Re: Non-Object errors

On 4/28/09 11:23 AM, "Jan G.B." <ro0ot.w00t@googlemail.com> wrote: 2009/4/28 Miller, Terion <tmiller@springfi.gannett.com>: > Can someone help with how to make this work, trying to get the number of > rows but am getting the "trying to get property of non-object" > > Code----------------- > > $query = "select blah, blah, blah from table where ".$type." like > '%".$name."%'"; > > $result = mysql_query($query); > > $num_results = $result->num_rows; <---------------this is the line with > error > Hi Terion, you're mixing normal mysql_* with mysqli. Mysqli is object oriented and the mormalö mysql_ functions aren't - thei're procedural. So you can do it either like $result = mysql_query("SELECT 1"); $num = mysql_num_rows($result); Or in OOP $res = $mysqlObj->query("SELECT 1"); // $mysqlObj should be initialized first ;) hint: new mysqli() $num = $res->num_rows(); Check ot this URL: http://php.net/mysqli byebye Hmm, well I did some looking at other codes etc and some tweaking but still get two problems: Changed my code to try and use an array because that has worked before when I have run into the object error, but I still get this error: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource Here is my code now: $query = "select name, age, warrant, bond, wnumber, crime FROM warrants where ".$warranttype." = ".$warranttype." OR ".$searchname." = ".$searchname." "; $result = mysql_query($query); $row = mysql_fetch_assoc($result); $num_results = mysql_num_rows ($result); Should this post of gone on the Db list maybe?