problem_id = problem_votes.problem_id
       LEFT JOIN problem_categories
         ON problems.problem_id = problem_categories.problem_id
       LEFT JOIN (SELECT COUNT(solution_name) AS Solution_Count,
                         problem_id
                  FROM   suggested_solutions
                  GROUP  BY problem_id) ss
         ON problems.problem_id = ss.problem_id
WHERE  problem_categories.category_id = 1
GROUP  BY problem_title;  

请注意，您的

WHERE problem\u categories.category\u id=1

将

problem\u投票

和

problems

之间的第一个

左连接

转换为

内部连接