如何在PHP中基于所选值及其中的数据创建多个下拉列表？

Question

So I am trying to create multiple dropdown boxes that shows data from a database. I am able to create multiple dropdown, but it only shows data in the 1st dropdown and not on the other created ones. I am new to this and would like some help from some of the pros. Here is the code:

<form action="#" method="post">
    <select id="aantalMaaltijden" name="aantalMaaltijden">
        <option value="1">1</option>
        <option value="2">2</option>
        <option value="3">3</option>
        <option value="4">4</option>
    </select>
    <input type="submit" name="submit" value="Bereken" />
</form>

<?php
$sql = "SELECT * FROM test";
$result = $conn->query($sql);

if(isset($_POST['submit'])){
    $selected_val = $_POST['aantalMaaltijden'];  // Storing Selected Value In Variable

    for($i=0; $i<$selected_val; $i++){
        echo "<select>";

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

        echo "</select>";                   
    }
}

?>

Thank you :)

Answer 1

Change your code like this and let me know

if (isset($_POST['submit'])) {

    $options = "";
    $selected_val = $_POST['aantalMaaltijden'];  // Storing Selected Value In Variable

    while ($row = $result->fetch_array()) {

        $options .= "<option>" . $row['name'] . "</option>";
    }

    for ($i = 0; $i < $selected_val; $i++) {

        echo "<select>";
        echo $options;
        echo "</select>";
    }
}

Edited

The issue was with $result->fetch_array() . You can't call $result->fetch_array() twice, against the same resource. The resource result you pass in to $result->fetch_array() is done by reference. You'll need to reset the position of the pointer before you can use $result->fetch_array() a second time. That's why your 1st dropdown gets filled with data not others.

Since, all your dropdowns consist same options, I added all the options in the variable $options and the same concatenated in all select .