Re: making FORM dissapear when successful login

nevermind...i figured it out. I will read up on sessions and test some things with what you posted and if i have anymore questions, i will post them. Thanks Jerry. If anyone else has any other suggestions, please feel free to share them with me. JHollis wrote:

Where do i need to start my session? Ive copied and pasted exactly what you have typed..and the form doesnt display unless i hard code the variable to "NO". Jerry Kita wrote:

I do exactly the same thing on my website. I use PHP's Session handling to determine if a user is actually logged on. You can see that I check a variable called $validlogon which I set to either YES or NO at the beginning of the script <?php if ($validlogon == "NO") { print "<p class=\"hnavbar\">Salkehatchie Members Login Here:<br>\n"; print "<form action=\"Salkehatchie_LoginScript.php\" method=\"POST\">\n"; print "<p class=\"login\"><small>USER ID:</small><br>\n"; print "<input type=\"text\" name=\"userid\"><br><br>\n"; print "<small>PASSWORD:</small><br>\n"; print "<input type=\"password\" name=\"password\"><br><br>\n"; print "<input type=\"submit\" value=\"Enter\"></p>\n"; print "</form>\n";

                          }

?> Near the beginning of my script I check for the existence of a valid session <?php if (isset($_SESSION[userid])) { $validlogon = "YES"; } else { $validlogon = "NO"; } ?> Setting the $validlogon variable allows me to do a number of useful things. For example, there are certain parts of the webpage that I reveal to logged in users. Checking $validlogon allows me to decide dynamically what to render to the browser. In the following I display a message showing the user as being logged in and give them a link that logs them out. if ($validlogon == "YES") { print "<table style=\"width: 100%\";>\n"; print "<tbody>\n"; print "<tr>\n"; print "<td width=\"50%\"><p class=\"login\"><small><b>YOU ARE

LOGGED     IN AS: $_SESSION[userid]</b></small></p></td>\n";
   print "<td width=\"50%\"><p class=\"login\"><small><b>CLICK <a href=\"LogoutScript.php\"> here</a> TO LOGOUT</b></small></p></td>\n";
            print "</tr>\n";
            print "</table>\n";
            }

Hope this is helpful Jerry Kita JHollis wrote:

I had this code working the way i wanted it to (as far as correct username and password allowing successful login)...but what i want to happen now is when a user successfully logs it it will make the login form disappear and just say successfully logged in or welcome user and a link below it so they can log off and make the form re-appear. Below is the code that i have where i tried to get it to disappear on successful login, but it stays disappeared all the time. Can someone please point out what im doing wrong. I have tried everything i can think of...and nothing works. Im a PHP newbie...so im sure some of you might get a laugh out of this...if it is real easy. <---snippet <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html> <head> <link href="style.css" rel="stylesheet" type="text/css" /> </head> <body> <div id="container"> <div id="top"> <h1>header</h1> </div> <div id="leftnav"> <p>

    <?php

    $username=$_POST['username'];
    $password=$_POST['password'];
    $db="user";
    $server="localhost";
    $db_username="root";
    $db_password="*******";
        $connect = mysql_connect($server,$db_username,$db_password);
    if (!$connect) {
            die ("could not connect to database");
                    }
    $select = mysql_select_db($db,$connect);
    if (!$select) {
            die ("could not select database $db");
                    }

/*username='$username'";*/

    $sql = "SELECT * FROM passwords, user_info where id=PID and username='$username'";
    $result = mysql_query($sql);

/*$num_rows = mysql_num_rows($result);*/

    while ($user = mysql_fetch_array($result))
        {
        $id = $user['id'];
        $username2 = $user['username'];
        $password2 = $user['password'];
        $firstname = $user['firstname'];
        $email = $user['email_address'];
                  IF ($username=="$username2" && $password=="$password2")
        {
        echo("\"Welcome, <b>$firstname</b>\"");?><br><?
        echo ("\"Your email address is <b>$email<b>\"");?></td><tr>
        <a href="<?$_SERVER['PHP_SELF']?>?username=""&?password=""">Logoff</a><?
        break;
        }
        else
        {
        ?>
                <FORM action="<?$_SERVER['PHP_SELF']?>" method="post">
                <INPUT type="hidden" name="id">
                <table>
                <td><b>*</b>Username:</td> <td><INPUT class="input" size="8" type="text" name="username" value="<?echo $username?>"></td><tr>
                <td><b>*</b>Password:</td> <td><INPUT class="input" size="8" type="password" name="password"></td><tr>
                <td class="xsmall"><b>* Case Sensitive</b></td>
                <td><INPUT type="submit" value="Login"></td><tr>
                <td>&nbsp </td>
                </table>
                </FORM>
                <?
               break;
            }
           }   //IF ($username != $username2 || $password != $password2) {//

          ?><br>
    <?
    if  ($username == "" && $password == "") {
        echo ("Please type in a Username and Password");}

    if  ($username != "" && $password == "") {
        echo ("Please type in a password");}

    if  ($username == "" && $password != "") {
        echo ("Please type in a username and password");}
         ?>
                           </p>

</div> <?if (($username2==$username && $password2==$password) && ($username2!="" || $password2!="")){?> <div id="rightnav" class="box"> <p> </p> </div> <?}?> <div id="content"> <h2>Subheading</h2> <p> </p> <p> </p> </div> <div id="footer"> <p> Today is <?php

    echo( date("F dS Y."));

?> </p> </div> </div> </body> </html> ---->snippet