How to load php file to dynamically load dropdown

Question

I have a PHP script that dynamically shows the 'course' options from my database

<?php


    $db_host = 'localhost';
    $db_user = 'root';
    $db_pass = '';
    $db_name = '';

    $con = mysqli_connect($db_host,$db_user,$db_pass, $db_name);
    if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
    }   


    $sql = "SELECT courseID, name FROM courses";

    $result = mysqli_query($con, $sql) or die("Error: ".mysqli_error($con));

    while ($row = mysqli_fetch_array($result))
    {
    echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>';
    }


    ?>

I have a dropdown in my HTML (scorecard.php) page.

    <form> <select id="selectCourse" > <option value = "">Select  Course</option></select></form>

I was wondering would anyone know a script or way of getting this data to display in my dropdown.

Thanks for any help

I had it using a JSON approach but was having problems displaying the course 'name' and then POST 'courseID' — Aaron Crawley
– Aaron Crawley, Commented Jul 21, 2015 at 16:11

Bhavin Solanki · Accepted Answer · 2015-07-21 16:07:40Z

0

You have to run your while loop inside the select tag

Here is updated code

<?php

$db_host = 'localhost';
$db_user = 'root';
$db_pass = '';
$db_name = '';

$con = mysqli_connect($db_host,$db_user,$db_pass, $db_name);
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}   


$sql = "SELECT courseID, name FROM courses";

$result = mysqli_query($con, $sql) or die("Error: ".mysqli_error($con));

?>  

<form> 
  <select id="selectCourse" >
    <option value = "">Select  Course</option>
     <?php while ($row = mysqli_fetch_array($result))
    {
       echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>';
    } ?>

 </select>

answered Jul 21, 2015 at 16:07

Bhavin Solanki

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Comments

Ehab Eldeeb · Accepted Answer · 2015-07-21 16:07:26Z

0

echo'<option value="'.$row['courseID'].'">'.$row['name'].'</option>';

change it to

$courses[] = '<option value="'.$row['courseID'].'">'.$row['name'].'</option>';

and where u want it to be displayed

foreach($courses as $c){
    echo $c;
}

answered Jul 21, 2015 at 16:07

Ehab Eldeeb

7284 silver badges12 bronze badges

1 Comment

Aaron Crawley Over a year ago

What do you mean by 'as $c)' would i place that last section of code in the HTML page?

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