我在一個簡單的 PHP MySQL 項目中遇到問題。當(dāng)我嘗試從選擇選項中獲取記錄時，它只提供 ID。我也希望傳遞來自選擇的值。<form class="form-style-9" method="POST" action="function.php?type=updatedistrict">``    <ul>    <li>ID:      <input type="hidden" name="dist_id"  class="field-style field-full align-none"  value="<?php echo $ids;?>" readonly />    </li>    <li>District Name:    <input type="text" name="name" class="field-style field-full align-none" value="<?php echo $names;?>" />    </li>    <li>Postal Code:    <input type="text" name="code" class="field-style field-full align-none" value="<?php echo $codes;?>"/>    </li>    <li>Province    <select name="province_id" class="form-control">      <option value="">Select Province</option>      <?php      $sql = mysqli_query($conn, "SELECT id, province From province");      $row = mysqli_num_rows($sql);      while ($row = mysqli_fetch_array($sql)){      echo "<option value='". $row['id'] ."'>" .$row['province'] ."</option>" ;      }      ?>      </select>    <li>    <input type="submit" value="Update" name="submit" />    </li>    </ul>    </form>這是更新功能代碼：case 'updateprovince':if (isset($_POST['submit'])) {          $id = $_POST['dist_id'];          $prov_id= $_POST['province_id'];                    $query=$conn->query("UPDATE province SET province='$prov_id' where id='$id'");          if ($query) {            echo '<script> window.location.replace("province.php");                  </script>';          }          else          {            echo "issue";          }      }break;