Question

我试图提供从2017年参加培训课程的人那里收集的数据，并对课程提供反馈。

我想提供2017年所有课程的列表，并列出每门课程的平均评分。

我正在使用的两个表是

courses
course_id   course_name
---------------------------
1         | Public speaking
2         | Social media skills


feedback
course_id   overall_rating
--------------------------
1         |  3
1         |  5
1         |  4
1         |  4
2         |  3
2         |  3
2         |  4

我可以用

获取课程列表

$yearscourses = "SELECT courseid,coursetitle FROM courses WHERE coursedate1 BETWEEN '2017-01-01' AND '2017-12-31'";
$yearsresult = mysqli_query($connect, $yearscourses);

平均评分

$avgrating = "SELECT AVG(overall_rating) FROM feedback WHERE courseid='courseid'";

但是我很难在使用HTML / PHP的表格中表达这一点。我尝试了以下但是它在每一行中重复相同的平均值。

<table cellspacing="0" table border="1" cellpadding="padding:5px;">
<tr style="font-size:20px; font-weight:bold; text-align:center;">
<td colspan="4">
2017
</td>
</tr><tr style="font-weight:bold;">
<td style="background-color:#AED6F1;">Course ID</td><td style="background-color:#AED6F1;">Course title</td><td style="background-color:#AED6F1;">Average rating</td>
</tr>

<?php

// Get a list of all courses for the year
$yearscourses = "SELECT courseid,coursetitle FROM courses WHERE coursedate1 BETWEEN '2017-01-01' AND '2017-12-31'";
$yearsresult = mysqli_query($connect, $yearscourses);


if (mysqli_num_rows($yearsresult) != 0) // Search has found results
{
$str = "\n";

}

while ($row = mysqli_fetch_array($yearsresult, MYSQLI_ASSOC)) {

// Get average rating for each course
$avgrating = "SELECT AVG(overall_rating) FROM feedback WHERE courseid='courseid'";
$avgresult = mysqli_query($connect, $avgrating);
foreach($avgresult as $row2) {

echo "<tr><td>" . $row['courseid']. "</td><td>" .$row['coursetitle']. "</td><td>".$row2['AVG(overall_rating)']."</td></tr>\n";
    $str .= " ";
}
}
$str .= "</table>\n</div>";
echo $str;


?>

Answer 1

使用别名作为平均值

 $avgrating = "SELECT AVG(overall_rating) as average_rating FROM feedback WHERE courseid='courseid'"; // use alias for avereage

和循环

echo "<tr><td>" . $row['courseid']. "</td><td>" .$row['coursetitle']. "</td><td>".$row2['average_rating']."</td></tr>\n";

Answer 2

正如另一篇文章所指出的，您可以使用别名轻松访问循环中的值。话虽这么说，你应该避免在循环中执行查询。您可以利用cx_oracle Documentation子句检索相同的信息。

以下是加入feedback表时的情况：

SELECT c.courseid, c.coursetitle, AVG(f.overall_rating) AS average_rating 
FROM courses c 
LEFT JOIN feedback f ON f.courseid = c.courseid
WHERE c.coursedate1 BETWEEN '2017-01-01' AND '2017-12-31'
GROUP BY f.courseid

请注意，对于此查询，未收到任何反馈的课程的NULL值为average_rating。

Answer 3

你只需要这样做

$avgrating = "SELECT AVG(overall_rating) FROM feedback WHERE courseid='".$row['courseid']."'";

而不是那个

$avgrating = "SELECT AVG(overall_rating) FROM feedback WHERE courseid='courseid'";

PHP / SQL JOIN 2表并得到平均分

3 个答案: