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Basically I have a live search suggestions which uses AJAX. I'm not familiar with AJAX or XML however i've managed to get it working without any errors.

If you take a look at the code file you can see how I would like the XML data to be generated which is the title and url tags to be populated by a database of products. Instead of displaying the database string the live search suggestions is displaying everything between the XML tags for example between the title tags: ".$aa." .

Can anyone please help me work around this issue. It's most likely a silly mistake or an easy solution.

<link>
<?php
header("Content-type: text/xml");
echo "<?xml version='1.0' encoding='UTF-8'?>";
$productdetails = mysql_query("SELECT * FROM products");

while($row = mysql_fetch_array($productdetails)){
 $aa = $row['product_name'];
 $bb = $row['link'];
 echo "<title>".$aa."</title>";
 echo "<url>".$bb."</url>";
 
}
?>
</link>
Xirlas
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  • check http://stackoverflow.com/questions/5112282/get-mysql-database-output-via-php-to-xml – Saty Dec 01 '15 at 13:35

1 Answers1

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There were link tags before and after the code but oin the php you tried sending a header - I think this might be more like what you are trying to do.

<?php
    header("Content-type: text/xml");
    $productdetails = mysql_query("SELECT * FROM products");

    echo "<?xml version='1.0' encoding='UTF-8'?>";
    while( $row = mysql_fetch_array( $productdetails ) ) {
        echo "
        <link>
            <title>".$row['product_name']."</title>
            <url>".$row['link']."</url>
        </link>";
    }
?>
Professor Abronsius
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  • Thank you for your response. I had the link tags inside however i was getting this error "Start tag expected, '<' not found in" so I put it in the beginning and it seemed to work. I tried your code and got the same error. – Xirlas Dec 01 '15 at 13:46