Question

I came up with a PHP version of a sitemap file for my dynamic site which only has index.php. Here is the result: http://www.oddsnavigator.eu/sitemap.php

Now I need a way to either export those results to sitemap.xml file, or to make this one XML and Google-friendly. Here’s the code:

<?PHP
$SQL="boring query";
$num_rows = mysql_num_rows(mysql_query($SQL));
echo $num_rows , "</BR>";
$result = mysql_query($SQL);
if($result === FALSE) { die(mysql_error());}

while($db_field=mysql_fetch_assoc($result)) {
$goto = "http://oddsnavigator.eu/?display=0&sport=" . $db_field['sportid'] . "&day=all&league=" . $db_field['stageid'];
$title = $db_field['sportname'] . " " . $db_field['tempname'] . " " . $db_field['stagename'] . " Sports Betting Odds Comparison - OddsNavigator.eu";

echo "<a href='$goto' title='$title'>" , $db_field['sportname'] , " " , $db_field['tempname'] , " " , $db_field['stagename'] , "</a>" , "</BR>";
}
?>

Any ideas?

Answer 1

In Your code $goto contains URLs, store all these url in an array called $urls and then use following function.
I expect the code goes in class, if not replace $this->variable with $variable.

//$urls = array();
$compress = true;

function save () 
{
    if (empty($this->urls)) return;
    $file = "sitemap.xml{$this->compress}";
    $xml = '<?xml version="1.0" encoding="UTF-8"?>' . "\n";
    $xml .= '<urlset xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">' . "\n";
    foreach ($this->urls as $url) 
    {
        $xml .= '  <url>' . "\n";
        if (is_array($url)) {
            foreach ($url as $key => $value) $xml .= "    <{$key}>{$value}</{$key}>\n";
        } 
        else {
            $xml .= "    <loc>{$url}</loc>\n";
        }
        $xml .= '  </url>' . "\n";
    }
    $xml .= '</urlset>' . "\n";
    $this->urls = array();
    if (!empty($this->compress)) $xml = gzencode($xml, 9);
    $fp = fopen(BASE_URI . $file, 'wb');
    fwrite($fp, $xml);
    fclose($fp);
}