Re: function not returning query

Okay, suddenly I got it to filter the results, but I still can't figure out where this part of the query is coming from, at the end of the query string in the URL, I have this "filter.x=0&filter.y=0". No where in my form do I have a field named filter.x or filter.y. I DO however, have 3 forms (I don't want to mess with AJAX), my set up looks like this: Filter by: User - [username dropdown v] Order by [database fields v] Asc/Desc [Ascend v] - Go School - [school dropdown v] Order by [database fields v] Asc/Desc [Ascend v] - Go Show - [show dropdown v] Order by [database fields v] Asc/Desc [Ascend v] - Go There are actually two order by fields, but this gives you the idea. Each of the three lines is a separate form, each with a unique name all with a "get" method, but all three Go buttons are named "filter", I didn't think to try changing it until now, but is this perhaps where the filter.x and filter.y are coming from? I have never seen this before in a query. Oh, now the filter that was working spontaneously gives me the error I have been getting all along, this is so frustrating. To those who asked, yes I am connected to the database; I forgot to mention that the else part of my if statement works, as long as I don't try to filter my results it works. Here is an example of the URL that my filter function (via one of the 3 forms) outputs: http://lpacmarketing.hostzi.com/afy/orders/default.php?filterby=school&schoolid=36&orderby1=order_id&asc_desc_order1=Descend&orderby2=pmt_recd_date&asc_desc_order2=Descend&filter.x=13&filter.y=8&filter=Go On Mon, Nov 23, 2009 at 8:03 PM, Philip Thompson <philthathril@gmail.com>wrote: > On Nov 23, 2009, at 6:22 PM, Allen McCabe wrote: > > > Hi, thanks for reading, I hope you can help: > > > > In my main file for an orders page I have the following code: > > > > > > if (isset($_GET['filterby'])) > > { > > $resultOrders = adminFilterQuery(); > > $numberOfOrders = mysql_num_rows($resultOrders); > > } > > else > > { > > $resultOrders = mysql_query("SELECT * FROM afy_order;") or > > die(mysql_error("Could not query the database!")); > > $numberOfOrders = mysql_num_rows($resultOrders); > > } > > You reduce this part by one line by putting the following after the else > statement and removing the other 2: > > $numberOfOrders = mysql_num_rows ($resultOrders); > > Also, these queries don't need a semi-colon (;) to end the query. PHP > handles this part. Remove them. > > > > adminFilterQuery() is a custom function that is supposed to return a > > mysql_query, here are the last few lines of this function: > > > > > > $query = "SELECT * FROM afy_order WHERE school_id = '{$school}' ORDER BY > > {$order_by_param};"; > > $result = mysql_query($query); > > return $result; > > > > l am getting this error when I try to filter my query using a form in > tandem > > with the quey building function: > > > > *Warning*: mysql_num_rows(): supplied argument is not a valid MySQL > result > > resource > > > > where the line is the one where I use the mysql_num_rows function. > > > > What am I missing here? > > > > Thanks! > > Do you get this warning with both queries? Make sure that your queries are > using a valid mysql connection. You may also consider using a database class > to perform the repetitive tasks so that you really only have to be concerned > with the queries you're writing...? > > <?php > class database { > public function query ($sql) { > $result = mysql_query ($sql); > if ($result === false) { > die ('Uh oh!'); > } > return $result; > } > > public function numRows ($result) { > return mysql_num_rows ($result); > } > } > $db = new database(); > $result = $db->query('SELECT * FROM afy_order'); > $numRows = $db->numRows($result); > ?> > > Of course this is just a simple example, but you get the idea. Hope that > stirs your brain! > > ~Philip