Parse error: syntax error, unexpected T_FUNCTION

[techrahul87]

HI everyone,

 

i m new to php. the code was working fine but when i include d.inc it is generating a error.

Parse error: syntax error, unexpected T_FUNCTION in every page of my application.

 

here is the sample of code from one page..........

<?php include(db.inc)
  if(isset($_REQUEST['submit']))
    {
   $user=$_POST["txtuser"];
   $pass=$_POST["txtpassword"];
       $res = 0;
   $con=mysql_connect($host1,$username1,$password1) or die(mysql_error());
   mysql_select_db($database1,$con) or die(mysql_error());
   $sql="select count(1) as c from userdata where username='".$user."' and password='".$pass."' and status=1 and now() < expireon";
   $result = mysql_query($sql);
   while($row = mysql_fetch_array($result))
	{
		$res = $row['c'];
	}

   if($res > 0)
    {
	  session_start();
	  $_SESSION['user'] = $user;
	  $_SESSION['pass'] = $pass;
	  header("Location:main.php");
	}
   else
    {
	  echo "Your username/password is either wrong or expired";
	}
	mysql_close($con);
}

 

error is generating on line where if is used.

thanks in advance

 

[Psycho]

You need to follow each line of code with a semi-colon. I would suggest never putting a line of code on the same line as the opening PHP tag for multi-line blocks of code - it impairs readability. The exception would be when echo'ing a value within the HTML content.

 

Also, that line won't work because the include file needs to be a string:

 

<?php
  include('db.inc');
  if(isset($_REQUEST['submit']))

 

EDIT: one last note. Since you are using a double-quoted string to define your query, you don't need to exit the string to include variables. Within double quoted strings variables will be interpreted as their assigned values - it makes the strings much more readable, IMHO. Also, when doing so, it is a good idea to enclose the variables within curly braces (i.e. {}) - that helps to prevent possible parsing errors.

 

 $sql="SELECT COUNT(1) AS c FROM userdata WHERE username='{$user}' AND password='{$pass}' AND status=1 AND NOW() < expireon";

[EdwinPaul]

Another thing: some browsers do not send the contents of the submit-button whent the user hits enter, so it might be dangerous to test for the contents of that button. Also: $_REQUEST is an array which consists of everything in the $_GET, $_POST and $_COOKIE -arrays. You'd better check:

if ($_SERVER['REQUEST_METHOD'] == "POST"){

in stead of

if(isset($_REQUEST['submit']))

Also: try to find something on the internet about sql-injection. Your script is vulnerable!

If I enter a user like  1' OR '1=1  with the same pasword, I can enter your site.

 

[techrahul87]

THANKS FOR HELPING ME OUT.............

THANK U VERY MUCH