看看$sql = mysql_query("select userid from user where userid=".$userid."",$db);这句报什么错。
$sql = mysql_query("select userid from user where userid=".$userid."",$db) or die(mysql_error())
$sql = mysql_query("select userid from user where userid=".$userid."",$db) or die(mysql_error())
mysql_select_db($dbname,$db); //链接BBS数据库$db2 = mysql_connect($dbhost, $dbuser, $dbpw);
mysql_select_db($dbname_1,$db2); //链接NEWS数据库$sql = mysql_query("select userid from user where userid=".$userid."",$db);
if($myrow = mysql_fetch_array($sql)){
echo $myrow['uid'];
}试试看。
mysql_select_db($dbname,$db); //链接BBS数据库$db2 = mysql_connect($dbhost, $dbuser, $dbpw);
mysql_select_db($dbname_1,$db2); //链接NEWS数据库$sql = mysql_query("select userid from news.user where userid=".$userid."",$db2);
if($myrow = mysql_fetch_array($sql)){
echo $myrow['userid'];
}
mysql_query1: 查询结构缓存。2:返回一个 resource 。。
你的提示是说:根本没有一个可用的resource 返回。
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ysql_fetch_array(): supplied argument is not a valid MySQL result resource
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