Display PHP variable in .php file displaying XML data

Question

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>

check stackoverflow.com/questions/5112282/…

Saty
– Saty

2015-12-01 13:35:14 +00:00
Commented Dec 1, 2015 at 13:35 — Saty
– Saty, Commented Dec 1, 2015 at 13:35

Professor Abronsius · Accepted Answer · 2015-12-01 13:40:50Z

0

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>";
    }
?>

answered Dec 1, 2015 at 13:40

Professor Abronsius

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1 Comment

Xirlas Over a year ago

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.

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