i did by my self the folowing to display on site castle status...but i got multiple times warning in page that
php Unable to jump to row 1 on MySQL result index X
here is the code
..if someone can see where is my own mistake and explain me how could i fix it would be great!!
and if my code is too big and its possible to be done the whole thing with less code dont kick my ass
yesterday i started learning about php
and thanks for you time
Code: Select all
<?php include "./configs/mysql.php"; $connect = @mysql_connect($mySQLserver, $mySQLuser, $mySQLpassword); @mysql_select_db($mySQLdefaultdb, $connect); if(!$connect) echo 'Can\'t connect to the database'; else $query = ("select id,name,taxPercent from castle order by id "); $result = mysql_query($query) or die(mysql_error()); $i=1; while($row = mysql_fetch_array($result)){ if($i=="1") { $FORM .="<tr> <th>Castle</th> <th>Owner</th> <th>Leader</th> <th>Tax Rate</th> </tr><br>"; } $FORM .= "<tr> <td>$row[1]</td>"; $castle=$row[0]; $query2 = sprintf("select clan_name,leader_id from clan_data where hasCastle='%s'",mysql_real_escape_string($castle)); $result2 = mysql_query($query2) or die(mysql_error()); $clan = mysql_result($result2, 0); if($tmp==NULL) $FORM .= "<td>NPC</td>"; else $FORM .= "<td>$clan</td>"; $leader = mysql_result($result2, 1); $query3 = sprintf("select char_name from characters where charid='%s'",mysql_real_escape_string($leader)); $result3 = mysql_query($query3) or die(mysql_error()); $leader = mysql_result($result3, 0); if($leader==NULL) $FORM .= "<td>None</td>"; else $FORM .= "<td>$leader</td>"; $FORM .= "<td>$row[2]</td></tr>"; $i++;} mysql_close(); ?>