Hello world! I need help!!! I'm at a dead end. I cannot deduce picture from database. I use Ajax(jQuery) with PHP & DB phpmyadmin Ajax request

Hello world! I need help!!! I'm at a dead end. I cannot deduce picture from database. I use Ajax(jQuery) with PHP & DB phpmyadmin Ajax request

Hello world!

I need help!!! I'm at a dead end.

I cannot deduce picture from database. I use Ajax(jQuery) with PHP & DB phpmyadmin
Ajax request

$(document).ready(function(){ 
  // code to get all records from table via select box
  $("#employee").change(function() {   
     var id = $(this).find(":selected").val();
     var dataString = 'empid='+ id;   
     $.ajax({
        url: 'getEmployee.php',
        dataType: "html",
        data: dataString,
        cache: false,
        success: function(employeeData) {
           if(employeeData) {
              $("#heading").show();      
              $("#no_records").hide();
              $("#emp_name").text(employeeData.employee_name);
              $("#emp_photo").load(employeeData.employee_photo);//!!!!! here photo

              $("#emp_salary").text(employeeData.employee_salary);
              $("#records").show();     
           } else {
              $("#heading").hide();
              $("#records").hide();
              $("#no_records").show();
           }     
        }
     });
  })
});

PHP Code

<?php
include_once("db_connect.php");
if($_REQUEST['empid']) {
  $sql = "SELECT id, employee_name, employee_salary, employee_photo FROM employee WHERE id='".$_REQUEST['empid']."'";
  $resultset = mysqli_query($conn, $sql) or die("database error:". mysqli_error($conn));
 
  $data = array();
  while( $rows = mysqli_fetch_assoc($resultset) ) {
     $data = $rows;
  }
  echo json_encode($data);
} else {
  echo 0;   
}


bootstrap 

<div class="row" id="records">
            <td>  <div class="col-sm-3"><select id="employee">
                        <option value="" selected="selected">Select Employee Name</option>
                        <?php
                        $sql = "SELECT id, employee_name, employee_salary, employee_age FROM employee";
                        $resultset = mysqli_query($conn, $sql) or die("database error:". mysqli_error($conn));
                        while( $rows = mysqli_fetch_assoc($resultset) ) {
                            ?>
                            <option value="<?php echo $rows["id"]; ?>"><?php echo $rows["employee_name"]; ?></option>
                        <?php }   ?>
                    </select></div></td>
            <td>  <div class="col-sm-3" id="emp_name"></div></td>
            <td> <div class="col-sm-3" id="emp_age"></div></td>
            <td> <div class="col-sm-3" id="emp_salary"></div></td>
        </div>
<div class="row" id="no_records"><div class="col-sm-4">Plese select employee name to view details</div></div










plz help me!