cant display image from database outside of table in php

Question

Hello,

I have this code

<?php
session_start();
if (isset($_GET["cmd"]))
  $cmd = $_GET["cmd"];
else
  die("You should have a 'cmd' parameter in your URL");
 $pk = $_GET["pk"];
$con = mysql_connect("localhost","root","geheim");
if(!$con)
{
die('Connection failed because of' .mysql_error());
}
mysql_select_db("ebay",$con);
if($cmd=="GetAuctionData")
{
echo "<table border='1' width='100%'>
<tr>
<th>Username</th>
<th>Start Date</th>
<th>Description</th>
</tr>";
$sql="SELECT * FROM Auctions WHERE ARTICLE_NO ='$pk'";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
echo "<tr>
<td>".$row['USERNAME']."</td>
<td>".$row['ARTICLE_NO']."</td>
<td>".$row['ARTICLE_NAME']."</td>
<td>".$row['SUBTITLE']."</td>
<td>".$row['CURRENT_BID']."</td>
<td>".$row['START_PRICE']."</td>
<td>".$row['BID_COUNT']."</td>
<td>".$row['QUANT_TOTAL']."</td>
<td>".$row['QUANT_SOLD']."</td>
<td>".$row['ACCESSSTARTS']."</td>
<td>".$row['ACCESSENDS']."</td>
<td>".$row['ACCESSORIGIN_END']."</td>
<td>".$row['USERNAME']."</td>
<td>".$row['BEST_BIDDER_ID']."</td>
<td>".$row['FINISHED']."</td>
<td>".$row['WATCH']."</td>
<td>".$row['BUYITNOW_PRICE']."</td>
<td>".$row['PIC_URL']."</td>
<td>".$row['PRIVATE_AUCTION']."</td>
<td>".$row['AUCTION_TYPE']."</td>
<td>".$row['ACCESSINSERT_DATE']."</td>
<td>".$row['ACCESSUPDATE_DATE']."</td>
<td>".$row['CAT_1_ID']."</td>
<td>".$row['CAT_2_ID']."</td>
<td>".$row['ARTICLE_DESC']."</td>
<td>".$row['COUNTRYCODE']."</td>
<td>".$row['LOCATION']."</td>
<td>".$row['CONDITIONS']."</td>
<td>".$row['REVISED']."</td>
<td>".$row['PAYPAL_ACCEPT']."</td>
<td>".$row['PRE_TERMINATED']."</td>
<td>".$row['SHIPPING_TO']."</td>
<td>".$row['FEE_INSERTION']."</td>
<td>".$row['FEE_FINAL']."</td>
<td>".$row['FEE_LISTING']."</td>
<td>".$row['PIC_XXL']."</td>
<td>".$row['PIC_DIASHOW']."</td>
<td>".$row['PIC_COUNT']."</td>
<td>".$row['ITEM_SITE_ID']."</td>
<td>".$row['STARTS']."</td>
<td>".$row['ENDS']."</td>
<td>".$row['ORIGIN_END']."</td>
</tr>
<tr><td></td></tr>";

}
echo "</table>";
echo "<img src=".$row['PIC_URL'].">";
}
mysql_close($con);
?>

Here is the generated html:

<table border='1' width='100%'>
<tr>
<th>Username</th>
<th>Start Date</th>
<th>Description</th>
</tr><tr>
<td>fashionticker1</td>
<td>220288560247</td>
<td>Ed Hardy Herren Shirt Rock & Roll Weiss XXL Neu & OVP</td>
<td></td>
<td>0.00</td>
<td>49.00</td>
<td>0</td>
<td>1</td>
<td>0</td>
<td>1.10.2008 16:22:09</td>
<td>6.10.2008 16:22:09</td>
<td>6.10.2008 16:22:09</td>
<td>fashionticker1</td>
<td>0</td>
<td>1</td>
<td>0</td>
<td>59.00</td>
<td>http://storage.supremeauction.com/flash/ebay2/10/49/76/10497654/13895964e.jpg&lt;/td&gt;
<td>0</td>
<td>1</td>
<td>6.10.2008 16:21:47</td>
<td>6.10.2008 16:28:31</td>
<td>32315</td>
<td>0</td>
<td><!-- +++++++++++++++++++++++++ Bitte ändern Sie im eigenen Interesse nichts an diesem Code! ++++++++++++++++++++++++ -->
<!-- +++++++++++++++++++++++++ Das kann massive Fehldarstellungen ihrer Auktion zur Folge haben! +++++++++++++++++++ -->
<!-- +++++++++++++++++++++++++ ++++++++++++++++++++++++++ Ihr Supreme Team +++++++++++++++++++++++++++++++++++++++++ -->
</td>
<td>
<br>
<br>
<style ty</td>
<td>float: right;
</td>
<td>margin: 0px;
</td>
<td>0</td>
<td>0</td>
<td>0</td>
<td>padding:5px; 
</td>
<td>0.00</td>
<td>0.00</td>
<td>0.00</td>
<td>0</td>
<td>0</td>
<td>font-size: 12px;
</td>
<td>color: #333333;
}
#h</td>
<td>0000-00-00 00:00:00</td>
<td>0000-00-00 00:00:00</td>
<td>0000-00-00 00:00:00</td>
</tr>
<tr><td></td></tr></table><img src=>

Whatever I do, I can not get an image to display, and when PIC_URL is empty, it only displays a placeholder image above the table, and I want it below.

Answer 1

It should be:

echo "<img src=\"".$row['PIC_URL']."\">";

Answer 2

<style ty</td> looks like the problem to me.

Answer 3

You are generating that img tag outside of your while loop, which ends when $row is false. In other words, you'll only reach that code when $row doesn't contain any information because you've finished reading rows!

Either output a cell for the image, or output a new row with a single cell for the image spanning the entire row.

Answer 4

while ($row = mysql_fetch_array($result))
{
*snip*
}
echo "</table>";
echo "<img src=".$row['PIC_URL'].">";

Because the loop finished, $row is false. You need to change it to something like.

while ($row = mysql_fetch_array($result))
{
*snip*
$lastImg = $row['PIC_URL'];
}
echo "</table>";
echo "<img src=\"$lastImg\">";

Im starting to doubt you have a valid PIC_URL returned by yer database query. You should try this, see if you get any valid pictures from the url's, and especially if the last url in the list is valid.

while ($row = mysql_fetch_array($result))
{
    *snip*
    $images[] = $row['PIC_URL'];
}
echo "</table>";
?><p>Are these valid urls you can open in yer browser?</p><pre><?php
    print_r($images);
?></pre><?php