Mark Hensler — Sat, 15 Dec 2001 05:36:39 +0000

No, that error tells you that $news is not valid.

This should give you more debuging help:

<?php
$query = \"SELECT * FROM news ORDER BY date DESC LIMIT 8\";
$result = mysql_query($query);
if (!$result) {
    echo \"\n\n\";
    echo \"<!--\n\";
    echo \"FILE: \" . __FILE__ . \"\n\";
    echo \"LINE: \" . __LINE__ . \"\n\";
    echo \"\n\";
    echo \"QUERY ERROR:\n\";
    echo mysql_error() . \"\n\";
    echo \"\n\";
    echo \"QUERY:\n\";
    echo $query . \"\n\";
    echo \"-->\n\";
    echo \"\n\";
}
while ($result && $row=mysql_fetch_array($result)) {
    // do something
}
?>

This should print the file and approximate line number where the faulty query was executed. It will also print the mySQL error and faulty query.

a_gajic — Fri, 14 Dec 2001 21:05:43 +0000

Yes news is valid.

Warning: Supplied argument is not a valid MySQL result resource in C:\apache\server\htdocs/agNews/read.php on line 8

that is the error

here is the line above the while command..

<?php
$news = mysql_query(\"SELECT * FROM news ORDER BY date DESC LIMIT 8;\") || die(\"Cannot select records\");
?>

Is it my query? It's the only other thing that I can think of..

Mark Hensler — Fri, 14 Dec 2001 17:04:18 +0000

What is the error?
Is $news a valid mySQL result?