如何查询星级评分系统投票的用户数量

时间:2016-01-22 09:49:26

标签: php mysql mysqli

我有一张桌子,我想要投票给5个评分,4个评分,3个评分,2评级和1个评分的用户数

id  user_id question_id rating
1   1        1          3   
2   2        1          3       
3   3        1          4   

我希望结果如下:for question_id 1

* * * * * (0)
* * * * (1)
* * * (2)
* * (0)
* (0)

这是我的代码:

$con=mysqli_connect("localhost","root","","db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"SELECT * FROM `survey_answers`");


for($i=5;$i>0;$i--){

    for($j=0;$j<$i;$j++){

        echo " * ";

    }
    echo "<br>";

}

3 个答案:

答案 0 :(得分:1)

这应该对您有用,只需将表名和问题ID替换为您想要的任何内容。

SELECT COUNT(*) AS vote_count, rating FROM ratings_table WHERE question_id = ? GROUP BY rating

如果您想立即获得所有问题:

SELECT COUNT(*) AS vote_count, rating FROM ratings_table GROUP BY rating, question_id

答案 1 :(得分:0)

这应该可以解决问题:

SELECT question_id
    , rating
    , COUNT(user_id) voteCount
    FROM survey_answers
    GROUP BY question_id, rating

我假设您收到所有问题及其各自的投票数。按设计,用户每个问题只能投票一次。

答案 2 :(得分:0)

这可能是解决问题的愚蠢方法。但它可以用纯sql完成。请试试这个:

select concat(tmp.star,' (',ifnull(cnt,0),')')as ratings from(
    select 1 as rating,'*' as star union all
    select 2, '**' union all 
    select 3, '***' union all 
    select 4, '****' union all 
    select 5, '*****'
) tmp
left join (
    select rating,count(1) cnt
    from survey_answers
    where question_id = 1
    group by question_id,rating
) as tbl on tbl.rating = tmp.rating
order by tmp.rating desc

结果应如下所示:

ratings
---------
***** (0)
**** (1)
*** (2)
** (0)
* (0)