PHP dropdown menu that concatenates 2 rows from MySQL data table

Question

Not experienced with creating forms in PHP.

I can get my form to produce a dropdown list that has one of my rows listed as an option, but as soon as I try to concatenate 2 rows together (from the same table) for option output...

a) It just doesn't work and I get errors

b) I get the first row as a single option, then my next row as a separate option.

I know there is a simple solution to this, but I am an online student just learning, and I can't seem to find a good example of the code to write it. I'm pretty sure it's an issue of quotes not being placed correctly.

MySQLTable Data:

Table Name: courses
Table Rows: course_id, course_name, max_enrolment
Sample Data: LO-COMP-8001, Intro to HTML, 20

function select_course(){
        global $open;
        $select = "SELECT * FROM courses";
        $result = mysqli_query($open, $select);
        return $result;
    }
    <form action="insert.php" method="post">
    <dl>
        <dt>Select Course</dt>
        <dd><select name="course_id">
           <?php   // CREATE dropdown menu
                 $result = select_course();  
                    while ($row = mysqli_fetch_assoc($result)) {
                    foreach ($row as $selection) {
                       echo "<option value=\"$selection\">$selection</option>";
                    }}
              ?>
        </select>
        </dd>
    </dl>

Then there are a few more form fields such as student name and student id afterwards...

Goal Output:

course_id course-name
"LO-COMP-8001 Intro to HTML" ... as a single connected dropdown option and other remaining courses in a dropdown menu

Current Output:

LO-COMP-8001 (as an option)
Intro to HTML (as another option! ... No good)
20 (must be hidden, I need to check if course is full in another function and either allow or deny a student to enrolled etc.)

I have tried:

// output is the one mentioned above..
echo "<option value=\"$selection\">$selection</option>";
// or alternatively...
echo '<option value="'.$row['course_id'].'">'.$row['course_id'].'</option>';

But the second option creates all kinds of weird results.

This is what I am experimenting with right now...

echo '<option value="'.$row['course_id'] $row['course_name']'">'.$row['course_id'] $row['course_name'].'</option>';

But there is a bunch of issues with quotes and square brackets, and I just don't know how to format it correctly for the output.

Any assistance is appreciated.

Your last effort is almost there. Try echo '<option value="' . $row['course_id'] . ' ' . $row['course_name'] . '">' . $row['course_id'] . ' ' . $row['course_name'] . '</option>'; — Nick
– Nick, Commented Apr 18, 2020 at 1:35
Thanks! Worked perfectly in combination with the answer from u/nelsonrakson. I took out the foreach loop and used the echo you provided. Perfect! — SteamCode666
– SteamCode666, Commented Apr 18, 2020 at 1:43

Nelson Rakson · Accepted Answer · 2020-04-18 01:42:42Z

2

$row holds the entire row as an associative array therefore no need for the 'foreach' loop.

function select_course(){
        global $open;
        $select = "SELECT * FROM courses";
        $result = mysqli_query($open, $select);
        return $result;
    }
    <form action="insert.php" method="post">
    <dl>
        <dt>Select Course</dt>
        <dd><select name="course_id">
           <?php   // CREATE dropdown menu
                 $result = select_course();  
                    while ($row = mysqli_fetch_assoc($result)) {?>
                       <option value="<?php echo $row["course_id"]; ?>"><?php echo $row["course_name"]; ?></option>
                    <?php }
              ?>
        </select>
        </dd>
    </dl>
    </form>

edited Apr 18, 2020 at 1:42

answered Apr 18, 2020 at 1:36

Nelson Rakson

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1 Comment

SteamCode666 Over a year ago

Thanks! As soon as I took the foreach out and changed the echo to: echo '<option value="' . $row['course_id'] . ' ' . $row['course_name'] . '">' . $row['course_id'] . ' ' . $row['course_name'] . '</option>'; It worked great. Appreciated!

SteamCode666 · Accepted Answer · 2020-04-18 02:05:52Z

0

I was able to come up with another solution as well:

Once the foreach loop was removed, I tried cleaning up the code some... I'm not sure if this is uncommon or 'bad' style, but it does work.

  $result = select_course();  
                    while ($row = mysqli_fetch_assoc($result)) {
                       $course_id     = $row['course_id'];
                       $course_name   = $row['course_name'];
                       echo  "<option value=\"$course_id\">$course_id $course_name</option>";

Results in: LO-COMP-8001 Intro to HTML as a single option, plus all my other courses in the database.

answered Apr 18, 2020 at 2:05

SteamCode666

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