Re: [PHP] WebMail client

2001-06-19 Thread Greg K

Try squirremail at http://www.squirrelmail.org


Rosen [EMAIL PROTECTED] wrote in message
9gnf7f$dsk$[EMAIL PROTECTED]">news:9gnf7f$dsk$[EMAIL PROTECTED]...
 Hi,
 I want to find some Web Mail script in PHP with possibilities to
 create mail accounts directly on Linux Mail server.

 Thanks
 Rosen Marinov






Re: [PHP] Parse error..help!

2001-05-18 Thread Greg K

Also take off the semicolon of the closing of the function }
Taline Makssabo [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 Here is another error i keep on getting, this is suppose to send me an
email
 each time someone enters in my website but i keep on gettong this error
 message:



 Parse error: parse error in
 /home/virtual/ppcu/home/httpd/html/php2/login.php on line 8
 I don't see anything wrong, please help.



 ?

 SetLogging(1);

 Function AccessHit
 (

 $NL = (\n);


 $H = getLastHost();


 $R = getLastRef();

 $To = [EMAIL PROTECTED];


 $Sub = Page Accessed;


 $D = Date(D d M y h:ia,time());

 $Msg = $H + $NL + $R + $NL + $D;

 mail($To,$Sub,$Msg);
 );
 AccessHit();

 ?




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[PHP] function login

2001-05-17 Thread Greg K

I am creating a function called login, I have a main file called index.php
.. which has my login form , it calls a login.php file, and in that
login.php there is file include statement which calls config.php and that is
where the function lie. This is my function , now if i don't set this as a
function and just put it my login.php it works fine. But when I name the
function in put it in my config.php no matter what I do all ways giving me
the error You have entered an invalid username or password. I dont know if
this has to do with passing of variables.. or what is going on but if
something can please help me I would appreciate it ..

Thank You

  function login() {
$query=Select uname from members where pass='$pass'and uname='$username';
$result=mysql_query($query) or die(You are not authorized to be here.);
$array= @mysql_fetch_array($result);
$uname = $array[0];
if($uname==)
{
echo You have entered an invalid username or password;
}
else

{
echo You are logged in;
$remote_address = getenv(REMOTE_ADDR);
echo Your IP address is $remote_address.;
}
}



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[PHP] resource id #2

2001-04-16 Thread Greg K

I am trying to run a query and in my log I am  getting a message the message
resource id #2.

$query=mysql_query("Select pass from members where uname='$username'");
$result = mysql_query($query)
or die("You are not authorized to be here.");


Can someone tell me what I am doing wrong and guide me in the right
direction



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