CagedApe

Thank you Pikachu! That works perfectly.

Thanks. That pointed out my error and now the select command isn't giving an error. However, now its telling me mysql_fetch_array(): supplied argument is not a valid MySQL result resource. Any ideas? $q = "SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND defender_id = $AttackedId AND attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()"; mysql_query($q) or die('<br />Query string: ' . $q . '<br />Produced error: ' . mysql_error() . '<br />'); $row = mysql_fetch_array($q) or die(mysql_error()); if($row['count'] == 3){ die("You have already attacked this person 3 times this hour."); }

This should probably be in Javascript help...

I've written out the following, but it returns a syntax error near the bolded portion. I don't see anything wrong with the syntax, but I'm sure theres something I missed. The script is supposed to scan the table to see if you have attacked your current target player 3 times in the last hour. $q = mysql_query("SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND defender_id = $AttackedID AND attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()") or die(mysql_error()); $row = mysql_fetch_array($q) or die(mysql_error()); if($row['count'] == 3){ die("You have already attacked this person 3 times this hour.");

I changed that but still get the same error. What else did you see that might be wrong?

I keep getting the error: mysql_fetch_array(): supplied argument is not a valid MySQL result resource. I can't see the problem in my code. Any ideas? $message = $_POST['message']; $senderid = $SESSION['SESS_MEMBER_ID']; $recipient = $_POST['recipient']; $q = mysql_query("SELECT * FROM members WHERE member_id = $senderid"); $row = mysql_fetch_array($q); ///error says this is the bad line $sender = $row['username'];

You'd do this: echo ' <tr> <td>'."<a href='profilepage.php?id=$ids'>$ids[$i]".'</a></td> Then your profilepage will use $_GET to find their username like this: if (isset($_GET['id']) && $_GET['id']!='') { $UserId=$_GET['id']; }

Are you simply asking how to put a link in your result? If so this is how you'd do it. Change echo ' <tr> <td>'.$ids[$i].'</td> to echo ' <tr> <td>'."<a href='profilepage.php'>$ids[$i]".'</a></td>

I have a page called attack.php. Everything works except when I try to get $defender_name. I have a feeling there's something wrong with my second sql query, but I'm still a noob so I can't see it. I would appreciate some help from the resident experts session_start(); include "config.php"; //Get AttackId from url if (isset($_GET['id']) && $_GET['id']!='') { $AttackId=$_GET['id']; } //Select attack info(defender's id, winner's id) $q=mysql_query("SELECT * FROM attack WHERE attackid = $AttackId"); $row = mysql_fetch_array($q); $defender_id = $row['defender_id']; // Find defender's username $q1="SELECT login FROM members WHERE member_id = $defender_id"; $row2 = mysql_query($q1); $defender_name = $row2['login']; //Show result if ($_SESSION['SESS_MEMBER_ID'] = $row['winner_id']){ echo "You have defeated ". "$defender_name"; } else{ echo "You have been defeated by ". "$defender_name"; }

Sorry I guess i should have explained a bit better. $row['login'] returns a username. In any event that worked for me, thank you very much! I'm sure I'll be back to ask more questions, but for now thanks and good luck.

I'm trying to add a hyperlink to the results of an array I'm pulling. This is the code I have: $query=("SELECT * FROM members"); $result=mysql_query($query); while($row = mysql_fetch_array($result)){ echo $row['login']; echo "<br />" ; } ?> How would I edit echo $row['login'] to display the result as a hyperlink? Thanks in advance.