On Tue, 2009-11-24 at 02:11 -0800, Allen McCabe wrote: > I am! Will these extra query variables cause any problems or should I > use standard submit inputs? > > Thanks Ashley! > > > On Tue, Nov 24, 2009 at 1:10 AM, Ashley Sheridan > <ash@xxxxxxxxxxxxxxxxxxxx> wrote: > > > On Mon, 2009-11-23 at 21:53 -0800, Allen McCabe wrote: > > > 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@xxxxxxxxx>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 > > > > > My guess would be that you're submitting the form using an > image button, which would send the x and y coordinates of the > click within the button. > > Thanks, > Ash > http://www.ashleysheridan.co.uk > > > > > The only time they'll cause a problem is if you use some sort of loop to translate all the form inputs into something that is used in your code. For example, if you looped through all the form inputs to create your filter, regardless of what the inputs were called, then you would be running into all sorts of problems. This is not something you should try and 'fix' on the client-side but the server side, as everything that comes from the client is not to be trusted, ever! Saying that though, I have seen some systems (HSBC payment system) reject inputs containing x and y coordinates from image buttons and cause the whole form to fail. In your case that won't happen, but it's something to keep in mind in the future maybe? If you want to change the button, you could use a regular submit button and style it up with css: #submit_button_id { border: 0px none; background-image: url('button.png'); background-repeat: no-repeat; width: 100px; height: 25px; } Thanks, Ash http://www.ashleysheridan.co.uk
