Load blank image if nothing found...

[roldahayes]

(last question for this project...!)

 

This code loads and image from a URL stored in a database.

 

Some of the results pages wont have an image stored in them and I don't want it to load the "image not found" icon.  Therefore I have uploaded a transparent .gif file and would like that to load if no result was found in the database.

 

I am using IF and ELSE but this doesn't seem to work correctly - Can anyone advise on what I am doing wrong please.

 

Thanks

 

 

<?php

$van_query = "SELECT Image_Van  FROM products WHERE Car_Make= '$strMake' AND Car_Model = '$strModel'";                                                             
					  
// the result of the query
$van_result = mysql_query($van_query) or die("Invalid query: " .mysql_error());


$pic = mysql_fetch_array($van_result);

// show the image


if ($van_query !== "")
echo "<img src='images/product_images/".$pic['Image_Van']."'/>";

else
echo "<img src=\"images/transparent.gif/>";

?>

[litebearer]

try...

<?php
$van_query = "SELECT Image_Van  FROM products WHERE Car_Make= '$strMake' AND Car_Model = '$strModel'";                                                             
// the result of the query
$van_result = mysql_query($van_query) or die("Invalid query: " .mysql_error());
$pic = mysql_fetch_array($van_result);
// show the image
$my_image = trim($pic['Image_Van']);
if ($my_image != "") {
echo "<img src='images/product_images/" . $my_image ."'/>";
}else{
echo "<img src=\"images/transparent.gif/>";
}
?>

the above will show only 1 record, to show more you will need to incorporate a WHILE loop

[roldahayes]

I only need it to show the one image,

 

That is still not working though, it is not loading the transparant.gif when it has nothing else to show...

[litebearer]

Oops..

change this

	echo "<img src=\"images/transparent.gif/>";

to this

echo "<img src='images/transparent.gif>'";

 

 

[roldahayes]

Excellent - Thanks for your help