php banner rotater problem - php

Question

i want to create dynamic banner rotater wih php ajax i want to pass the mysql_fetch_array() to an array to create a new array() to create xml response..........
here is my code
$sql = mysql_query("SELECT * FROM ads");
header('Content-type: text/xml');
echo '<?xml version="1.0" ?>';
while($row = mysql_fetch_array($sql)){
$title = $row['title'];
$img = $row['file'];
$body = $row['body'];
$ban = '<b>'.$title.'</b><br><br><img src="ads/'.$img.'"><br><br>'.$body;
$banners = array(
$ban,
);
$html = $banners[array_rand($banners)];
}
<banner>
<content><?php echo htmlentities($html); ?></content>
<reload>3000</reload>
</banner>
but it is return only one ad not return multiple ads how can i fix that

Answer 1

The proble is here: $banners = array($ban);. What you're trying to do is include all the ads in the $banners array as an entry but you are failing to achieve that.
The correct code for including an entry in an array would be $banner[] = $ban. That way each ad that comes as a result from your query will be stored as an individual entry.
So the correct code would be:
$sql = mysql_query("SELECT * FROM ads");
$banner = array(); //Define the array before trying to add elements.
header('Content-type: text/xml');
while($row = mysql_fetch_array($sql))
{
$title = $row['title'];
$img = $row['file'];
$body = $row['body'];
$ban = '<b>'.$title.'</b><br><br><img src="ads/'.$img.'"><br><br>'.$body;
$banner[] = $ban; //Adding a new entry at the end.
$html = array_rand($banner); //Getting a random entry.
}

Answer 2

Use this in the while loop:
$banners[] = $ban
Instead of
$banners = array(
$ban,
);
and
$html = array_rand($banners);
instead of
$html = $banners[array_rand($banners)];

Answer 3

And to what bzabhi said, define your
$banners = array();
before the while loop, and the randomization part has to go after the loop.