Hi Team,
Below is my php code.
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<TITLE> Items </TITLE>
</HEAD>
<BODY>
<form method="post">
<select name="item">
<option> Select... </option>
<option> Flower Item</option>
<option> Choclate Item</option>
<option> Gift Item</option>
</form>
<?php
$item=$_POST['item'];
if ($item == "Flower")
{
include('Flower.php');
}
elseif ($place == "Choclate")
{
include('Choclate.php');
}
elseif ($place == "Gift")
{
include('Gift.php');
}
?>
</BODY>
</HTML>
What i want to do is, if user select choclate, then php should open choclate.php file and place all codes after form. So that, whenever user select either of choclate, flower of gift, he should able to see the table coded in respective file.
The choclate.php contains a table code, with images in each colum like below.
<BR><BR><BR>
<table id="customers" width="200" border="0" align="center" cellpadding="0" cellspacing="0">
<tr>
<th>Flower</th>
</tr>
</table >
<table id="customers" align="center" >
<tr >
<td align="center"><img src="images/1.jpg" name="home_0" border="0" width="100" height="120" hspace="0" vspace="0" ><br><b>Price=20</b></td>
<td align="center"><img src="images/2.jpg" name="home_0" border="0" width="100" height="120" hspace="0" vspace="0" ><br><b>Price=50</b></td>
<td align="center"><img src="images/3.jpg" name="home_0" border="0" width="100" height="120" hspace="0" vspace="0" ><br><b>Price=60</b></td>
<td align="center"><img src="images/4.jpg" name="home_0" border="0" width="100" height="120" hspace="0" vspace="0" ><br><b>Price=80</b></td>
<td align="center"><img src="images/5.jpg" name="home_0" border="0" width="100" height="120" hspace="0" vspace="0" ><br><b>Price=100</b></td>
</tr>
</table>
But above code is not working. Could you please tell me how to do it?
