$strQuery = "SELECT iNewsID, iNewsTitle FROM tb_news_CH ORDER BY
iNewsID DESC";
$result = mysql_query($strQuery);
$i = NUM;
while(($row = mysql_fetch_array($result )) && $i > 0)
{
$array[] = array("NewsID", substr($row["iNewsID"], 0, 40),
"NewsTitle", substr($row["vcNewsTitle"], 0, 40));

$i--;
}
$smarty->assign("News_CH", $array);
unset($array);
mysql_free_result();

出现错误

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in E:\www\index.php on line 33

哪里有错？谢谢