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Posted by Steve on 09/10/06 21:18
> Steve,
>
> The code you've posted shouldn't do that. Mind posting ALL of you
> rcode? For instance - where do you fetch the data and retrieve the
> category?
>
Here you go.
<?
if (isset($_POST['parts_choice']))
{
$parts_choice = $_POST['parts_choice'];
if (substr($parts_choice,0,6) != "Choose")
{
if (db_connect())
{
$query2 = "select parts.category, parts.part_number,parts.years, parts.description, parts.price, parts.graphic, categories.description
AS Category";
$query2 = $query2." FROM parts INNER JOIN categories on parts.category = categories.code";
$query2 = $query2." where parts.category = '$parts_choice'";
//echo("<p class=\"text\" align=\"center\">\$query2= $query2</p>");
$result2 = mysql_query($query2) or die(mysql_error());
$num_results2 = mysql_num_rows($result2) or die(mysql_error());
if ($_POST['get_parts'] == "Display Parts")
{
$category_name = mysql_result($result2,0,6);
?>
<center>
<h2><?php echo $category_name ?></h2>
<table border="1" cellpadding="5">
<tr>
<th>Part Number</th>
<th>Years</th>
<th>Description</th>
<th>Price</th>
<th>Picture</th>
</tr>
<?
}
else
{
echo("<p class=\"text\" align=\"center\">There was an error.</p>");
}
for ($i = 0;$i < $num_results2; $i++)
{
$row2 = mysql_fetch_array($result2);
if (!$row2)
{
echo "No parts found for this category.";
}
?>
<tr>
<td><?php echo $row2["part_number"]?></td>
<td><?php echo $row2["years"]?></td>
<td><?php echo $row2["description"]?></td>
<td>$<?php echo $row2["price"]?></td>
<?php
if (strlen($row2["graphic"]) > 0)
{
?>
<td align="center"><img src="graphics/<?php echo $row2["graphic"].".jpg"?>" border="0"></td>
<?php
}
else
{
?>
<td>No picture available</td>
<?php
}
?>
</tr>
<?
}
?>
</table>
<br>
<br>
<?
}
}
}
do_html_footer();
?>
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