the problem i get is that its say no database selected Anthony
"Chris" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > It looks like mysql_query is failing, add these two lines to your code, and > see what message pops up: > > ... > $dbdo = mysql_query($query,$dbconnect); > if(false === $dbdo) echo mysql_errno(),': ',mysql_error(); > else echo 'The query worked, $dbdo value is:',$dbdo; > while($row= mysql_fetch_array($dbdo)){ > ... > > > Chris > > -----Original Message----- > From: tony [mailto:[EMAIL PROTECTED] > Sent: Saturday, December 27, 2003 4:18 PM > To: [EMAIL PROTECTED] > Subject: [PHP] newbie mysql and php > > > hello > > I'm new with php just learning and i have just a problem with the following > code > > $dbconnect = mysql_connect("localhost", "prog_tony","PASSWORD"); > mysql_select_db("prog_dealer", $dbconnect); > $stop = 0; > $counter = 1; > while(!$stop){ > $query = "SELECT name FROM category WHERE id = $counter"; > $dbdo = mysql_query($query,$dbconnect); > while($row= mysql_fetch_array($dbdo)){ > if($row[0]){ > $counter++; > }else{ > $stop=1; > } > $row[0] = ""; > } > > } > > > > I'm getting an error with mysql_fetch_array() which is line 14 because I > didn't show the other lines since it is not relevant. > here is the error > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource in /home/prog/public_html/php/cateadded.php on line 14 > > Thank you > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
