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What I want is that, I have two buttons in same form. But I want them to post values on pages where they are linked, given that they are linked to two different pages.
If I am unable to explain my problem please refer following codes.
<?php
session_start();
echo "<center><table width='75%'>";
echo "<tr><td><div align='left'><h3>Welcome ".$_SESSION['userName']."!</h3></div></td>";
echo "<td><div align='right'><a href='StoreSignOut.php'>Sign Out</a></div></td></tr>";
echo "</table></center>";
include_once('DatabaseConnection.php');
$order=$_POST['store'];
$sql="SELECT * from ".$order;
$query = mysql_query($sql);
if($query)
{
echo "<br />if ".mysql_error();
}
else
{
echo "<br />else ".mysql_error();
}
echo "<form method='post' name='myform' id='myform'>";
echo "<center>";
$_SESSION['order']=$order;
echo "Order No. <b>".$order."</b><br />";
echo "<table border='20' cellspacing='0'>";
echo "<tr>";
echo "<th><center>Product Order</center></th>";
echo "<th><center>Quantity</center></th>";
echo "</tr>";
while($result = mysql_fetch_assoc($query))
{
echo "<tr><td>".$result['medname']."</td>";
echo "<td>".$result['quantity']."</td></tr>";
}
echo "</table><br /><br />";
echo "<input type='submit' value='Order Completed' name='submit' id='submit' onclick='myform.action='OrderComplete.php'; return true;')></input>";
echo "<input type='submit' value='Order Given' name='received' id='received' onclick='myform.action='OrderReceived.php'; return true;'></input></center>";
echo "</form>";
?>
In the line where
echo "<input type='submit' value='Order Completed' name='submit' id='submit' onclick='myform.action='OrderComplete.php'; return true;')></input>";
echo "<input type='submit' value='Order Given' name='received' id='received' onclick='myform.action='OrderReceived.php'; return true;'></input></center>";
These are two buttons. I want them to submit different actions when clicked.
It will be so nice if anyone can help. Thank You...
</div>