show another image if image not found

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james19

Hi
ive created a webpage which gets images from a database. I need to tell it to to display another image (which says no image available) if an image is not found. How would I do this????

Your suggestions would be very much appreciated

ellisgl

Can we see the code to see how you pull the images?

scragar

if your pulling images from the DB then it wont return any rows when looking, so use mysql_num_rows() to check the number of rows is 1 or more, otherwise redirect or load your default fall back image.

james19

the code is shown below ellisgl

<HTML>
<HEAD>
<TITLE>
All Shoes
</TITLE>
</HEAD>

<BODY>
<H1>All Shoes</H1>

<p>All shoes available for sale. You can search for certain sizes
by using the search function.</p>

<br>

<TABLE CELLSPACING=1 CELLPADDING=4 bgcolor="#FFFFFF" WIDTH=666 BORDER=0>

<?php

#connect to MySQL
$conn= pg_connect( "host=pgdbs user=wwwuser dbname=dwt password=wwwuser")
	 			or die( "Err:Conn" );


$query = "SELECT * FROM shoes";
  $result = pg_Exec ($conn, $query);

  $rows = pg_Numrows($result);

  if ((!$result) || ($rows < 1)) {
              echo "<H1>ERROR - no rows returned</H1><P>";
              exit;  //exit the script
              }

 for ($i=0; $i < $rows; $i++)
{
$code = pg_result($result,$i,"code");   	 
$colour = pg_result($result,$i,"colour");
    $name = pg_result($result,$i,"name");
$sex = pg_result($result,$i,"sex");
$min_size = pg_result($result,$i,"min_size");
$max_size = pg_result($result,$i,"max_size");
$price = pg_result($result,$i,"price");
$image_url = pg_result($result,$i,"image_url");
?>

<TR>
<TD colspan="7" align=center>
<IMG src= "<?php echo $image_url ?>" width="100" height="100">
</TD>
<TD>
No: <?php echo $code ?> , <?php echo $name ?>, Colour: <?php   	      echo $colour ?>, Min Size: <?php echo $min_size ?>,
	Max Size: <?php echo $max_size ?>, Price: £<?php echo 		$price ?>.00
</TD>

<?php
	}
?>

</BODY>
</HTML>

thasnks for the advice scragar, im very new to php which is why I dont fully know how to do what you have suggested lol but I am grateful for the advice and am currently looking into how I would do what you have suggested.
Thanks ellisgl and scragar

ellisgl

if(empty($image_url)))
{
$image_url = 'http://www.whatever.com/defaultimage.jpg';
}

scragar

try replacing:

<IMG src= "<?php echo $image_url ?>" width="100" height="100">

with something like:

<IMG src= "<?php
echo ($image_url=="")?"DEFAULT IMAGE URL GOES HERE":$image_url;
?>" width="100" height="100">

you will need to fill in the default image URL and maybe change the empty quotes to something else if your DB using something other than a blank string is no URL is input.

james19

The code you gave works. Thanks for the help ellisgl and scragar