Variable drop down menu in php - php

Question

I want to have a dropdown menu with as many options as a number in my database.
This is what i have but doesnt work:
<?php
// set the pointer back to the beginning
mysqli_data_seek($result, 0);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo '<div class="w-clearfix">
<div class="price">' .$row['Price']. '€</div>
//some code
<select class="w-select selectorplatos" id="numeroplatos" name="numeroplatos" data-name="numeroplatos">';
$capacity= "SELECT Capacity FROM Meals WHERE Meal_ID = '".$meal."';";
$resultcap = mysqli_query($conn, $capacity) or die("Couldn't execute query. ". mysqli_error($conn));
$i = 1;
// output data of each row
while($row1 = $resultcap->fetch_assoc()) {
echo '<option value="' .$i. '">' .$i. "</option>";
$i = $i+1;
}
echo '</select>
<input class="w-button pedirya" type="submit" value="¡Pedir ya!" data-wait="Espera por favor..." wait="Espera por favor...">
</form>
// more code here
mysqli_close($conn);
?>
Any clue whats wrong?
The command to display the drop down list cannot be executed.
Cheers!

Answer 1

Try changing your code to this:
echo '<select class="w-select selectorplatos" id="numeroplatos" name="numeroplatos" data-name="numeroplatos">';
$capacity = "SELECT Capacity FROM Meals WHERE Meal_ID = '".$meal."';";
$resultcap = mysqli_query($conn, $capacity) or die("Couldn't execute query. ". mysqli_error($connection));
$i = 1;
while($i < $resultcap) {
echo '<option value="' .$i. '">' .$i. '</option>';
$i = $i+1;
};
echo '</select>';
You weren't properly concatenating you MySQL query. It's also good practice to add or die("Couldn't execute query. ". mysqli_error($connection)) to get feedback on what went wrong when executing your query.
I also added echo because, assuming all of this is wrapped inside <?php ?> tags, well you need to echo HTML.

Answer 2

Please try this. I have tested it:
<select class="w-select selectorplatos" id="numeroplatos" name="numeroplatos" data-name="numeroplatos">'
<?php
$sql = "SELECT Capacity FROM meals";
$result = $conn->query($sql);
$i = 1;
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<option value="' .$i. '">' .$i. "</option>";
$i = $i+1;
}
?>
</select>

Answer 3

Ok after a lot of research and inspiration from comments, i got it working the following way:
<?php
mysqli_data_seek($result, 0);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo '<div class="w-clearfix">
<div class="price">' .$row['Price']. '€</div>
<div class="priceindic">Precio por plato</div>
<div class="w-form numplatosselect">
<form class="w-clearfix" id="numerodeplatos" name="numerodeplatos" data-name="numerodeplatos" action="https://rest.nexmo.com/sms/json?api_key='
<label class="labelselector" for="numeroplatos-2">Numero de platos:</label>
<select class="w-select selectorplatos" id="numeroplatos" name="numeroplatos" data-name="numeroplatos">';
for ($i = 1; $i <= $row['Capacity']; $i++) {
echo '<option value="' .$i. '">' .$i. "</option>";
}
?>
a simple for loop does the trick using fetch_assoc from above and including Capacity in the original query