davidcriniti Posted November 9, 2010 Share Posted November 9, 2010 Hi, I've used this code before, but have had to make some modifications and am now getting a mysql fetch array error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/tesolcla/public_html/test/results10k201010.php on line 134 If anyone can help, it would be much appreciated. <?php $dbcnx = @mysql_connect('localhost', 'MYUSERNAME', 'MYPASSWORD'); //$dbcnx = @mysql_connect('localhost', 'root', 'mysql'); if (!$dbcnx) { exit('<p>Unable to connect to the ' . 'database server at this time.</p>'); } if (!@mysql_select_db('MYDATABASE')) { //if (!@mysql_select_db('rac')) { exit('<p>Unable to locate the results ' . 'database at this time.</p>'); } $asc_on = '<img src="images/results_sorting/Asc.gif" border="0" />'; $asc_off= '<img src="images/results_sorting/AscOff.gif" border="0" />'; $desc_on = '<img src="images/results_sorting/Desc.gif" border="0" />'; $desc_off= '<img src="images/results_sorting/DescOff.gif" border="0" />'; $sortfield = isset($_GET['sort']) ? $_GET['sort'] : '4'; $sorttype = isset($_GET['type']) ? $_GET['type'] : '1'; for($i=1; $i<5; $i++) { if($i==$sortfield) { if ($sorttype==1) $srt[$i] = $asc_on.'<a href="?sort='.$i.'&type=2">'.$desc_off.'</a>'; else $srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a>'.$desc_on; } else { $srt[$i] = '<a href="?sort='.$i.'&type=1">'.$asc_off.'</a><a href="?sort='.$i.'&type=2">'.$desc_off.'</a>'; } } $fields = array("firstname", "lastname", "time", "position"); $sorts = array("ASC", "DESC"); $field = $fields[$sortfield-1]; $sort = $sorts[$sorttype-1]; $field = $field=="" ? $fields[4] : $field; $sort = $sort=="" ? $sorts[0] : $sort; $sql = mysql_query("SELECT firstname, lastname, time, position FROM 10k_results ORDER BY $field $sort"); echo "<table border='1' align='center' bordercolor='#000000' CELLPADDING=5 cellspacing='0' STYLE='font-size:13px'>"; echo "<tr bgcolor='#008000' STYLE='color:white'> <td>*</td><td><H3>First name $srt[1]</h3></td> <td><H3>Lastname $srt[2]</H3></td> <td><H3>Time $srt[3]</H3></td> <td><H3>Position $srt[4]</H3></td></tr>"; // keeps getting the next row until there are no more to get $row_counter = 1; //create row counter with default value 0 // Print out the contents of each row into a table while ($row = mysql_fetch_array($sql)) { // Print out the contents of each row into a table echo "<tr>\n"; echo "</td><td>"; echo $row_counter++; echo "</td>"; echo "<td>{$row['firstname']}</td>\n"; echo "<td>{$row['lastname']}</td>\n"; echo "<td>{$row['time']}</td>\n"; echo "<td>{$row['position']}</td>\n"; echo "</tr>\n"; } echo "</table>"; ?> Quote Link to comment Share on other sites More sharing options...
revraz Posted November 9, 2010 Share Posted November 9, 2010 That error indicates your query failed. Use mysql_error after the query to see why and also echo your query to see what is being passed. Quote Link to comment Share on other sites More sharing options...
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