Question

我目前正在运行以下查询，该查询返回一组作业及其类别和发布该作业的人的用户名。

SELECT job_id, user_id, title, profiles.user_name
FROM (jobs)
JOIN profiles ON jobs.user_id = profiles.user_id 
JOIN job_categories ON jobs.cat_id = job_categories.cat_id 
JOIN job_sub_categories ON jobs.sub_cat_id = job_sub_categories.sub_cat_id 
WHERE `status` = 'open' 
ORDER BY post_date desc 
LIMIT 5

我有一个名为反馈的表格，根据以前的交易为特定雇主提供大量反馈（很像ebay）。

我希望能够根据雇主当前的反馈评级对结果进行排序和过滤，我不知道如何将其添加到我的查询中。好像我必须在查询中做一些数学运算或者运行一个子查询？或者我应该修改我的反馈表以包含另一个字段，例如为特定评级提供的总反馈？

非常感谢任何帮助。

评分是根据加在一起的所有反馈分数除以行数除以数字4计算的，因为有4个评分字段（性能，质量，可用性和通信），因此feedback_avg = (feedback_total/num_rows)/4

Answer 1

让我试一试。我假设您只有两个表，employers: [id, name]和feedback: [id, employer_id, score]。

首先，得分子查询：

SELECT employer_id, SUM(score) AS total_score, COUNT(*) AS num_rows
       FROM feedback GROUP BY employer_id;

现在主要查询：

SELECT name, total_score/num_rows AS avg_score
       FROM employers JOIN ([subquery]) AS sq ON(employers.id = sq.employer_id)
       WHERE avg_score > 0.5;

将整个子查询粘贴到指定的位置。

提示：观看次数

如果您愿意，可以将子查询作为永久视图，并在主查询中使用它：

CREATE VIEW score_tally AS
SELECT employer_id, SUM(score) AS total_score, COUNT(*) AS num_rows
       FROM feedback
       GROUP BY employer_id;

SELECT name, total_score/num_rows AS avg_score
       FROM employers JOIN score_tally ON(employers.id = score_tally.employer_id)
       WHERE avg_score > 0.5;

提示（再次）：以上提示是愚蠢的，我们应该使用内置的AVG：

CREATE VIEW score_tally AS
SELECT employer_id, AVG(score) AS avg_score
       FROM feedback
       GROUP BY employer_id;

SELECT name, avg_score
       FROM employers JOIN score_tally ON(employers.id = score_tally.employer_id)
       WHERE avg_score > 0.5;

让我们猜猜您的完整查询可能是什么样的：

SELECT job_id,
       user_id,
       title,
       profiles.user_name AS user_name,
       avg_score

       FROM jobs
            JOIN profiles ON(jobs.user_id = profiles.user_id)
            JOIN job_categories ON(jobs.cat_id = job_categories.cat_id)
            JOIN job_sub_categories ON(jobs.sub_cat_id = job_sub_categories.sub_cat_id)
            JOIN (SELECT employer_id, AVG(score) AS avg_score FROM feedback GROUP BY employer_id) AS sq
                 ON(employers.id = sq.employer_id)

        WHERE status = 'open' AND avg_score > 0.5

        ORDER BY post_date desc
        LIMIT 5

如何将反馈分数整合到搜索结果中

1 个答案: