[SOLVED] Session and ID ....

[Jay2391]

I created a Log in with password and all that good stuff

and You can log on to the web site and autheticate.... That is all cool now here is the issue....

I can Log in and I carry a variable saying display the name "John  Doe"..

Then I click and  go to another page ... and i can test the session is active ...

session(): that is okay ....BUT !!!!!!!!

The session does not test to show that the session is "JOHN DOE" session....



Example:

1.Log in page

JOHN DOE AUTHENTICATES

2.Confirmation or welcome page...

        Session is True
        JOHN DOE YOU ARE LOG IN

3.Page X

        Session is True
        "Blank - Nothing displays" you are log in



So i can not probe in that page that John Doe is the one log in ????

can some one help i have check all the books and I can find a explanation ....
I also try passing a variable but i guess i did it wrong.

[trq]

Post some related code.

[HuggieBear]

Try posting your code for the login page and the page that's not working, this will give us something to work with.

Regards
Huggie

[Jay2391]

okay here is my code...

the Authetication page is HTML

Name and password you know thw basic stuff...

then you get this cheking if you indded logged-in sucessfully

CODE:
[code]
<?php
session_start();
include("LOGINFO.php");

///////////////Variables///////////////////////////////////////////////////////
$table = 'md_user_main';
$email = $_POST['email'];
$password = md5($_POST['password']);
$wtd = "relogin";

///////////////////////////////database connection and array//////////////////
 
    $tc = mysql_connect ($host, $user, $pass);
    $db = mysql_select_db ($database, $tc);

$query = "SELECT * FROM $table WHERE (email=\"$email\") ";
$result = mysql_query($query);
$row = mysql_fetch_array($result);

                $fname= $row['first_name'];

//////////////////////////Re Log In///////////////////////////////////////////// 

if($_SESSION['session_var'] == "skipLogin"){
    echo "$fname is Log-In";
echo "<br>If you like to end the session, Click here to Log-Off";
  }else if($wtd == "relogin"){
 
    if($password == $row['password']){
          echo "You are loged In $fname";
      $_SESSION['session_var'] = "skipLogin";
    }else{
          echo "You are not log-In try again";
//session_destroy();
//unset($_SESSION);
    }
    }
?>
[/code]
Then I created this test to check if the session and user are log-in in a 3rd page ...here is where the session is active but I can't ID the user....
before i try passing  a variable but it didn't work so this is what i have now...

what i will like is to say who is Loged In

[code]
<?php
session_start();
 
  if($_SESSION['session_var'] == "skipLogin"){
  echo "You are loged-In";
echo "<br>Session still active";
  }else{
    echo "You stinky";
  }

?>
[/code]

[trq]

A simple example.

p1.php
[code]
<?php
  session_start();
  $_SESSION['u'] = "foo";
  echo "<a href='p2.php'>click</a>";
?>
[/code]

p2.php
[code]
<?php
  session_start();
  echo $_SESSION['u'];
?>
[/code]

Does that work? 

[HuggieBear]

You need to put the name into a session variable too...

[code=php:0]
if($password == $row['password']){
  echo "You are loged In $fname";
  $_SESSION['session_var'] = "skipLogin";
  $_SESSION['fname'] = $fname; // I added this line
}else{
  echo "You are not log-In try again";
  //session_destroy();
  //unset($_SESSION);
}[/code]

Then when you echo [code=php:0]$_SESSION['fname'][/code] you get their name.

Regards
Huggie

[Jay2391]

Good Stuff !!!! Yeah I was doing my passing variable wrong

;D

[Jay2391]

one more ?


I put this on my session test ... but it won't show the name,  if I put the $_SESSION['fname'] in a variable

if i do           """"       echo $_SESSION['fname'];      """""  that display the name
but when i do a variable like below it just show """""   You are loged In """""

<?php
session_start();
   
    if($_SESSION['session_var'] == "skipLogin"){
           $_SESSION['fname'] = $fname;
           echo "You are loged In $fname";      
   
    }else{
          echo "You stinky";
    }

?>

[Jay2391]

can you check my preavious post ...one small issue with the variable and session

[KevinM1]

[quote author=Jay2391 link=topic=120690.msg495465#msg495465 date=1167761750]
one more ?


I put this on my session test ... but it won't show the name,  if I put the $_SESSION['fname'] in a variable

if i do           """"       echo $_SESSION['fname'];      """""  that display the name
but when i do a variable like below it just show """""   You are loged In """""

<?php
session_start();
   
    if($_SESSION['session_var'] == "skipLogin"){
           $_SESSION['fname'] = $fname;
           echo "You are loged In $fname";      
   
    }else{
          echo "You stinky";
    }

?>

[/quote]

You have to remember how variable assignment works.  The left side of the statement is assigned the value of the right side.  So, if you haven't defined $fname in your script, you're merely overwriting your session variable with an empty local variable, hence no output.  Since it looks like you're trying to assign a local variable the contents of your session variable, switch the statement around.  In other words:

$fname = $_SESSION['fname'];

[Jay2391]

I tough i had try that and it didn't work ... but it seems to work now ....anyways thanks and sorry for the bugging with this...

makes me mad i tought i try that maybe next time i should close my browser