Mysql计算所选行的百分比(Mysql Calculate Percentage Of Selected Rows)

 如果我的问题无关紧要，我很抱歉。 因为我不是mysql的专家。 
 
 我在我的网站上有调查。 如果有人对我的调查进行投票，我想向选民展示调查结果。 
 
 这是DB结构; 
 
 surveys 表 （包括调查名称描述和survey_id ） 
 
  
 
 survey_choices 表 （包括调查chocies以及与survey_id相关的调查表） 
 
  
 
 我正在尝试仅计算可见的调查选择％百分比。 
 
 如果我只有1个调查，它正在计算正确的结果。 但是，如果我有超过1个调查，mysql计算所有表的整个survey_vote_count值。 
 
 这是我的MySQL查询; 
 
SELECT 
survey_vote_id, survey_vote_name, survey_vote_count, survey_id, 
survey_vote_count * 100 / t.s AS survey_percentage
FROM survey_votes
CROSS JOIN (SELECT SUM(survey_vote_count) AS s FROM survey_votes) t
WHERE visibility = :visibility
AND survey_id = :surveyId
ORDER BY `survey_votes`.`order_id` ASC
 
 我如何计算例如，仅survey_id = 1 and visibility = 1 ％percentage =％100？ 
 
 任何帮助都会大大增加。 

I'm sorry if my question irrelevant. Because I'm not expert of mysql.
 
I have surveys in my website. If someone votes my survey, i want to show survey results to the voter.
 
Here is the DB Structure;
 
surveys Table (Includes Survey Name Description and survey_id)
 
 
survey_choices Table (Includes Survey Chocies and related with survey_id to Surveys Table)
 
 
I'm trying to calculate only visible survey choices %percentage. 
 
If i have only 1 survey, it's calculating correct results. But if i have more than 1 survey, mysql calculating whole survey_vote_count values of all table.
 
Here is my MySQL query;
 
SELECT 
survey_vote_id, survey_vote_name, survey_vote_count, survey_id, 
survey_vote_count * 100 / t.s AS survey_percentage
FROM survey_votes
CROSS JOIN (SELECT SUM(survey_vote_count) AS s FROM survey_votes) t
WHERE visibility = :visibility
AND survey_id = :surveyId
ORDER BY `survey_votes`.`order_id` ASC
 
How can i calculate for eg only survey_id = 1 and visibility = 1 %percentage = %100?
 
Any help will greatly appricated.

原文：https://stackoverflow.com/questions/41198739