如何从输入中获取价值并将其用于PHP查询中？

Question

I want to show in option values that datas where the id equal to the role ID. For example the Role1 has classification1 and classification2 and in this case it just show the classification1 and classification2 in the option value. My problem is that if I get the ID (It's in the input), then I write the query, the option value is empty, but in the text field, the ID is good. I fetch the data's id in another PHP file to the input. How can I solve this to show the classifications where the id equal to the input value?

index.php

<?php
$id = $_POST["RoleID"];

$testQuery = $conn->prepare("SELECT classification.class_id AS id, classification.name AS name
    FROM classification
    INNER JOIN role_classification ON classification.class_id = role_classification.class_id
    WHERE role_classification.role_id = ?");
$testQuery->bind_param("i", $id);
$testQuery->execute();
$testResult = $testQuery->get_result();
$testQuery->close();
?>

<form method="post" id="insert_form"> 
   <select name="classification[]" id="classification" multiple>
      <?php 
      while($row = $testResult->fetch_array()) {
      ?>
      <option value="<?php echo $row['id'];?>"><?php echo $row['name'];?></option>
      <?php
      }
      ?> 
   </select>
   <input type="hidden" name="RoleID" id="RoleID" /> 
</form>

Answer 1

Add a value to your input : <input type="hidden" name="RoleID" id="RoleID" value=" "/>

It may be <input type="hidden" name="RoleID" id="RoleID" value="<?php echo $row['id'];?"/>