Question

我使用以下查询来计算并显示某些测验的用户平均分

$query = 'SELECT AVG(answer) FROM quiz_answers WHERE id_user = '.$id_user.'';
$averagescore = mysql_query($query, $conn) or die(mysql_error());
$row = mysql_fetch_array($averagescore);

echo number_format($row['AVG(answer)'],3);

这部分工作正常，但在这个个人信息旁边，我还希望显示我们所有网站用户的前3个平均分数，比如一个小名人堂......

我尝试过以下代码：

$queryall =  'SELECT * FROM quiz_answers WHERE id_user IN (SELECT id_user FROM quiz_answers GROUP BY id_user HAVING COUNT(id_user) > 1)'

$groupresultsbyuser = mysql_query($query, $conn) or die(mysql_error());
$rows = mysql_fetch_array($groupresultsbyuser);
$numberofrowsperuser = mysql_num_rows($groupresultsbyuser);

我被困在这里..请帮助我找到一种方法让它发挥作用。

Answer 1

假设您想要显示某些用户（例如用户名）以及高分，您可以按用户提取平均分数，然后加入您的用户表（为了论证，我' m调用users并假设它有一个user_id字段）

SELECT u.*, scores.avgScore
FROM users u
    JOIN (
        SELECT id_user, AVG(answer) AS avgScore
        FROM quiz_answers
        GROUP BY id_user
        ORDER BY avgScore DESC LIMIT 3
    ) scores ON u.id_user = scores.id_user

DEMO

Answer 2

试试这个：

SELECT id_user,AVG(answer)
FROM quiz_answers
GROUP BY id_user
ORDER BY AVG(answer) DESC
LIMIT 3

来自所有网站用户的前3个平均分数

2 个答案: