PHP SCRIPT
#1

Код:
<?php
$variable1 = $_POST['var1'];
$variable2 = $_POST['var2'];
$variable3 = $_POST['var3'];
$variable4 = $_POST['var4'];
 
 
$keycodee = $variable3;
 
$hwid = $variable3;
$uname = $variable1;
$passworda = $variable2;
 
$server = "";
$username = "";
$password = "";
$dbname = "";
 
 
 
 
 
 
$pwrd = sha1(strtolower($uname) . $passworda);
// Create connection
$conn = new mysqli($server, $username, $password, $dbname);
 
// Check connection
if ($conn->connect_error) die("Connection failed: " . $conn->connect_error);
 
//Declare SQL statement
$sql = "SELECT Username, Password FROM login WHERE Username= '$uname'";
 
//Catch result
$result = $conn->query($sql);
 
//Check for result
if ($result->num_rows > 0) {
 
    // output data of each row
  while($row = $result->fetch_assoc()) {
 
$checkuname = $row["Username"];
$checkpwrd = $row["Password"];
$hwiddd = $row["HWIDMAC"];
 
//echo "id: " . $checkgroup. " - Name: " . $checkuname. " " . $checkpwrd . "<br>";
   }
    }
   
if ($uname == $checkuname && $pwrd == $checkpwrd)
{
    echo "1";
 
}
else
{
echo "Pword/Uname is wrong";
}
 
//if ($uname != $checkuname) echo "Your username could not be found";
//if ($pwrd != $checkpwrd) echo "Your password was incorrect";
 
 
 
$conn->close();
 
 
 
 
?>
Always says echo "Pword/Uname is wrong";
Reply
#2

We need to see your <form> tags in HTML.

(Don't ask me why I'm helping you.)
Reply
#3

Код HTML:
<?php
$variable1 = $_POST['var1'];
$variable2 = $_POST['var2'];
$variable3 = $_POST['var3'];
$variable4 = $_POST['var4'];
 
 
$keycodee = $variable3;
 
$hwid = $variable3;
$uname = $variable1;
$passworda = $variable2;
 
$server = "";
$username = "";
$password = "";
$dbname = "";
 
 
 
 
 
 
$pwrd = sha1(strtolower($uname) . $passworda);
// Create connection
$conn = new mysqli($server, $username, $password, $dbname);
 
// Check connection
if ($conn->connect_error) die("Connection failed: " . $conn->connect_error);
 
//Declare SQL statement
$sql = "SELECT Username, Password FROM login WHERE Username= '$uname'";
 
//Catch result
$result = $conn->query($sql);
 
//Check for result
if ($result->num_rows > 0) {
 
    // output data of each row
  while($row = $result->fetch_assoc()) {
 
$checkuname = $row["Username"];
$checkpwrd = $row["Password"];
$hwiddd = $row["HWIDMAC"];
 
//echo "id: " . $checkgroup. " - Name: " . $checkuname. " " . $checkpwrd . "<br>";
   }
    }
   
if ($uname == $checkuname && $pwrd == $checkpwrd)
{
    echo "1";
 
}
else
{
echo "Pword/Uname is wrong";
}
 
//if ($uname != $checkuname) echo "Your username could not be found";
//if ($pwrd != $checkpwrd) echo "Your password was incorrect";
 
 
 
$conn->close();
 
 
 
 
?>
Always says echo "Pword/Uname is wrong";[/QUOTE]
Reply
#4

Quote:
Originally Posted by alexanderjb918
Посмотреть сообщение
Код HTML:
<?php
$variable1 = $_POST['var1'];
$variable2 = $_POST['var2'];
$variable3 = $_POST['var3'];
$variable4 = $_POST['var4'];
 
 
$keycodee = $variable3;
 
$hwid = $variable3;
$uname = $variable1;
$passworda = $variable2;
 
$server = "";
$username = "";
$password = "";
$dbname = "";
 
 
 
 
 
 
$pwrd = sha1(strtolower($uname) . $passworda);
// Create connection
$conn = new mysqli($server, $username, $password, $dbname);
 
// Check connection
if ($conn->connect_error) die("Connection failed: " . $conn->connect_error);
 
//Declare SQL statement
$sql = "SELECT Username, Password FROM login WHERE Username= '$uname'";
 
//Catch result
$result = $conn->query($sql);
 
//Check for result
if ($result->num_rows > 0) {
 
    // output data of each row
  while($row = $result->fetch_assoc()) {
 
$checkuname = $row["Username"];
$checkpwrd = $row["Password"];
$hwiddd = $row["HWIDMAC"];
 
//echo "id: " . $checkgroup. " - Name: " . $checkuname. " " . $checkpwrd . "<br>";
   }
    }
   
if ($uname == $checkuname && $pwrd == $checkpwrd)
{
    echo "1";
 
}
else
{
echo "Pword/Uname is wrong";
}
 
//if ($uname != $checkuname) echo "Your username could not be found";
//if ($pwrd != $checkpwrd) echo "Your password was incorrect";
 
 
 
$conn->close();
 
 
 
 
?>
Always says echo "Pword/Uname is wrong";
[/QUOTE]


Again, That's the PHP,
We need the HTML, Where doyou have this PHP code? Give us either the whole html page or only the <form> tag.
Reply
#5

Also, Why do you have strltolower?
Basically-> You're turning all letters down to lowercased, then hash it with sha1.
Why?
Reply
#6

Leave him be, He's new to PHP it looks like.

If you want more assistance i've added your skype.

- Alexander
Reply
#7

I am new to PHP coding as you see the code is 'aids' this is a post login script via a c++ application that gets posts/requests then logs the client in if the information is correct.
Reply
#8

Quote:
Originally Posted by minijackc
Посмотреть сообщение
Leave him be, He's new to PHP it looks like.

If you want more assistance i've added your skype.

- Alexander
No offence but I did not mean to harm anyone, I just needed to know why he did so.
____________________________________
PHP код:
<?php
$conn 
= new mysqli($server$username$password$dbname);
if (
$conn->connect_error)
    die(
"Connection failed: " $conn->connect_error);
if(isset(
$_POST["submit"])) {
    if(!
is_null($_POST["username"])) {
        if(!
is_null($_POST["password"])) {
            
$username $conn->real_escape_string(stripslashes($_POST['username']));
            
$password sha1($conn->real_escape_string(stripslashes($_POST['password'])));
            
$sql "SELECT Username, Password FROM login WHERE Username= '$username' AND Password= '$password'";
            
$result $conn->query($sql);
            if(
$result->num_rows 0) {
                echo 
"logged in.";
            }
        }
        else
            echo 
"Don't leave the password empty.";
    }
    else
        echo 
"Don't leave the username empty.";
}
$conn->close();
?>
Now it wont work, cause you need to get the name="" in the <input> in order to handle a form request.
So that's why we can't help you without the html.
Reply


Forum Jump:


Users browsing this thread: 1 Guest(s)