What this message means?
What it is the error?
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
resource in f:\.....\none.php on line 42
This is the code.
Thanks very much.
Marcelo
<?php
if (eregi("block-Concurso.php", $PHP_SELF)) {
Header("Location: index.php");
die();
}
$usemarquee = 1;
$scrolldirection = "left";
global $prefix, $multilingual, $currentlang, $dbi;
$sql = "DELETE FROM ".$prefix."_anuncios WHERE datafim<NOW()";
sql_query($sql, $dbi);
if ($multilingual == 1) {
$querylang = "WHERE (alanguage='$currentlang' OR alanguage='')";
} else {
$querylang = "";
}
$result = "SELECT categoria, nombre, datafim FROM ".$prefix."_anuncios ORDER
BY datafim LIMIT 0,30";
sql_query($result, $dbi);
$content = "<table width=\"100%\" border=\"0\">";
$content .="<center> <STYLE=\"text-decoration: none\"><font
color=\"#666666\"><b>�ltimos 10 Concursos</b></center>";
$content .= "<Marquee Behavior=\"Scroll\" Direction=\"$scrolldirection\"
$width=\"100%\" ScrollAmount=\"3\" ScrollDelay=\"90\"
onMouseOver=\"this.stop()\" onMouseOut=\"this.start()\">";
\\ line 42
while (list($categoria, $nombre, $datafim, $counter) =
mysql_fetch_row($result, $dbi)) {
$content .= "<img src=blocks/images/diamond.gif> <a
href=\"modules.php?name=Concurso\"><font color=\"#096C88\"><b>$categoria
<br> $nombre</b></a> ";
}
$content .= "</table>";
?>
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