Jump to content

Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in C:\wam


Reece112233

Recommended Posts

Having a problem with a php code.

Im new to php and have only been learning for one day but have had previous experiances with php in which i have copied and pasted and edited to suit my needs .

 

This is my code

<?php

   $username = $_POST["user"];
   $password = $_POST["pass"];
   $login = $_GET["login"];
   
   setcookie("username","$username",time()+86400);
   
   if($login=="yes") {
      
      $con = mysql_connect("localhost","root","arceye");
      
      mysql_select_db("login");
      
      $get = mysql_query("SELECT count(id) FROM login WHERE user='$username' and pass='$password'");
      
      $result = mysql_result($get, 0);
      
      if($result!=1) {
         echo "Invalid Username Or Password";
      } else {
         echo "Login Successful"
         $_SESSION["username"] = $username;
      }
   }

 

i named that login.php

 

im running a wamp server thinge on localhost.

 

when i try to run it on http://localhost/login.php i get the error

Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in C:\wamp\www\login.php on line 23

 

can anyone help

i had 5 other error but managed to use common sense to fix them this one i cant seem to work out. :(

 

Thanks in advance

 

MOD EDIT: code tags added.

Link to comment
Share on other sites

try this:

 

<?php
session_start();
$username = trim($_POST["user"]);
$password = trim($_POST["pass"]);
$login = trim($_GET["login"]);
setcookie("username","$username",time()+86400);
if($login == "yes"){
$con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
mysql_select_db("login",$con) or die('Cannot select that database...');
$get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());      
$res = mysql_fetch_assoc($get);
@mysql_close($con);
$result = $res['total'];
if($result !=1 ){
	echo "Invalid Username Or Password";
	exit();
}else{
	echo "Login Successful"
	$_SESSION["username"] = $username;
}
}
?>

Link to comment
Share on other sites

Nope it dosent work although i like this code better it now says

Parse error: syntax error, unexpected T_VARIABLE in C:\wamp\www\login.php on line 3

and what is a typo

this is the code i wrote i dont see any differance in the 3rd line

<?php
session_start
$username = trim($_POST["user"]);
$password = trim($_POST["pass"]);
%login = trim($_GET["login"]);
setcookie("username","$username",time()+86400)
if($login == "yes"){
   $con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
   mysql_select_db("login",$con) or die ('Cannot select that database...');
   $get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());      
   $res = mysql_fetch_assoc($get);
   @mysql_close($con);
   $result = $res['total'];
   if($result !=1 ){
      echo "Invalid Username Or Password";
      exit();
   }else{
      echo "Login Successful"
      $_SESSION["username"] = $username;
   }
}
?>

thanks...

 

MOD EDIT: code tags added.

Link to comment
Share on other sites

another problem. got one on  line 5 but fixed by changing % to $ a typo.. (i know what a typo is now) and line 7 the line b4 it was wrong fixed that now im onto line 19 error is

Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in C:\wamp\www\login.php on line 19

 

this is my code (well my code improved by you) so far

<?php
session_start();
$username = trim($_POST["user"]);
$password = trim($_POST["pass"]);
$login = trim($_GET["login"]);
setcookie("username","$username",time()+86400);
if($login == "yes"){
   $con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
   mysql_select_db("login",$con) or die ('Cannot select that database...');
   $get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());      
   $res = mysql_fetch_assoc($get);
   @mysql_close($con);
   $result = $res['total'];
   if($result !=1 ){
      echo "Invalid Username Or Password";
      exit();
   }else{
      echo "Login Successful"
      $_SESSION["username"] = $username;
   }
}
?>

thanks once again

Link to comment
Share on other sites

oh i see you know your very clever. i added a @ before the $username stuff on the  6 , 7 , 8 line of the code and that sorted out the database not found error. i also added another user and pass to my database . the user is john and the pass is 1234 the id is 2

 

im now getting a blank webpage after i login nothing happens .. the website url is http://localhost/login.php?username=Admin&password=oliver

i thought there might be an error and its not showing up so i edited the code to do error reporting but thats not working i just get a blank screen..

 

i have decided to include my html code in this bit because the part thats not working is a visual part usually handled by html .

 

the html code is

<html>
<body>

   <form action="login.php" method"post">
      Username: <input type="text" name="username" />
      <br>
      Password <input type="password" name="password" />
      <br>
      <input type="submit" value="Login" />
   </form>
</body>
<html>

 

and the new php code is

<?php
error_reporting(E_ALL);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
session_start();
@$username = trim($_POST["user"]);
@$password = trim($_POST["pass"]);
@$login = trim($_GET["login"]);
setcookie("username","$username",time()+86400);
if($login == "yes"){
   $con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
   mysql_select_db("login",$con) or die ('Cannot select that database...');
   $get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());      
   $res = mysql_fetch_assoc($get);
   @mysql_close($con);
   $result = $res['total'];
   if($result !=1 ){
      echo "Invalid Username Or Password";
      exit();
   }else{
      echo "Login Successful";
      $_SESSION["username"] = $username;
   }
}
?>

 

 

 

NOTE: these are two separate files login.html and login.php both being stored in the www directory of my wamp server and can be accesses by typing

 

localhost/login.html

or

localhost/login.php

 

thanks

 

MOD EDIT: code tags added.

Link to comment
Share on other sites

As I understand it, you are not being sent back to login.html after processing form on login.php.  Try

<?php
error_reporting(E_ALL);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
session_start();
@$username = trim($_POST["user"]);
@$password = trim($_POST["pass"]);
@$login = trim($_GET["login"]);
setcookie("username","$username",time()+86400);
if($login == "yes"){
   $con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
   mysql_select_db("login",$con) or die ('Cannot select that database...');
   $get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());     
   $res = mysql_fetch_assoc($get);
   @mysql_close($con);
   $result = $res['total'];
   if($result !=1 ){
      echo "Invalid Username Or Password";
      exit();
   }else{
      echo "Login Successful";
      $_SESSION["username"] = $username;
header("location: login.html");
   }
}
?>

Link to comment
Share on other sites

@ is bad coding practice in my opinion.  Do it right, and save yourself some trouble in the long run.

 

@$username = trim($_POST["user"]);
@$password = trim($_POST["pass"]);
@$login = trim($_GET["login"]);

Should be:

$username = (!emtpy($_POST['user'])) ? trim($_POST["user"]) : NULL;
$password = (!empty($_POST['pass'])) ? trim($_POST["pass"] : NULL;
$login = (isset($_GET['login'])) ? trim($_GET["login"]) : NULL;

Link to comment
Share on other sites

ok ive edited my html code to this

<html>
<body>

   <form action="login.php if($login == "yes"){" method"post">
      Username: <input type="text" name="username" />
      <br>
      Password <input type="password" name="password" />
      <br>
      <input type="submit" value="Login" />
   </form>
</body>
<html>

 

 

and my php file to this

<?php
error_reporting(E_ALL);
ini_set('display_errors', TRUE);
ini_set('display_startup_errors', TRUE);
session_start();
$username = (!emtpy($_POST['user'])) ? trim($_POST["user"]) : NULL;
$password = (!empty($_POST['pass'])) ? trim($_POST["pass"] : NULL;
$login = (isset($_GET['login'])) ? trim($_GET["login"]) : NULL;
setcookie("username","$username",time()+86400);
if($login == "yes"){
   $con = mysql_connect("localhost","root","arceye") or die('Cannot connect to server...');
   mysql_select_db("login",$con) or die ('Cannot select that database...');
   $get = mysql_query("SELECT count(`id`) as total FROM `login` WHERE `user`='$username' and `pass`='$password'",$con) or die(mysql_error());      
   $res = mysql_fetch_assoc($get);
   @mysql_close($con);
   $result = $res['total'];
   if($result !=1 ){
      echo "Invalid Username Or Password";
      exit();
   }else{
      echo "Login Successful";
      $_SESSION["username"] = $username;
   header("location: login.html");
   }
}
?>

ok?

 

MOD EDIT: code tags added.

Link to comment
Share on other sites

im now getting an error saying

 

Parse error: syntax error, unexpected ':' in C:\wamp\www\login.php on line 7

 

i tried changing the : to a ; which just said

 

Parse error: syntax error, unexpected ';' in C:\wamp\www\login.php on line 7

 

i tried removing it completely and the screen went blank..

 

so i just clicked undo and left it as it is with the @ although if anyone would to come up with a solution to that problem i would use the person whos name begins with j (soz cant see name from the post reply screen) idea.. thanks

Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.