javascript - Why is my second <select> tag is not showing up?

I am trying to display <select id="second_choice"> tag after I choose one of the options in <select id="first_choice">. Both are selected from my database.

technician.php

<select name="cntNum" id="first_choice" class="form-control" onchange="fetch_select(this.value);" required>
  <?php 
    include('config.php');
    $query = "SELECT cntNum FROM contract";
    $result = mysqli_query($link,$query);
    while($row = mysqli_fetch_array($result)) {
      echo "<option>".$row{'cntNum'}."</option>";
    }                                                                                       
</select>
<select name="cntNum" id="second_choice" class="form-control" required></select>

technician_fetch.php

require_once( 'config.php' );

$choice = mysqli_real_escape_string($_GET['choice']);
$query = "SELECT * FROM equipment WHERE cntNum = $choice";
$result = mysqli_query($link,$query);

while($row = mysqli_fetch_array($result)){
    echo "<option>" . $row{'cntNum'} . "</option>";
}

jQuery

<script>
  $("#first_choice").change(function () {
    $("#second_choice").load("technician_fetch.php?choice=" + $( "#first_choice").val());
  });
</script>

Solution:

You need to close your PHP before writing anymore HTML.

<select name="cntNum" id="first_choice" class="form-control" onchange="fetch_select(this.value);" required>
  <?php 
    include('config.php');
    $query = "SELECT cntNum FROM contract";
    $result = mysqli_query($link,$query);
    while($row = mysqli_fetch_array($result)) {
      echo "<option>".$row{'cntNum'}."</option>";
    }
  ?>
<!-- PHP must be closed if you then continue with HTML -->                                                                                       
</select>
<select name="cntNum" id="second_choice" class="form-control" required></select>

If you are just writing PHP in a file then you don't need to close it but if you have a different language after it (HTML) you must close it.