Warning:mysql_fetch_array():_PHP教程

$connect=mysql_connect("localhost","root","123") or die("无法连接数据库".mysql_error());

mysql_select_db("cmstest") or die("无法选择数据库".mysql_errno());

$sql="select * from article where pid=1";
$result=mysql_query($sql,$connect) or die("无法查询sql语句".mysql_error());
$nums=mysql_num_rows($result);
echo "nums is:".$nums;
while($res=mysql_fetch_array($result)){
echo $res[0];
}

原来是$result重复了.

http://www.bkjia.com/PHPjc/632485.htmlwww.bkjia.comtruehttp://www.bkjia.com/PHPjc/632485.htmlTechArticleWarning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in $connect=mysql_connect(localhost,root,123) or die(无法连接数据库.mysql_error()); mysq...