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
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