adamh91 Posted March 1, 2007 Share Posted March 1, 2007 Hey, I'm getting this error in my code: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/r3v/public_html/scripts/includes/emails.php on line 40 I was wondering if its an error in my code, i cannot see any, or i have chmodded it incorrectly. <? include 'config.php'; $i = 1; //---Display Emails---\\ $conn = mysql_connect($db_host, $db_user, $db_pass) or die ('Error: '. mysql_error()); mysql_select_db($db_name); $sql = "SELECT email FROM emails"; $res = mysql_query($sql); while($row = mysql_fetch_array($res)){ if ($i == 1) { echo '<tr valign="top">'; echo '<td width="100%" bgcolor="#0098CC" class="alt1">'.$row['email'].'</td>'; echo '</tr>'; $i = 2; } else { echo '<tr valign="top">'; echo '<td width="100%" bgcolor="#00A6DA" class="alt1">'.$row['email'].'</td>'; echo '</tr>'; $i = 1; } } mysql_free_result($res); mysql_close($conn); //---End Display---\\ ?> Quote Link to comment Share on other sites More sharing options...
monk.e.boy Posted March 1, 2007 Share Posted March 1, 2007 $sql = "SELECT email FROM emails"; $res = mysql_query($sql); do: if( $res ) { } looks like your sql is wrong, so $res is FALSE. monk.e.boy Quote Link to comment Share on other sites More sharing options...
simcoweb Posted March 1, 2007 Share Posted March 1, 2007 The code you posted.. that's not from emails.php as it references 'line 40'. There's not 40 lines there. Quote Link to comment Share on other sites More sharing options...
adamh91 Posted March 1, 2007 Author Share Posted March 1, 2007 Thanks guys It was a silly mistake in config.php, the $db_name has a spelling mistake Quote Link to comment Share on other sites More sharing options...
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