cleeclee Posted November 16, 2012 Share Posted November 16, 2012 (edited) Hi i'm trying to display a table thru mysql. Just learnt to do so and i haven't got a clue what is wrong with the code. In line 19 the error is that 'mysql_numrows() expects parameter 1 to be resource'. Can i get some help as to what to correct it with? <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Untitled Document</title> </head> <body> <?php $host="localhost"; $username="root"; $password=""; $database="energy"; mysql_connect($host,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM tablename"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); ?> <table border="0" cellspacing="2" cellpadding="2"> <tr> <td><font face="Arial, Helvetica, sans-serif">Value1</font></td> <td><font face="Arial, Helvetica, sans-serif">Value2</font></td> <td><font face="Arial, Helvetica, sans-serif">Value3</font></td> <td><font face="Arial, Helvetica, sans-serif">Value4</font></td> <td><font face="Arial, Helvetica, sans-serif">Value5</font></td> </tr> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"field1"); $f2=mysql_result($result,$i,"field2"); $f3=mysql_result($result,$i,"field3"); $f4=mysql_result($result,$i,"field4"); $f5=mysql_result($result,$i,"field5"); ?> <tr> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f1; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f2; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f3; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f4; ?></font></td> <td><font face="Arial, Helvetica, sans-serif"><?php echo $f5; ?></font></td> </tr> </table> <?php $i++; } ?> </body> </html> Edited November 16, 2012 by cleeclee Quote Link to comment Share on other sites More sharing options...
Barand Posted November 16, 2012 Share Posted November 16, 2012 And do you have a table in your database called "tablename"? Quote Link to comment Share on other sites More sharing options...
AyKay47 Posted November 16, 2012 Share Posted November 16, 2012 (edited) Debug the query using mysql_error Edited November 16, 2012 by AyKay47 Quote Link to comment Share on other sites More sharing options...
cleeclee Posted November 16, 2012 Author Share Posted November 16, 2012 (edited) Hi i've decided to keep it simple. Here's what i did. Without the while loop i'm only able to display 1 line of data. I'd like to display the other lines too, so i used the loop and i have the same error msg as in my original post. "Warning: mysql_fetch_array() expects parameter 1 to be resource" Edited November 16, 2012 by cleeclee Quote Link to comment Share on other sites More sharing options...
Pikachu2000 Posted November 16, 2012 Share Posted November 16, 2012 And do you have a table in your database called "tablename"? Well? Do you? Quote Link to comment Share on other sites More sharing options...
cleeclee Posted November 16, 2012 Author Share Posted November 16, 2012 Got it figured out! thanks it was a small mistake in my codes. sorry to trouble you guys. Thank you! Quote Link to comment Share on other sites More sharing options...
AyKay47 Posted November 16, 2012 Share Posted November 16, 2012 Got it figured out! thanks it was a small mistake in my codes. sorry to trouble you guys. Thank you! Great, what was the issue? Quote Link to comment Share on other sites More sharing options...
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