bravo14 Posted June 7, 2008 Share Posted June 7, 2008 Hi folks Newbie to this forum, but hoping there is someone there who can help I am trying to display the results of a MySQL query on a page using PHP, the query works in phpmyadmin see code below <?php $con = mysql_connect('213.171.219.88', 'yardleyheroes', 'heroes', true); if(!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db('yardleyheroes', $con); $result = mysql_query('SELECT Sum(fines.fine_amount) AS SumOffine_amount, profiles.name FROM fines INNER JOIN profiles ON fines.profile_id = profiles.profile_id GROUP BY profiles.name ORDER BY Sum(fines.fine_amount); LIMIT 0, 30 ', $con); if(mysql_num_rows($result) < 1) { echo('<tr><td colspan="2">Sorry, no results were found.</td></tr>'); } else { while($row = mysql_fetch_array($result)) { echo('<tr><td valign="top">' . $row['name'] . '</td>'); echo('<td> £' . $row['SumOffine_amount'] .'.00</td></tr>'); } } mysql_close($con); ?> I get the following message on the page Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/fhlinux147/y/yardleyheroes.co.uk/user/htdocs/fines.php on line 52 Line 52 is if(mysql_num_rows($result) < 1) Any ideas, the page can be seen at www.yardleyheroes.co.uk/fines.php Link to comment https://forums.phpfreaks.com/topic/109172-php-mysql/ Share on other sites More sharing options...
GingerRobot Posted June 7, 2008 Share Posted June 7, 2008 That would meant that your query is failing. Try this: $result = mysql_query('SELECT Sum(fines.fine_amount) AS SumOffine_amount, profiles.name FROM fines INNER JOIN profiles ON fines.profile_id = profiles.profile_id GROUP BY profiles.name ORDER BY Sum(fines.fine_amount); LIMIT 0, 30 ', $con) or die(mysql_error()); Also, next time please put tags around your code. P.S. Welcome to the forum Link to comment https://forums.phpfreaks.com/topic/109172-php-mysql/#findComment-560037 Share on other sites More sharing options...
.josh Posted June 7, 2008 Share Posted June 7, 2008 try taking out the ; on Sum(fines.fine_amount); Link to comment https://forums.phpfreaks.com/topic/109172-php-mysql/#findComment-560065 Share on other sites More sharing options...
masonite85 Posted June 8, 2008 Share Posted June 8, 2008 So I have a mysql database that is working and can display the results correctly. My next step is after it prints the "state" array in the code below for each result I want it to create a button or a link to display the full information of that record. I could make each record a separate link and page but I would rather be able to have it pass information based on the link clicked to a single page that can pull the information . Not sure exactly how to go about this or where to start looking. if($results) { while($array= mysql_fetch_assoc($results)){ echo '<br>'; print $array['School Name']; echo " "; print $array['City']; echo ","; print $array['State']; echo " "; } } Link to comment https://forums.phpfreaks.com/topic/109172-php-mysql/#findComment-560245 Share on other sites More sharing options...
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