Re: Non-Object errors

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On 4/28/09 11:23 AM, "Jan G.B." <ro0ot.w00t@xxxxxxxxxxxxxx> wrote:

2009/4/28 Miller, Terion <tmiller@xxxxxxxxxxxxxxxxxxxx>:
> 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?

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