[Resolved] Dynamic select list... how to pass selected ID value?

[sunshine66]

Hi,

I am dynamically creating a select list from a MySQL database (employeeProfile table) so that when the user selects an option and clicks PRINT, they are redirected to a page that displays the entire employee Profile.

Problem is that I can't pass the EmployeeID variable so that the next page knows which employee profile to display.

I am attempting 2 methods and neither works. Assuming either one of these is close to the correct way to do this, I don't know which one is the better method.

Method 1: Form action method
Method 2: HREF method

With the HREF method, I am passing a variable, but it is not the employeeID of the employee I select !!

Can someone help me fix this?

Also, why do I have to repeat the same query for each instance I want the dynamic select list? Is there a way around this?

<?php

session_start();

if ($_SESSION['valid_log'][0] != $_SERVER['REMOTE_ADDR'] || 
$_SESSION['valid_log'][1] != $_SESSION['User'])
{
//redirect to error page if session is invalid
echo '<META HTTP-EQUIV="Refresh" Content="1; URL=/error_access.html">';
exit();
}

$link = mysql_connect("localhost", "root", "password") or die("Could not connect. Please try again later.");
mysql_select_db("database",$link) or die("Could not select database. Please try again later.");

// Get Company ID for user logged into system
    $query = "SELECT * FROM userProfiles,companyProfile WHERE userProfiles.CompanyID=companyProfile.CompanyID";
    $result = mysql_query($query);	
    $row = mysql_fetch_array($result);
     if($row) {
        $CoID = $row["CompanyID"];
        $id = $row["ProfileID"];
    }

// Find employees profiles for Company ID 
    $query2 = "SELECT * FROM employeeProfile WHERE CompanyID=$CoID ORDER BY EmployeeLast,EmployeeFirst"; 
    $result2 = mysql_query($query2);	

   $query3 = "SELECT * FROM employeeProfile WHERE CompanyID=$CoID ORDER BY EmployeeLast,EmployeeFirst"; 
    $result3 = mysql_query($query3) or die("Error with query 3.");	
/* ??????????????????????????????????????
Why do I have to repeat the same query for each instance I want the dynamic select list? 
????????????????????????? */

// create select list with Print as the form function
echo '<FORM NAME="form1" METHOD="post" ACTION="f-A_initialnotice.php"> 
	<SELECT name="initialnotice"> ';

while($line = mysql_fetch_array($result2))  
	{  
	$lastname = $line["EmployeeLast"];
	$firstname = $line["EmployeeFirst"]; 
	$id = $line["EmployeeID"];   
	echo "<OPTION VALUE='$id'>$lastname, $firstname</OPTION>";  
}

echo'</SELECT> 
	<input type=hidden name="id" value="$id">
	<INPUT class="small" TYPE="submit" NAME="print" VALUE="Print"> 
	</FORM>';


// create select list with Print as the HREF function

?> 

<SELECT NAME="terminations" SIZE="1"> 
<?

// Find employees profiles for Company ID 

while($line = mysql_fetch_array($result3))  
{  
$lastname = $line["EmployeeLast"];
$firstname = $line["EmployeeFirst"]; 
$id = $line["EmployeeID"];   
echo "<OPTION VALUE='$id'>$lastname, $firstname</OPTION>";  
}
?> </SELECT> <A HREF="f-A_initialnotice.php?id=<?php echo $id ?>">PRINT</A>

[SD-Syndicate]

Hi Tracey,

I hope this will be of some help.

Form Method:
When you create your select list, Call it say id and make the Label: The Employer name etc And the value: The Employers id.

and on your form action, make it goto the page where you want the profile to display.

Then on that page add

if (!$_GET['id']) { $id=$_POST['id']; }

At the very top of the page. Then all you need to do is Select from the database WHERE id=$id.

HREF Method:

This is probably simpler.

On the Print Button Just Add

 f-A_initialnotice.php?id=$id

Hopefully that should work.

[sunshine66]

For the HREF method, I am already using the
f-A_initialnotice.php?id=$id.

The ID value it passes is always the last ID in the employee list.
This is because of :

$id = $line["EmployeeID"];

I think I need to move $line["EmployeeID"] directly into the OPTION VALUE parameter, but I'm having trouble with this:

echo "<OPTION VALUE='$line["EmployeeID"] >$lastname, firstname</OPTION>";

Am I on the right track?

[sunshine66]

I finally figured it out...
You cannot use a SELECT box without the FORM tags.
The form action needs to contain the value of the SELECT NAME.

<FORM ACTION="f-B_group.php?id=<?php echo $id ?>" METHOD="POST"> 
<SELECT NAME="id" SIZE="1"> 
<?
// Find employees profiles for Company ID 
    $query3 = "SELECT * FROM employeeProfile WHERE CompanyID=$CoID ORDER BY EmployeeLast,EmployeeFirst"; 
    $result3 = mysql_query($query3) or die("Error with query 3.");	

// Find employees profiles for Company ID 
	while($row3 = mysql_fetch_array($result3))  

	{  

	$lastname = $row3["EmployeeLast"];
	$firstname = $row3["EmployeeFirst"]; 
	$EEid = $row3["EmployeeID"];   

	echo "<OPTION VALUE=$EEid>$lastname, $firstname</OPTION>";  

	}
?> 
</SELECT>
<INPUT TYPE="submit" NAME="Print" VALUE="Print" class="small"> 
</FORM>