Redirect logging in users to different pages

jusjus7 · January 29, 2012

I have this code which logs in a user to my site but it only directs them to the login success page. i need to direct users to 3 different pages depending on their access level eg levels 1,2,4 but i dont know how to do this, can any one help or point me in the right direction??

thanks :confused:

//Form

<strong>Member Login </strong>

Username: <input name="UsersID" type="text" id="UsersID" />

Password: <input name="U_Password" type="password" id="U_Password" />

<?php

$host="localhost"; // Host name

$username=""; // username

$password=""; // password

$db_name="test"; // Database name

$tbl_name="members"; // Table name

// Replace database connect functions depending on database you are using.

mysql_connect("$host", "$username", "$password");

mysql_select_db("$db_name");

//submitting query

// username and password sent from form

//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!

$mUsersIDe=mysql_real_escape_string($_POST['UsersID']);

$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT * FROM $Users WHERE UsersID='$UsersID' and U_Password='$U_Password'";

$result=mysql_query($sql);

//checking results

// Replace counting function based on database you are using.

$count=mysql_num_rows($result);

// If result matched $myusername and $mypassword, table row must be 1 row

//Direct Userbased on result

if($count==1){

// Register $UsersID, $U_Password and redirect to file "login_success.php"

session_register("UsersID");

session_register("U_Password");

header("location:login_success.php");

} else {

echo "Wrong Username or Password";

}

17443_.txt

jusjus7 · January 29, 2012

I have been able to get a user to login to one login page but i need the site to direct users to different login pages depending on their access level. I think i just need to add an If statement to my code but i do not know how to do that!

Can anyone help?

<?php
$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");


//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT * FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  session_register("UsersID");
  session_register("U_Password");
  header("location:login_success.php");
} else {

  echo "Wrong Username or Password";
}
?>

Pikachu2000 · January 29, 2012

I've merged your duplicate topics. In the future, just post your corrections in the same thread instead of creating a new one.

jusjus7 · January 29, 2012

oh ok thanks for that!

Drummin · January 29, 2012

I wouldn't save a password to session. As far as your question, where you have the header going to login_success.php, just check the user level and based on this, send the user to appropriate page with the header. A few simple IF statements should do the job.

jusjus7 · January 29, 2012

ok thanks i will try that out

jusjus7 · January 29, 2012

i have entered this and it doesnt seem to work any idea where i have gone wrong?

//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  session_register("UsersID");
  if("U_YearID" = "1")  
  header("location:login_success.php");
  if("U_YearID"="2")
  header("location:login_success2.php");
  if("U_YearID"="4")
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}

Drummin · January 29, 2012

U_YearID is plain text and not defined as string and when comparing two values use two == signs. The code posted is not tested and assumes U_YearID is a table field in your users table.

<?php
session_start();
$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");


//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT UsersID,U_YearID FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

jusjus7 · January 29, 2012

I changed the code with this code just now

//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  session_register("UsersID");
  session_register("U_Password");
  if('U_YearID'=="1")  
  header("location:login_success.php");
  if('U_YearID'=="2")
  header("location:login_success2.php");
  if('U_YearID'=="4")
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

i keep getting this errors

Deprecated: Function session_register() is deprecated in C:\xampp\htdocs\Site1\test\checklogin.php on line 39

Deprecated: Function session_register() is deprecated in C:\xampp\htdocs\Site1\test\checklogin.php on line 40

ie something is wrong with the session_register

any ideas anyone? (srry really new at this) lol

jusjus7 · January 29, 2012

with your code i get this error

Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0

:s

U_YearID is plain text and not defined as string and when comparing two values use two == signs. The code posted is not tested and assumes U_YearID is a table field in your users table.

<?php
session_start();
$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");


//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT UsersID,U_YearID FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

Drummin · January 29, 2012

I don't know about this error but it seems like it might be a htaccess issue or related to the version of php you are using.

jusjus7 · January 29, 2012

i am using these versions

Apache/2.2.21 (Win32) mod_ssl/2.2.21 OpenSSL/1.0.0e PHP/5.3.8

jusjus7 · January 29, 2012

I don't know about this error but it seems like it might be a htaccess issue or related to the version of php you are using.

I was able to disable that error but i get this error now, and i have checked and the username and password and level i am using is inside the database

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Site1\test\checklogin1.php on line 30

Wrong Username or Password

spiderwell · January 29, 2012

echo out your sql statement it will help you solve that

jusjus7 · January 29, 2012

echo out your sql statement it will help you solve that

sorry i know i am sounding really dumb but how do i do that? where to i type echo by the select by the from or at the where end?

spiderwell · January 29, 2012

$sql="SELECT UsersID,U_YearID FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

jusjus7 · January 29, 2012

$sql="SELECT UsersID,U_YearID FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

Nope :/ i get this error now

SELECT UsersID, U_YearID FROM users WHERE UsersID='aa101' and U_Password='000123456'

Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\Site1\test\checklogin1.php on line 31

Wrong Username or Password

jusjus7 · January 29, 2012

This is all the code i am using right now

<?php
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);

$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT * FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

spiderwell · January 29, 2012

you have removed the code for connecting to the database before you execute your query

jusjus7 · January 29, 2012

i have put it bk in

<?php
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);

$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name"); 

//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT * FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

and i get this error

jusjus7 · January 29, 2012

i have put it bk in

<?php
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);

$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name"); 

//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT * FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row


//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

and i get this error

SELECT UsersID,U_YearID FROM users WHERE UsersID='aa101' and U_Password='000123456'

Pikachu2000 · January 29, 2012

That isn't an error. It's the output from echo $sql;.

You need to include session_start() at the beginning of the script.

The header() redirects will never work if there is has already output sent to the browser, such as echo $sql;

jusjus7 · January 29, 2012

That isn't an error. It's the output from echo $sql;.

You need to include session_start() at the beginning of the script.

The header() redirects will never work if there is has already output sent to the browser, such as echo $sql;

Still nothing i have added session_start() at the top of the php code and still getting that :/

session_start();
ini_set('session.bug_compat_42',0);
ini_set('session.bug_compat_warn',0);
$host="localhost"; // Host name
$username="phpuser"; // username
$password="phpuser"; // password
$db_name="phpsite"; // Database name
$tbl_name="users"; // Table name

// Replace database connect functions depending on database you are using.
mysql_connect("$host", "$username", "$password");
mysql_select_db("$db_name");

//submitting query
// username and password sent from form
//NEVER Remove the mysql_real_escape_string. Else there could be an Sql-Injection!
$UsersID=mysql_real_escape_string($_POST['UsersID']);
$U_Password=mysql_real_escape_string($_POST['U_Password']);

$sql="SELECT UsersID,U_YearID FROM users WHERE UsersID='$UsersID' and U_Password='$U_Password'";
echo $sql;
$result=mysql_query($sql);

//checking results
// Replace counting function based on database you are using.
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row

//Direct Userbased on result
if($count==1){
  // Register $UsersID, $U_Password and redirect to file "login_success.php"
  $_SESSION['UsersID']=$result[0];
  $_SESSION['U_YearID']=$result[1];
  if($_SESSION['U_YearID']==1)  
  header("location:login_success.php");
  if($_SESSION['U_YearID']==2)
  header("location:login_success2.php");
  if($_SESSION['U_YearID']==4)
  header("location:login_success4.php");
} else {

  echo "Wrong Username or Password";
}
?>

Drummin · January 29, 2012

As Pikachu2000 pointed out, you can't send anything to the browser before setting any sessions. This includes echoing out the $sql. Comment out that line or remove it.

jusjus7 · January 29, 2012

As Pikachu2000 pointed out, you can't send anything to the browser before setting any sessions. This includes echoing out the $sql. Comment out that line or remove it.

I commented it out and i dont get that sql "error" any more but all i get is a blank page...any ideas would it could be and why its not taking me to any page?

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