PHP / SQL JOIN 2表并获得平均分

Question

I'm trying to present data collected from people who have attended training courses in 2017 and have given feedback on the course.

I want to present a list of all the courses that ran in 2017 and next to each course give the average rating.

The two tables that I'm working with are

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

I can get the list of courses with

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

And the average rating with

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

But I'm struggling to present this in a table using HTML / PHP. I tried the following but it just repeats the same average in every row.

<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

use alias for average

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

and in loop

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

Answer 2

As pointed out by the other post, you can use an alias to easily access the value in your loop. That being said, you should avoid performing queries inside your loop. You can retrieve the same info by taking advantage of the GROUP BY clause.

Here is how it would go when joining the feedback table:

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

Note that with this query, courses that did not receive any feedback will have a NULL value for the average_rating .

Answer 3

you just need to do this

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

instead of that

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