.load into DIV with PHP content - php

Question

I am trying to .load a script called 'refreshImages.php'. Inside that script is a while loop pulling from the database. I have got it to load a single echo function but it wont load anything inside the while loop I have on the script... this is what the php file has...
<?php
include 'includes/config.php';
$pimages = mysql_query("SELECT * FROM property_images WHERE pid='$pid'");
//Cant Post Images So Leaving The Echo Content Out//
while($img = mysql_fetch_array($pimages)){
$image = $img['image'];
$image_alt = $img['image_alt'];
echo "<li>$img</li>";
}?>
I am using .load('refreshImages.php') on the page I need it to show up on.
Any explanation I am not seeing?

Answer 1

Your $img is an array, not a string. You will get output like <li>Array</li>, if you have stuff coming from the database. Is that what you mean? Or are you getting an empty result?
If empty - what does your mysql_num_rows tell you when ran against the result resource?

Answer 2

try changing this:
echo "<li>$img</li>";
to
echo "<li><img src=\"{$image}\" alt=\"{$image_alt}\" /></li>";

Answer 3

You may not be getting any results from the database. Try using this code which will display a message if there is something wrong with your sql query.
<?php
include 'includes/config.php';
$pimages = mysql_query("SELECT * FROM property_images WHERE pid=" . $pid );
if (mysql_num_rows($pimages) > 0) { // checks to see if you are getting results from db
while($img = mysql_fetch_array($pimages)){
$image = $img['image'];
$image_alt = $img['image_alt'];
echo '<li><a class="thumb" href="{$image}"><img src="{$image}" width="50px" height="50px" alt="{$image_alt}"></a></li>';
}
} else {
echo "no results returned from database";
} // end of mysql_num_rows check
?>

Answer 4

You might be better off concatenating all the images and then echo-ing it out rather than echo-ing each one e.g
$htmlOutput = '';
while($img = mysql_fetch_array($pimages)){
$image = $img['image'];
$image_alt = $img['image_alt'];
$htmlOutput .= "<li><img src=\"{$image}\" alt=\"{$image_alt}\" /></li>";
}
echo $htmlOutput ;