Multiple dropdown values to create SQL query PHP

Question

I am currently tring to create a form wherein there is three dropdown option boxes and one submit button. All three of the dropdown boxes are populated from the database and I would like the selected options to be included into a new query and printed. This example only shows one dropdown

PHP Code

// Create connection
$con=mysqli_connect('', '', '', '');

// Check connection
if (mysqli_connect_errno($con))
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$course_dropdown ="";
$query_course = "SELECT * FROM course";
$result_course = mysqli_query($con,$query_course) or die(mysqli_error());

while($row = mysqli_fetch_assoc($result_course))
{
    $course_dropdown .= "<option value='{$row['CourseName']}'{$row['CourseName']}           </option>";
}

Above is the code that is used to create the dropdown lists

HTML

<form="index.php" method="post">
<select name="Course"><?php echo $course_dropdown; ?></select>
<input name="button" value="Submit" type="submit">

I am at a loss over what way to proceed, I have tried various different techniques but cannot come up with an answer.

Latest attempt

$course = mysqli_real_escape_string($con, $_POST['Course']);
$query = mysqli_query($con,"SELECT * FROM course_module WHERE CourseName = $course"); 

this brought an error

Notice: Undefined index: Course in C:\\Users\\seanin\\Desktop\\xampp\\htdocs\\index.php on line 33

So have edited as suggested and stil have an error, may be missing something small.

// Check connection
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$course_dropdown ="";
$query_course = "SELECT * FROM course";
$result_course = mysqli_query($con,$query_course) or die(mysqli_error());

while($row = mysqli_fetch_assoc($result_course))
{
$course_dropdown .= "<option value='{$row['CourseName']}'>{$row['CourseName']}</option>";
} 

if ($_POST['button'] == 'Submit') {
$course = mysqli_real_escape_string($con, $_POST['Course']);
$query = mysqli_query($con,"SELECT * FROM course_module WHERE CourseName = $course"); 
}

Still have this error

Notice: Undefined index: button in C:\\Users\\seanin\\Desktop\\xampp\\htdocs\\index.php on line 30

submit button issue

---

Nearly done, thanks for all the help so far.

What do I need to do to get the results and print them???

Answer 1

Please read about SQL injections . They can destroy your life.

I reckon that you are trying to access 'Course' in the following line and it is not defined:

$course = mysqli_real_escape_string($con, $_POST['Course']);

Are you able to submit the page? There is an error in your HTML form: <form="index.php" is not a valid HTML tag so you are not able to submit the page, that is if you posted the exact code you are using. Your form should be:

<form action="index.php" method="post">
<select name="Course"><?php echo $course_dropdown; ?></select>
<input name="button" value="Submit" type="submit">
</form> <!-- and don't forget the closing tag -->

You can check whether the page was submitted or not by doing something like this:

if ($_POST['button'] == 'Submit') {
    $course = mysqli_real_escape_string($con, $_POST['Course']);
     // please note the missing single quotes, and please read the first line of my answer
    $query = mysqli_query($con,"SELECT * FROM course_module WHERE CourseName = '$course'");
}

There is also an invalid HTML syntax in the following line:

$course_dropdown .= "<option value='{$row['CourseName']}'{$row['CourseName']}           </option>";

The format for <option> is: <option value="value">label</option> .

Answer 2

您的下拉列表不起作用的原因是缺少“>”，用此替换while循环内的行

$course_dropdown .= "<option value='{$row['CourseName']}'>{$row['CourseName']}</option>";