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Home/ Questions/Q 7999637
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T15:33:41+00:00 2026-06-04T15:33:41+00:00

I know I am making rookie errors here, and that I need to make

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I know I am making rookie errors here, and that I need to make a variable to pass forward – though I am not sure how to approach this one, and searching online tutes is not helping. If someone can steer me in the right direction I would be really grateful.

I would like to have the results echo into an array. That is; have ‘xmlurl’ row field from the ‘xml’ table display in the $rss array. I hope that makes sense.

// Get URLs from Database
$result = mysql_query("SELECT * FROM xml");
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
printf("'%s',", $row["xmlurl"]);
}
mysql_free_result($result);

// Take the URLs (xmlurl) and place them in an array
$rss = array(
'http://www.xmlurl.com.au',
'http://www.xmlurl.com.au'
);
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  1. Editorial Team
    Editorial Team
    2026-06-04T15:33:42+00:00Added an answer on June 4, 2026 at 3:33 pm

    Try this:

    $rss = array();
    
    $result = mysql_query("SELECT * FROM xml");
    while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        $rss[] = $row["xmlurl"];
    }
    mysql_free_result($result);
    
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