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this is my code that gets the info from the database but something is wrong, it goes not redirect me to either google or yahoo if my loggin details are correct or not, can anyone help me find the problem please...

 

<?php
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

if(isset($_POST['Login'])){
// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword'];
$mycompany=$_POST['mycompany']; 

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mycompany = stripslashes($mycompany);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mycompany = mysql_real_escape_string($mycompany);
//$_SESSION['myusername']=$dbusername;

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword' and company='$mycompany'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword and $mycompany, table row must be 1 row

if($count==1){
// Register $myusername, $mycompany and $mypassword and redirect to file "profile.php"
session_register("myusername");
session_register("mypassword");
session_register("mycompany"); 
header("location:http://www.yahoo.com");
}
else {
header("location: http://www.google.com");
}
}
?>

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https://forums.phpfreaks.com/topic/170452-solved-logging-in/
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Rather than use header() to see what is happing just use echo.

 

if($count==1){
echo 'User is logged in';
}
else {
echo 'User is NOT logged in';
}

 

Also you should not be using session_register for creating session variables. This function is depreciated. You should instead use the the $_SESSION superglobal when defining/using session variables.

 

So instead of

session_register("myusername");
session_register("mypassword");
session_register("mycompany"); 

 

You'd this instead

$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$_SESSION['mycompany'] = $mycompany;

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Yes. That is now the correct way for using sessions.

 

What does your script output now? It should display user is logged in when using the correct username/password. For the wrong username/password the message user is NOT logged in will be displayed.

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ok im still getting an error, this is my newest code..

<?php
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

if(isset($_POST['Login'])){
// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword'];
$mycompany=$_POST['mycompany']; 

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mycompany = stripslashes($mycompany);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mycompany = mysql_real_escape_string($mycompany);
//$_SESSION['myusername']=$dbusername;

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword' and company='$mycompany'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword and $mycompany, table row must be 1 row

if($count==1){
$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$_SESSION['mycompany'] = $mycompany;
echo 'User is logged in';
}
else {
echo 'User is NOT logged in';
}
}
?>

 

and this is my html form...

<form name="form1" method="post" action="checklogin.php">
   <input type="text" id="myusername"                            style="font-size:15px;"><br><br>
   <input type="text" id="mycompany"                             style="font-size:15px;"><br><br>
   <input type="password" id="mypassword"                        style="font-size:15px;"><br><br>
   <input type="submit" id="indexsearch" name="log in" value="Log In"  style="height:35px;">
</form>

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The name of your submit button should be named as Login

 

ok im still getting an error

Which is what? Post the error you're getting here. Or are you just getting a blank page? If its a blank page then it could mean your script is failing.

 

 

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i am just getting a blank page, what does it mean if i am getting a blank page and it is failing?

It means there could be an error which is preventing the script from running. You'll get a blank screen is errors are disabled.

 

To enable errors, at the top of your script add the following after the opening <?php tag

error_reporting(E_ALL);
ini_set('display_errors', 'on');

 

If there are any errors they should be displayed. Alternatively you can always check your servers error log

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i am still getting a blank page with this...

<?php
error_reporting(E_ALL);
ini_set('display_errors', 'on');
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

if(isset($_POST['Login'])){
// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword'];
$mycompany=$_POST['mycompany']; 

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mycompany = stripslashes($mycompany);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mycompany = mysql_real_escape_string($mycompany);
//$_SESSION['myusername']=$dbusername;

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword' and company='$mycompany'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword and $mycompany, table row must be 1 row

if($count==1){
$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$_SESSION['mycompany'] = $mycompany;
echo 'User is logged in';
}
else {
echo 'User is NOT logged in';
}
}
?>

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Your first three form fields don't have any name="..." attributes either. After you add them, post your current form because it is likely that the actual name="..." attribute of the submit button is not what your php code is expecting.

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still getting a black screen...  :shrug: :shrug:

 

<form name="form1" method="POST" action="checklogin.php">
   <input type="text" id="myusername"      name="myusername"    style="font-size:15px;"><br><br>
   <input type="text" id="mycompany"       name="mycompany"     style="font-size:15px;"><br><br>
   <input type="password" id="mypassword"  name="mypassword"    style="font-size:15px;"><br><br>
   <input type="submit" id="Login" name="Login" value="Log In"  style="height:35px;">
</form>

 

<?php
error_reporting(E_ALL);
ini_set('display_errors', 'on');
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name=""; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

if(isset($_POST['Login'])){
// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword'];
$mycompany=$_POST['mycompany']; 

// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$mycompany = stripslashes($mycompany);
$myusername = mysql_real_escape_string($myusername);
$mypassword = mysql_real_escape_string($mypassword);
$mycompany = mysql_real_escape_string($mycompany);
//$_SESSION['myusername']=$dbusername;

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword' and company='$mycompany'";
$result=mysql_query($sql);

// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
// If result matched $myusername and $mypassword and $mycompany, table row must be 1 row

if($count==1){
$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$_SESSION['mycompany'] = $mycompany;
header("location:about.php");
}
else {
echo 'User is NOT logged in';
}
}
?>

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There was no

session_start();

,

 

Here i cleaned it up aswell.

 

<?php session_start();

error_reporting(E_ALL);

ini_set('display_errors', 'on');

$host="localhost"; 
$username="";
$password=""; 
$db_name=""; 
$tbl_name="members";

mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB");

if(isset($_POST['Login'])){

$myusername = mysql_real_escape_string(stripslashes($_POST['myusername'])); 

$mypassword = mysql_real_escape_string(stripslashes($_POST['mypassword']));

$mycompany = mysql_real_escape_string(stripslashes($_POST['mycompany'])); 


$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword' and company='$mycompany'";

$result = mysql_query($sql);

if(mysql_num_rows($result) == '1'){

$_SESSION['myusername'] = $myusername;
$_SESSION['mypassword'] = $mypassword;
$_SESSION['mycompany'] = $mycompany;

echo 'User is logged in';

} else {

echo 'User is NOT logged in';

  }
}
?>

 

James.

 

 

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