[Resolved] Textbox & Dropdown List HELP!!

[wendymelon]

Is there any problem with this code? This code will display a list of Emp_ID in a dropdown list.Then when the EmpID is choosen, it will get the corresponding Emp_Name to be displayed in the textbox next to it.

Please help..

 
include('connect2db.php'); 
$query = "SELECT Emp_ID FROM employee ORDER BY Emp_ID;"; 
$result = mysql_query($query); 
while($row = mysql_fetch_array($result,MYSQL_NUM)){ 
echo '<option>'.$row[0].'</option>'; 

} 
echo '</select> 
<input type="text" style="width: 250px; name="EmpName" id="EmpName"></input>'; 

if(isset($_POST['EmpID'])){ 
$query2 = "SELECT Emp_Name FROM employee WHERE Emp_ID = '".$_POST['EmpID']."';"; 
$result2 = mysql_query($query2); 
while($row2 = mysql_fetch_array($result2,MYSQL_NUM)){ 
?><script type="text/javascript" language="Javascipt"> 
document.HODApplyfrm.EmpName.value = </script><?php $row[0]; 

} 
}

[DonChris]

include('connect2db.php');

$query = "SELECT Emp_ID FROM employee ORDER BY Emp_ID";
$result = mysql_query($query);

echo "<select name='EmpID'>\n";

while($row = mysql_fetch_array($result,MYSQL_NUM)) {
	echo '<option>'.$row[0].'</option>\n';
}
echo "</select>\n";

$EmpID=$_POST['EmpID'];
if( isset($EmpID) && is_numeric($EmpID) ) {
	$query2 	= "SELECT Emp_Name FROM employee WHERE Emp_ID = '".$EmpID;
	$result2 	= mysql_query($query2);
	$row2 		= mysql_fetch_array($result2,MYSQL_NUM);
	$EmpName	=$row2[0];
}
echo "<input type='text' style='width: 250px;' name='EmpName' id='EmpName' value='".$EmpName."'>\n";

[wendymelon]

It gives me an error of

Undefined index: EmpID.... how can i solve it?

[wendymelon]

I managed to solve it using onChange="javascript"

This solution will help in the situation where....the user clicks on the option(ID), and it will show the corresponding (NAME) in the textbox in the same page.

include('connect2db.php');
     $query = "SELECT employee.Emp_ID, employee.Emp_Name FROM employee,supervisor WHERE employee.Supervisor_ID = supervisor.Supervisor_ID AND supervisor.Supervisor_ID = '".$_SESSION['EmpID']."' ORDER BY employee.Emp_ID;";
     $result = mysql_query($query);
     echo '<select name="emID" onChange="javascript:EName.value = emID.options[emID.selectedIndex].value;" id="emID" style="width: 100px;">';
     while($row = mysql_fetch_array($result,MYSQL_NUM)){
         echo '<option value="'.$row[1].'">'.$row[0].'</option>';

 }
 echo '</select>';


 echo '<input type="text" style="width: 250px; name="EName" id="EName" ></input>';