Question

我有这个正常工作的查询：

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suggested_solutions表和sum（column_in_suggested_solutions_table）上添加另一个连接以查看有问题的建议解决方案的数量时，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的行数，但原始的sum（）函数返回一个膨胀的计数，而suggest_solutions的count函数也返回一个膨胀的数字。

知道我在这里做错了吗？

谢谢！

Answer 1

特定suggested_solutions有多个problem_id行，这意味着在原始查询中汇总的行会被夸大。

您可以在派生表中执行group by，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_categories.category_id = 1也会将LEFT JOIN和problem_votes之间的第一个problems转换为INNER JOIN

向现有查询添加连接和计数会使sums和count函数输出膨胀

1 个答案: