[Resolved] [Resolved] Select Option Affect the Input(Text) ??

wendymelon

When the user click on the Select OPtion, it will display the corresponding information in the INPUT Text.

How can I do this?Can anybody help me out?

<?php
     include('connect2db.php');
     $query = "SELECT employee.Emp_ID from employee,supervisor WHERE employee.Supervisor_ID = supervisor.Supervisor_ID and supervisor.Supervisor_ID ='". $_SESSION['Department']."'";
     $result = mysql_query($query);
     echo '<font size="4"><b>Employee :&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>';
     echo '<select name="emID" id="emID" style="width: 100px;">';
     echo '<option></option>';
     while($row = mysql_fetch_array($result,MYSQL_NUM)){
         echo '<option value="'.$row[0].'">'.$row[0].'</option>';

 }
 echo '</select>';


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

In the codes above.....the Employee ID that is selected will display the NAME in the Input text.

TheDefender

This is a JavaScript issue and not PHP, but I have something in my snippet that will help you.

<script language="JavaScript">
<!-- 
function onChange() {
var Current = document.form.emID.selectedIndex;
document.form.EName.value = document.form.emID.options[Current].text;
}
//--></script>

<?php 
// I didn't see your form tag on your posted code, so I added this part.
echo '<FORM name="form" onSubmit="return false;">';


 include('connect2db.php'); 
 $query = "SELECT employee.Emp_ID from employee,supervisor WHERE employee.Supervisor_ID = supervisor.Supervisor_ID and supervisor.Supervisor_ID ='". $_SESSION['Department']."'"; 
 $result = mysql_query($query); 
 echo '<font size="4"><b>Employee :&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>'; 
 echo '<select name="emID" id="emID" style="width: 100px;" onChange="onChange()">'; 
 echo '<option></option>'; 
 while($row = mysql_fetch_array($result,MYSQL_NUM)){ 
     echo '<option value="'.$row[0].'">'.$row[0].'</option>'; 

 } 
 echo '</select>'; 


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

// Added this part too
echo '</FORM>'; 
?>

P.S. You can find this info here

wendymelon

after inserting the codes that you provided, (SHOWN BELOW)..

it gives me an ERROR of :
Stack Overflow at LINE 26

Line 26 : after the line of <form> tag

Why does this occur??

<script language="JavaScript">

function onChange()
{
         document.HODApplyfrm.EName.value = document.HODApplyfrm.emID.options[document.HODApplyfrm.emID.selectedIndex].value;
}
</script>

<html>
<head>
       <title>Title here!</title>
</head>
<body bgcolor="#D6D6AD" link="#0000FF" text="#000000" alink="#FF0000" vlink="#FF00FF">

<table bgcolor="#800000" width="100%" border="" bordercolor="#000000">
<tr height="50"><td>
<p><font size="6" color="#FFFFFF"><b>Employee Overtime Application Form</b></font></p>
</td></tr></table>


<form method="POST" action ="HODApply.php" name=HODApplyfrm>

<?php
     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['Department']."'";
     $result = mysql_query($query);
     echo '<font size="4"><b>Employee :&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>';
     echo '<select name="emID" id="emID" style="width: 100px;" onchange="onchange()">';
     echo '<option></option>';
     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; value="" name="EName" id="EName" ></input>';
 echo '<br>';
 echo '<font size="4"><b>OT Rate : &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>';
 echo '<select name="OT_Rate" id="OT_Rate">';
 $query3 = "SELECT * FROM otcode;";
 $result3= mysql_query($query3);
 while($rate = mysql_fetch_array($result3,MYSQL_NUM)){
     echo '<option>'.$rate[0].'</option>';
 }
 echo '</select>'
?>
............
.........
............
........
</table>
</form>
</body>
</html>

wendymelon

I managed to solve the problem without using the javascript function....i've added the javascript code to the ONCHANGE property.

<html>
<head>
<title>Title here!</title>
</head>
<body bgcolor="#D6D6AD" link="#0000FF" text="#000000" alink="#FF0000" vlink="#FF00FF">

<table bgcolor="#800000" width="100%" border="" bordercolor="#000000">
<tr height="50"><td>
<p><font size="6" color="#FFFFFF"><b>Employee Overtime Application Form</b></font></p>
</td></tr></table>
<form method="POST" action="HODApply.php" name="HODApplyfrm">
<?php
     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['Department']."'";
     $result = mysql_query($query);
     echo '<font size="4"><b>Employee :&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>';
     echo '<select name="emID" id="emID" style="width: 100px;" onchange="\javascript:document.HODApplyfrm.EName.value = document.HODApplyfrm.emID.options[document.HODApplyfrm.emID.selectedIndex].value">';
     echo '<option></option>';
     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; value="" name="EName" id="EName" ></input>';
 echo '<br>';
 echo '<font size="4"><b>OT Rate : &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</b></font>';
 echo '<select name="OT_Rate" id="OT_Rate">';
 $query3 = "SELECT * FROM otcode;";
 $result3= mysql_query($query3);
 while($rate = mysql_fetch_array($result3,MYSQL_NUM)){
     echo '<option>'.$rate[0].'</option>';
 }
 echo '</select>'
?>