จะให้ select menu เลือก option จากฐานข้อมูลโดยเช็คจาก textbox

<script type="text/javascript">
$(document).ready(function(){

	$("#txtCustomerID").change(function(){
			$.ajax({ 
				url: "return_emp_data.php" ,
				type: "POST",
				data: 'sCusID=' +$("#txtCustomerID").val()
			})
			.success(function(result) { 

				var obj = jQuery.parseJSON(result);
				
					if(obj == '')
					{
						
					}
					else
					{
						  $.each(obj, function(key, inval) {

							   $("#txtCustomerID").val(inval["emp_no"]);
							   $("#emp_name1").val(inval["emp_name"]);
							   $("#emp_Lname1").val(inval["emp_lname"]);
							   $("#department").val(inval["emp_dep"]);

						  });
					}

			});

		});
	});
</script>


       <div class="col-md-3">
      <label for="inputdefault">รหัสพนักงาน</label>
      <input id="txtCustomerID" name="emp_id" type="text"  required class="form-control" >
    </div>
    
    	<div class="col-md-4">
      <label for="inputdefault">ชื่อ</label>
      <input name="emp_name" type="text"  required class="form-control" id="emp_name1">
    </div>
    
   	 <div class="col-md-5">
      <label for="inputlg">นามสกุล</label>
      <input name="emp_lname" type="text" required class="form-control " id="emp_Lname1">
    </div>
    
      <div class="col-md-6">
      <label for="inputsm">หน่วยงาน</label>
      <select name="department" type="text" class="form-control " id="department" required>
      <option value="" > <strong >กรุณาเลือกหน่วยงาน</strong></option>
      <?
		
		$sq="SELECT * FROM tb_department ORDER BY dep_id";
		$p1=mysql_query($sq);
		while($pf=mysql_fetch_array($p1)){
		?>
        <option value="<?=$pf['dep_id']?>"><?=$pf['dep_name']?></option>
        <?
		}
		?>
      </select>
      </div>

<?
	include ('module/con_db.php');
?>

<?php
	$strSQL = "SELECT * FROM tb_employee WHERE emp_no = '".$_POST["sCusID"]."' ";
	$objQuery = mysql_query($strSQL) or die (mysql_error());
	$intNumField = mysql_num_fields($objQuery);
	$resultArray = array();
	while($obResult = mysql_fetch_array($objQuery))
	{
		$arrCol = array();
		for($i=0;$i<$intNumField;$i++)
		{
			$arrCol[mysql_field_name($objQuery,$i)] = $obResult[$i];
		}
		array_push($resultArray,$arrCol);
	}
		
	echo json_encode($resultArray);
?>