WatsonN

Pings just fine. I use Godaddy Hosting Control Center I dont think they allow it.

I tried that and pointed it to the ip just like admin is set up and and it 403s me. You don't have permission to access / on this server.

Same thing in latest Safari

Oh well If I had of read it right Sorry for the waste of time

In the forums when someone post long code. it posts and expands the width of all the elements and looks ugly(TBH) is this because of an update?

Hey yall Ive got an issue with a mysql INSERT Query string: INSERT INTO `YBK_Ads` (`BSN`, `CNF`, `CNL`, `ADD`, `CITY`, `STATE`, `ZIP`, `ToA`, `AS`, `PT`, `CN`, `BY`, `ACI`, `TAN`, `VM`, `Date`) VALUES ('Business/Service Name', 'First', 'Last', 'Address', 'City', '38002', 'Business', '1/8', 'Bill', 'check', 'No', 'Ad Copy • Artwork/Photo Instructions', 'Yes', 'No', 'October 6, 2010, 7:15 pm') Produced an error: Column count doesn't match value count at row 1 where is the code mysql_real_escape_string($insert = "INSERT INTO `YBK_Ads` (`BSN`, `CNF`, `CNL`, `ADD`, `CITY`, `STATE`, `ZIP`, `ToA`, `AS`, `PT`, `CN`, `BY`, `ACI`, `TAN`, `VM`, `Date`) VALUES ('{$e1}', '{$e71}', '{$e72}', '{$e2}', '{$e3}', '{$e5}', '{$e10}', '{$e13}', '{$e11}', '{$e6}', '{$e12}', '{$e8}', '{$e15}', '{$e16}', '{$date}')"); mysql_query($insert) or die( 'Query string: ' . $insert . '<br />Produced an error: ' . mysql_error() . '<br />' );

Fixed change setcookie(Errors, $error, time()+20); to setcookie('Errors', $error, time()+20);

In my login form I'm getting the error Notice: Use of undefined constant Errors - assumed 'Errors' in /home/content/n/a/t/nathanwatson/html/admin/YBK/post.php on line 91 Warning: Cannot modify header information - headers already sent by (output started at /home/content/n/a/t/nathanwatson/html/admin/YBK/post.php:91) in /home/content/n/a/t/nathanwatson/html/admin/YBK/post.php on line 91 Warning: Cannot modify header information - headers already sent by (output started at /home/content/n/a/t/nathanwatson/html/admin/YBK/post.php:91) in /home/content/n/a/t/nathanwatson/html/admin/YBK/post.php on line 92 //if the name exists it gives an error if ($check2 != 0) { //Line 85 $error="<span style="; $error .="color:red"; $error .=">"; $error .= "Sorry, the username is already in use."; $error .="</span>"; setcookie(Errors, $error, time()+20); //Line 91 header('Location: /YBK/'); //Line 92 exit; }

Is there a way to set up a sub sub domain such as iphone.admin.mysite.com ? If so how do I accomplish this? Thanks in advanced

edit it to your liking <? ... mysql cnx code ... $sql="SELECT id, thing FROM table"; $result=mysql_query($sql); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["id"]; $thing=$row["thing"]; $options.="<OPTION VALUE=\"$id\">".$thing; } ?> ... html code ... <SELECT NAME=thing> <OPTION VALUE=0>Choose <?=$options?> </SELECT> ...

I fixed it by changing it to exit;

Changed to header('Location: ./?p=UCP'); but now instead of die() it continues with the script -edit- changed to exit and halts -edit-

In my post.php file i have the following code // checks if the username is in use if (!get_magic_quotes_gpc()) { $_POST['username'] = addslashes($_POST['username']); } $usercheck = $_POST['username']; mysql_real_escape_string($usercheck); $check = mysql_query("SELECT username FROM users WHERE username = '$usercheck'") or die(mysql_error()); $check2 = mysql_num_rows($check); //if the name exists it gives an error if ($check2 != 0) { $error="<span style="; $error .="color:red"; $error .=">"; $error .= "Sorry, the username is already in use."; $error .="</span>"; setcookie(Errors, $error, time()+20); header('Location ./?p=UPC'); die(); } The problem is it always 500s if the username is already in use.

Yes. Like put your default CSS in there if you want to give them a base and edit it from there.

Well you could have a file for each user like username.css and username.html and use fopen()

Thank Ya Pawn, That worked perfectly

Howdy I'm trying to figure out to get a file to call a page on another domain(diffrent server) and check if the activation code matches the one on the activation server. So on the user's server it has a cron job or on load will call my server and run the code in their update.php file against my activation.php file and see if it matches. If($usr == $actv){ //DO THIS }else{ //run disable script }

bump

use <A NAME="about-us"> and <A NAME="Services"> then at the top link to them with <A HREF="#about-us">About Us</A> and <A HREF="#Services">Services</A>

It was displaying the table with the original code but I just want to know how to break each row into its own set of variables

They sure are

There was but now its gone and the table isn't outputting any data

I did that and now I get this error Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /home/content/n/a/t/nathanwatson/html/admin/LOGINS.php:1) in /home/content/n/a/t/nathanwatson/html/admin/timer.php on line 3 This is timer.php <?php $error = 'Your session timed out.'; session_start(); if($_SESSION['session_count'] == 0) { $_SESSION['session_count'] = 1; $_SESSION['session_start_time']=time(); } else { $_SESSION['session_count'] = $_SESSION['session_count'] + 1; } $session_timeout = 1800; $session_duration = time() - $_SESSION['session_start_time']; if ($session_duration > $session_timeout) { session_unset(); session_destroy(); session_start(); session_regenerate_id(true); setcookie("Key_WatsonN", 0, time()-3600); setcookie("Errors", 0, time()-3600); setcookie("UID", 0, time()-3600); setcookie("PWD", 0, time()-3600); setcookie(Errors, $error, time()+120000); $_SESSION["expired"] = "yes"; header("Location: /"); // Redirect to Login Page } else { $_SESSION['session_start_time']=time(); } ?>

I changed it to <?php $result = mysql_query(sprintf("SELECT * FROM Logins WHERE UID = %d", $_COOKIE['UID_WatsonN'])); //check login table against cookie if(($num = mysql_num_rows($result)) > 0){ mysql_close(); ?> <table border="0" cellspacing="2" cellpadding="2"> <tr> <th>ID</th> <th>UID</th> <th>Site</th> <th>Uname</th> <th>Pass</th> </tr> <center> <?php while ($row = mysql_fetch_array($result, MYSQL_NUM){ $f1 = $row['ID']; //Line 107 $f2 = $row['UID']; $f3 = $row['Site']; $f4 = $row['Uname']; $f5 = $row['Pass']; ?> <tr> <td><?php echo $f1; ?></td> <td><?php echo $f2; ?></td> <td><?php echo $f3; ?></td> <td><?php echo $f4; ?></td> <td><?php echo $f5; ?></td> </tr> <?php } } ?> and got the error Parse error: syntax error, unexpected '{' in /home/content/n/a/t/nathanwatson/html/admin/LOGINS.php on line 107

I have a table and the structure is ID, UID, Site, Uname, Pass I already have this <?php $result = mysql_query(sprintf("SELECT * FROM Logins WHERE UID = %d", $_COOKIE['UID_WatsonN'])); //check login table against cookie if(($num = mysql_num_rows($result)) > 0){ mysql_close(); ?> <table border="0" cellspacing="2" cellpadding="2"> <tr> <th>ID</th> <th>UID</th> <th>Site</th> <th>Uname</th> <th>Pass</th> </tr> <center> <?php $i=0; while ($i < $num) { $f1=mysql_result($result,$i,"ID"); $f2=mysql_result($result,$i,"UID"); $f3=mysql_result($result,$i,"Site"); $f4=mysql_result($result,$i,"Uname"); $f5=mysql_result($result,$i,"Pass"); ?> <tr> <td><?php echo $f1; ?></td> <td><?php echo $f2; ?></td> <td><?php echo $f3; ?></td> <td><?php echo $f4; ?></td> <td><?php echo $f5; ?></td> </tr> <?php $i++; } } ?> which only gets the data from that one person. What is in the table is usernames and passwords for diffrent sites and i want to take all the usernames and passwords and put them into variables to pass on to the login form.