[Resolved] assistance requested please -- problem with php coding.

brailleschool

The code below is part of a login form, if the username is in a database called terminated_request then it will give Error1 when someone tries to sign in. If the username is in a database called terminated_request then it will give Error2. If the username is not in any of those two databases, then it will log the user in.

It all works but it gives me an error when the user can login.

The code is below, and so is the error message. Please assist if you can.

<?
include("config.php");
$connection = mysql_connect("$server", "$db_user", "$db_pass");
$db = mysql_select_db("$database", $connection);
$query = "SELECT * FROM terminated_accounts where username='$username'";
$result = mysql_query($query, $connection);
$exterminated=mysql_numrows($result);
?>

<?
include("config.php");
$connection = mysql_connect("$server", "$db_user", "$db_pass");
$db = mysql_select_db("$database", $connection);
$query = "SELECT * FROM terminated_request where username='$username'";
$result = mysql_query($query, $connection);
$exterminationrequest=mysql_numrows($result);
?>

<?
include("config.php");
if($loged=='yes')
{
 echo "You are already logged in!";
 echo "<a href=\"logout.php\" accesskey=\"l\">$lang[logout]</a>";
}
else 
{
$connection = mysql_connect("$server", "$db_user", "$db_pass");
$db = mysql_select_db("$database", $connection);
$query = "SELECT * FROM login where username='$username'";
$result = mysql_query($query, $connection);

$rows = mysql_fetch_array($result);
$pass = $rows[password];
$email = $rows[email];
$name = $rows[fname];
}
if($password=='' OR $username=='')
{
echo "$lang[useloginform]";
}
elseif($exterminated=='1')
{
echo "$lang[paragraph_one_tos_violation]";
}
elseif($exterminationrequest=='1')
{
echo "$lang[member_request_termination]";
}
elseif($password==$pass)
 {
 setcookie("loged","yes",time()+3600);
 setcookie("username","$username",time()+3600);
 setcookie("name","$name",time()+3600);
 setcookie("email","$email",time()+3600);
 echo "$lang[thanksforlogin]";
 include("welcomepage.php");
 }
else 
{
echo "$lang[wrongpass]";
}
?>

Error message:

Warning: Cannot modify header information - headers already sent by (output started at /home/paddy/public_html/login.php:10) in /home/paddy/public_html/login.php on line 54

Warning: Cannot modify header information - headers already sent by (output started at /home/paddy/public_html/login.php:10) in /home/paddy/public_html/login.php on line 55

Warning: Cannot modify header information - headers already sent by (output started at /home/paddy/public_html/login.php:10) in /home/paddy/public_html/login.php on line 56

Warning: Cannot modify header information - headers already sent by (output started at /home/paddy/public_html/login.php:10) in /home/paddy/public_html/login.php on line 57

LINE 54:

setcookie("name","$name",time()+3600);

LINE 55:

setcookie("email","$email",time()+3600);

LINE 56:

echo "$lang[thanksforlogin]";

LINE 57:

include("welcomepage.php");

devinemke

first of all:
1. you don't need to include("config.php") 3 times
2. you don't need to call mysql_connect 3 times
3. you don't need to call mysql_select_db 3 times
you are adding a lot of unecessary overhead to your script

secondly:
you need to strip out all whitespace from your script prior to calling setcookie

pablodelapena

aftera brief look it looks like this area of code is your problem:


<? 
include("config.php"); 
$connection = mysql_connect("$server", "$db_user", "$db_pass"); 
$db = mysql_select_db("$database", $connection); 
$query = "SELECT * FROM terminated_accounts where username='$username'"; 
$result = mysql_query($query, $connection); 
$exterminated=mysql_numrows($result); 
?> 

<? 
include("config.php"); 
$connection = mysql_connect("$server", "$db_user", "$db_pass"); 
$db = mysql_select_db("$database", $connection); 
$query = "SELECT * FROM terminated_request where username='$username'"; 
$result = mysql_query($query, $connection); 
$exterminationrequest=mysql_numrows($result); 
?> 

<? 
include("config.php"); 
if($loged=='yes') 
{ 
 echo "You are already logged in!"; 
 echo "<a href=\"logout.php\" accesskey=\"l\">$lang[logout]</a>"; 
}

Headers are information sent to the browser before anything else. Here you are sending out put to the browser in the form of blank spaces and returns.

Try removing the gaps between the mysql_connect areas (why do you need 2? connect just the once, its always avaliable...) so basically your new code should look like this:


<? 

//connect to database ONCE!

include("config.php"); 
$connection = mysql_connect("$server", "$db_user", "$db_pass"); 
$db = mysql_select_db("$database", $connection); 


$query = "SELECT * FROM terminated_accounts where username='$username'"; 
$result = mysql_query($query, $connection); 
$exterminated=mysql_numrows($result); 



$query = "SELECT * FROM terminated_request where username='$username'"; 
$result = mysql_query($query, $connection); 
$exterminationrequest=mysql_numrows($result); 



if($loged=='yes') 
{ 
 echo "You are already logged in!"; 
 echo "<a href=\"logout.php\" accesskey=\"l\">$lang[logout]</a>"; 
}

Just make sure that the VERY FIRST line of your file is <?php and not an empty space. See if that sorts you out.

brailleschool

thanks -- that worked and did not produce any errors.

i didnt know about the include being once regardless. i will have to remember that -- will make it so much easier in the future.

thanks so much for the help.

i am still learning php/mysql and learn something new all the time.

robert_w_ark

I sometimes get those warnings too if no one is logged in.. Just at the top of your page add this

error_reporting(E_OFF);

or to turn it on

error_reporting(E_ALL);

Maybe this will help.

devinemke

Originally posted by robert_w_ark
I sometimes get those warnings too if no one is logged in.. Just at the top of your page add this
error_reporting(E_OFF);
[/B]

this is a bad idea. you should always do your coding and testing with error_reporting = E_ALL so you can see all errors that your code is throwing.

brailleschool

I agree with you devine. Because if there is an error and no errors are being reported. No members can tell me what is going on so that I can fix it/reproduce the errors etc.