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session problem

Discussion in 'PHP' started by Lpspider, Nov 8, 2006.

0
Page 1 of 2
  1. #1
    Okay I've been trying to get the sessions to work but this is what I get:


    Here's the error I get:

    
Warning: session_start() [function.session-start]: Cannot send session cookie - headers already sent by (output started at /var/www/html/webdev/uploads/index.php:6) in /var/www/html/webdev/uploads/index.php on line 7

Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at /var/www/html/webdev/uploads/index.php:6) in /var/www/html/webdev/uploads/index.php on line 7
You left the username or password field blank.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/webdev/uploads/index.php on line 59
Welcome
    PHP:
    Here's the code:
    
<?php
session_start(); 

   echo '<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">';
   echo "<head><title> {$title} </title>";

   echo '<style type="text/css">

         #forms { position:absolute;top:75px;left:200px;align:right;padding-bottom:5px;}
         #text { position:absolute; top:75px; left:125px; align:left;} 
         .text2 { padding-top:8px;} 
         .forms2 { padding-bottom:6px;} 
         #login {position:absolute;top:100px;left:75%;} 
         .nav{position:absolute; top:75px; left:10%; right:10%; width:80%;background-color:#0099FF;text-align:right;a}
         .input{width:100px;} </style>
          </head><body>';
   

$logout = '<a href="logout.php">Logout</a>';

   echo '<div class="nav"><a href="index.php">Home</a> |', $logout  , ' | <a href="register.php">Register</a> | <a href="links.php">Links</a> | <a href="faq.php">FAQ</a></div>';


   echo "<div id=\"login\"><form action=\"index.php\" method=\"post\">
         Login:<br>
         Username: <input type=\"text\" name=\"username\" cols=\"60\" class=\"input\"><br>
         Password: <input type=\"password\" name=\"password\" cols=\"60\" class=\"input\"><br>
         <input type=\"submit\"></form></div>";




mysql_connect(localhost,username,pass);


//check to see if sessions are set if not set them 
if(!isset($_SESSION['user_name']) || !isset($_SESSION['user_password'])){ 
if(!empty($_POST['username']) || !empty($_POST['userpassword'])){ 
$_SESSION['username'] = $_POST['username']; 
$_SESSION['password'] = $_POST['password']; 
}
else{ 
echo'You left the username or password field blank.'; 
} 
} 

$sql = "SELECT * FROM user WHERE User = '$_SESSION[username]' && Password='$_SESSION[password]'"; 
$result = mysql_query($sql); 
$row = mysql_fetch_array($result); 
//compare sessions to those in table 
//Incorrect username and password 
if($_SESSION['username'] != $row['username'] && $_SESSION['password'] != $row['password']){ 
$_SESSION = array(); 
if (isset($_COOKIE[session_name()])) { 
setcookie(session_name(), '', time()-42000, '/'); 
}
 
@session_destroy(); 
echo'Incorrect Username or password'; 
} 
//correct username and password 
else{ 
echo'Welcome '.$_SESSION['username']; 
} 




    echo '</body></html>';




?>
    PHP:
    Basically what I need the session to remember the username/password that it calls from the database for the user to log in.
     
    Lpspider, Nov 8, 2006 IP
  2. mad4

    mad4 Peon

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    #2
    You are trying to set the session after you output something to the browser. This cannot be done unless you use something called output buffering.
     
    mad4, Nov 8, 2006 IP
  3. SoKickIt

    SoKickIt Active Member

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    #3
    1. Delete everything before "<?php" (spaces, tabs, new lines...)
    2. You don't need sessions in this case...
     
    SoKickIt, Nov 8, 2006 IP
  4. Lpspider

    Lpspider Well-Known Member

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    #4
    1. Might be my problem. I have a space before <?php

    2. I know. I have it working without sessions - I want to add the ability to change data in the database - and I need sessions for that. ;)
     
    Lpspider, Nov 8, 2006 IP
  5. Lpspider

    Lpspider Well-Known Member

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    #5
    Okay, I got rid of the dead space. Now I get this error:

    Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/webdev/uploads/index.php on line 50
Welcome
    PHP:
     
    Lpspider, Nov 9, 2006 IP
  6. nico_swd

    nico_swd Prominent Member

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    #6
    Replace this
    
$result = mysql_query($sql);
    PHP:
    with this and see what error you get
    
$result = mysql_query($sql) OR die( mysql_error() );
    PHP:
     
    nico_swd, Nov 9, 2006 IP
  7. SoKickIt

    SoKickIt Active Member

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    #7
    Replace:

    $sql = "SELECT * FROM user WHERE User = '$_SESSION[username]' && Password='$_SESSION[password]'";
    Code (markup):
    with:

    $sql = "SELECT * FROM user WHERE User = '" . mysql_real_escape_string($_SESSION[username]) . "' AND Password='" . mysql_real_escape_string($_SESSION[password]) . "'";
    Code (markup):
     
    SoKickIt, Nov 9, 2006 IP
  8. Lpspider

    Lpspider Well-Known Member

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    #8
    Okay, I did some of the things recommeneded. The error result I get is that there is no database selected. But there should be.

    <?php
session_start(); 

mysql_connect(localhost,username,pass);

   echo '<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">';
   echo "<head><title> {$title} </title>";

   echo '<style type="text/css">

         #forms { position:absolute;top:75px;left:200px;align:right;padding-bottom:5px;}
         #text { position:absolute; top:75px; left:125px; align:left;} 
         .text2 { padding-top:8px;} 
         .forms2 { padding-bottom:6px;} 
         #login {position:absolute;top:100px;left:75%;} 
         .nav{position:absolute; top:75px; left:10%; right:10%; width:80%;background-color:#0099FF;text-align:right;a}
         .input{width:100px;} </style>
          </head><body>';
   

$logout = '<a href="logout.php">Logout</a>';

//menu
   echo '<div class="nav"><a href="index.php">Home</a> |', $logout  , ' | <a href="register.php">Register</a> | <a href="links.php">Links</a> | <a href="faq.php">FAQ</a></div>';

//login form
   echo '<div id="login"><form action="index.php" method="post">
         Login:<br>
         Username: <input type="text" name="username" cols="60" class="input"><br>
         Password: <input type="password" name="password" cols="60" class="input"><br>
         <input type="submit"></form></div>';



$connection = mysql_connect("localhost", "username", "pass");

if (!connection)
{
die('Could not connect to database:' . mysql_error());
}


//check to see if sessions are set if not set them 
if(!isset($_SESSION['username']) or !isset($_SESSION['password'])){ 
if(!empty($_POST['username']) or !empty($_POST['password'])){ 
$_SESSION['username'] = $_POST['username']; 
$_SESSION['password'] = $_POST['password']; 
}
else{ 
echo'You left the username or password field blank.'; 
} 
} 


$sql = "SELECT User, Password FROM user WHERE User = '$_SESSION[username]' AND Password='$_SESSION[password]'"; 
$result = mysql_query($sql) OR die( mysql_error() ); 
$row = mysql_fetch_array($result, MYSQL_NUM); 
//compare sessions to those in table 
//Incorrect username and password 
if($_SESSION['username'] != $row['username'] AND $_SESSION['password'] != $row['password']){ 
$_SESSION = array(); 
if (isset($_COOKIE[session_name()])) { 
setcookie(session_name(), '', time()-42000, '/'); 
}
 
@session_destroy(); 
echo'Incorrect Username or password'; 
} 
//correct username and password 
else{ 
echo'Welcome '.$_SESSION['username']; 
} 




    echo '</body></html>';




?>
    PHP:
     
    Lpspider, Nov 10, 2006 IP
  9. nico_swd

    nico_swd Prominent Member

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    #9
    I don't see where you select a database. I see you connecting to it twice, but not actally selecting a database.

    
mysql_select_db($dbname);
    PHP:
     
    nico_swd, Nov 10, 2006 IP
    Lpspider likes this.
  10. Lpspider

    Lpspider Well-Known Member

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    #10
    Ah. Of course.

    What I did was modify this code from existing stuff I had... must have edited it out in the process I guess. Thanks.
     
    Lpspider, Nov 10, 2006 IP
  11. Lpspider

    Lpspider Well-Known Member

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    #11
    Here is the slightly updated code. However, it still spits out an error.

    <?php
session_start(); 

   echo '<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">';
   echo "<head><title> {$title} </title>";

   echo '<style type="text/css">

         #forms { position:absolute;top:75px;left:200px;align:right;padding-bottom:5px;}
         #text { position:absolute; top:75px; left:125px; align:left;} 
         .text2 { padding-top:8px;} 
         .forms2 { padding-bottom:6px;} 
         #login {position:absolute;top:100px;left:75%;} 
         .nav{position:absolute; top:75px; left:10%; right:10%; width:80%;background-color:#0099FF;text-align:right;a}
         .input{width:100px;} </style>
          </head><body>';
   

$logout = '<a href="logout.php">Logout</a>';

//menu
   echo '<div class="nav"><a href="index.php">Home</a> |', $logout  , ' | <a href="register.php">Register</a> | <a href="links.php">Links</a> | <a href="faq.php">FAQ</a></div>';

//login form
   echo '<div id="login"><form action="index.php" method="post">
         Login:<br>
         Username: <input type="text" name="username" cols="60" class="input"><br>
         Password: <input type="password" name="password" cols="60" class="input"><br>
         <input type="submit"></form></div>';

$connection = mysql_connect("localhost", "username", "password");

if (!connection)
{
die('Could not connect to database:' . mysql_error());
}
mysql_select_db("database", $connection); 

//check to see if sessions are set if not set them 
if(!isset($_SESSION['username']) or !isset($_SESSION['password'])){ 
if(!empty($_POST['username']) or !empty($_POST['password'])){ 
$_SESSION['username'] = $_POST['username']; 
$_SESSION['password'] = $_POST['password']; 
}
else{ 
echo'You left the username or password field blank.'; 
} 
} 


$sql = "SELECT User, Password FROM user WHERE User = '$_SESSION[username]' AND Password='$_SESSION[password]'"; 
$result = mysql_query($sql) OR die( mysql_error() ); 
$row = mysql_fetch_array($result, MYSQL_NUM); 
//compare sessions to those in table 
//Incorrect username and password 
if($_SESSION['username'] != $row['username'] AND $_SESSION['password'] != $row['password']){ 
$_SESSION = array(); 
if (isset($_COOKIE[session_name()])) { 
setcookie(session_name(), '', time()-42000, '/'); 
}
 
@session_destroy(); 
echo'Incorrect Username or password'; 
} 
//correct username and password 
else{ 
echo'Welcome '.$_SESSION['username']; 
} 

    echo '</body></html>';

?>
    PHP:
    Error:

    Warning: mysql_connect(): Access denied for user: 'username@localhost' (Using password: YES) in /var/www/localhost/htdocs/webdev/uploads/index.php on line 31

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in /var/www/localhost/htdocs/webdev/uploads/index.php on line 37
No Database Selected
    PHP:
     
    Lpspider, Nov 14, 2006 IP
  12. nico_swd

    nico_swd Prominent Member

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    #12
    You have to replace username and password with your actual username and password for the database access.

    
mysql_connect("localhost", "username", "password");
    PHP:
     
    nico_swd, Nov 14, 2006 IP
  13. Lpspider

    Lpspider Well-Known Member

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    #13
    ^ I know. I did. I just changed them for this post.
     
    Lpspider, Nov 14, 2006 IP
  14. nico_swd

    nico_swd Prominent Member

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    #14
    The error you're gettig says the login details are wrong.
     
    nico_swd, Nov 15, 2006 IP
  15. mad4

    mad4 Peon

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    #15
    You can't echo any content to the browser before you set the cookie. You are echoing right at the top of the script.
     
    mad4, Nov 15, 2006 IP
  16. Lpspider

    Lpspider Well-Known Member

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    #16
    ugh. *sigh* Still not getting it. I get errors whenever I do it any other way. Any suggestions?
     
    Lpspider, Nov 26, 2006 IP
  17. Luke

    Luke Peon

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    #17
    Just to test, replace

    $connection = mysql_connect("localhost", "username", "password");
if (!connection){die('Could not connect to database:' . mysql_error());}
mysql_select_db("database", $connection);
    PHP:
    with

    
$db=mysql_connect("localhost","yourusername","yourpassword");
mysql_select_db("yourdatabasename", $db);
    PHP:
    and see if it works, if so, check your database base is definately correct and that the log in name and password you are using definately has access to said database.
     
    Luke, Nov 27, 2006 IP
  18. Lpspider

    Lpspider Well-Known Member

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    #18
    The reason I was getting database connection errors is that I recently had a server move. It will connect now, though it still doesn't login/work with sessions.

    Here's what I have: 
    <?php
session_start(); 



$db=mysql_connect("localhost","username","password");
mysql_select_db("database", $db);



//check to see if sessions are set if not set them 
if(!isset($_SESSION['username']) or !isset($_SESSION['password'])){ 
if(!empty($_POST['username']) or !empty($_POST['password'])){ 
$_SESSION['username'] = $_POST['username']; 
$_SESSION['password'] = $_POST['password']; 
}
else{ 
echo'You left the username or password field blank.'; 
} 
}


$_POST['username'] = $_SESSION['username']; 
$_POST['password'] = $_SESSION['password']; 

$sql = "SELECT User, Password FROM user WHERE User = '$_SESSION[username]' AND Password ='$_SESSION[password]'"; 
$result = mysql_query($sql) OR die( mysql_error() ); 
$row = mysql_fetch_array($result, MYSQL_NUM); 
//compare sessions to those in table 
//Incorrect username and password 
if($_SESSION['username'] !== $row['username'] AND $_SESSION['password'] !== $row['password']){ 
$_SESSION = array(); 
if (isset($_COOKIE[session_name()])) { 
setcookie(session_name(), '', time()-42000, '/'); 
}
 
@session_destroy(); 
echo'Incorrect Username or password'; 

//debug info
echo $_SESSION['username'] ."and ". $_SESSION['password'];

} 

//correct username and password 
else{ 
echo'Welcome '. $_SESSION['username']; 
} 




   echo '<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">';
   echo '<head><title> {$title} </title>';

   echo '<style type="text/css">

         #forms { position:absolute;top:75px;left:200px;align:right;padding-bottom:5px;}
         #text { position:absolute; top:75px; left:125px; align:left;} 
         .text2 { padding-top:8px;} 
         .forms2 { padding-bottom:6px;} 
         #login {position:absolute;top:100px;left:75%;} 
         .nav{position:absolute; top:75px; left:10%; right:10%; width:80%;background-color:#0099FF;text-align:right;a}
         .input{width:100px;} </style>
          </head><body>';
   

$logout = '<a href="logout.php">Logout</a>';

//menu
   echo '<div class="nav"><a href="index.php">Home</a> |', $logout  , ' | <a href="register.php">Register</a> | <a href="links.php">Links</a> | <a href="faq.php">FAQ</a></div>';

//login form
   echo '<div id="login"><form action="index.php" method="post">
         Login:<br>
         Username: <input type="text" name="username" cols="60" class="input"><br>
         Password: <input type="password" name="password" cols="60" class="input"><br>
         <input type="submit"></form></div>';




    echo '</body></html>';

?>
    PHP:
    Basically it spits out "welcome," and nothing. If I enter username and password (which is should call to database/compare with sessions) it says "" and "". It should say "useranme" and "password".

    What am I missing? Advice/guidance is appreciated.
     
    Lpspider, Nov 28, 2006 IP
  19. Lpspider

    Lpspider Well-Known Member

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    #19
    ...Anyone???
     
    Lpspider, Dec 4, 2006 IP
  20. krakjoe

    krakjoe Well-Known Member

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    #20
    remove $_SESSION = array(); your emptying the session array which by the way, is always an array, the above line is never needed.
     
    krakjoe, Dec 4, 2006 IP
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