PHP - Get Hardware Details

Hello, I am new to PHP and am trying to make a script that will ftp into the server via a cron of every hour or so and save the data to a mysql table. The first step is to draw variables out of information in the file via ftp. Its a ban list for a game server I run.

How could I get php to single out the ban reason, admin, etc.

the looks like this

Code: [Select]
"STEAM_0:0:42741441"
{
"name" "good mom"
"unban" "0"
"time" "1329161245"
"admin" "(Console)"
"reason" "no more trolling for you"
}
"STEAM_0:1:40563669"
{
"unban" "0"
"admin" "Headcrap(STEAM_0:1:42287608)"
"time" "1332855999"
"reason" ""
}
"STEAM_0:1:17693550"
{
"name" "ʎɖɑɱ"
"time" "1323055427"
"modified_time" "1323221308"
"unban" "0"
"modified_admin" "Strategos(STEAM_0:0:26662109)"
"admin" "Strategos(STEAM_0:0:26662109)"
"reason" "And so died the trolling"
}
Example: Code: [Select]
<?php
$bannedid = "STEAM_0:0:42741441";
$bannedby "(console)";
// and so on
?>

Hi all,

I am trying to write a script where the user can update their details.

The html form sends the details across to this php file below. All the php variables echo after the query however the values do not replace the current values in the mysql database.

Also for some reason the fname is the only one that changes no matter what the value is. It is entered in to mysql as '0'.

I think the problem will be in the php below.

As all the information below is echoed correctly, and because the value of 'fname' changes I know the connection is working fine.
I guess the error must be in the query I have written, however there is no error that comes up...

Hope you can point me in the right direction.

Thanks


<?php
include("../cxn.php");

$fname = $_POST['fname'];
$lname = $_POST['lname'];
$newemail = $_POST['newemail'];
$telephone = $_POST['telephone'];
$icao = $_POST['icao'];
$newpassword = $_POST['newpassword'];
$id = $_POST['id'];

$sql = "UPDATE Members SET fname='$fname' AND lname='$lname' AND email='$newemail' AND telephone='$telephone' AND password='$newpassword' AND icao='$icao' WHERE id='$id'";
$result = mysqli_query($cxn,$sql)
or die ("Couldn't execute query");

echo "Your new details a  <p>";
echo "$fname <br> $lname <br> $newemail <br> $telephone <br> $icao <br> $newpassword";

?>

Hi,

I am writing a little code that would populate the server name based on $_SERVER['HTTP_HOST']. My code and the error I encounter are given below. Can we ever use $_SERVER in a class?


class Get_server_name {

if ($_SERVER['HTTP_HOST'] == 'localhost'):
   $hostname = 'localhost';
else:
   $hostname = 'my remote server';
endif;

}

Error I get: Parse error: parse error, expecting `T_FUNCTION' in

Hello guys I have created a form and after entering details in the form user will click the generate button after that will get the qrcode. The problem I am facing is that when a user wants to change the qrcode details then it should upload the qrcode and change but how to do it I have no Idea any help will be appreciated.

The QRcode created using the PHP library. If you need some more information then  you can reply m.

i am working on a project in which i am facing a problem.
 my question is that

is it possible to get information/details of a user who is not logged in or who is not registered.

if a user logged in then we can find get details easily. but how can i do this. is it possible. 

Hi,
Im developing a program and it contains bank details in some of the fields stored in the MySQL database.
As part of the upload, I need to find bank details and then mask them by replacing them with XXXXXX for security purposes.

An example of a field could be "TRANSFER 540021 61782457" and I need to replace to "TRANSFER XXXXXX XXXXX457"

Any ideas to the best possible method?

Maybe I would mask blocks of numbers (ie numbers in blocks of 6 or 9) to cover this...im not sure what the best possible way is.

Thanks!

hello sir, I have created website where a user can create qr code by submitting data in form. I want users to edit qr code as well do you know how to do it ?

I have added qr code reader but it reads that data in qr code but I want the data to be edited by user.

Hello.  I just created a product gallery containing thumbnails of items that I will be selling.

When a user click on on of the thumbnails, I want a product details page to appear with more details about the chosen product

Could use some feedback if I am going about things the right way...

Step 1: User goes to my gallery located he   www.mysite.com/store/product-gallery.php

Step 2: System displays a page of thumbnails representing items that I am selling.

Step 3: User click on a given thumbnail which is a hyperlink to:  www.mysite.com/store/products/4571

Step 4: System rewrites the URL as such:  www.mysite.com/store/products.php?sku=4571

Step 5: System queries MySQL and retrieves record for sku 4571

Step 6: System displays product details for the chosen item in Step 3 which is sku 4571

 

How does that sound?

 

Perhaps this is a rather lay question, but, is there a way to gather specific connection details about an open MySQL connection in php?

Example:

<?php
    $connect = mysql_connect('localhost', 'username', 'password');
?>

Using the $connect variable, could I run a command that dumps the host, and username to a log file??

Thanks in advance, I'm still searching.

E

Code: [Select]
$sql= ("INSERT INTO custcards (username,Card_Number,Card_Type,Card_Name,Card_End) VALUES('$username','$cardnumber','$cardtype','$cardname','$cardend')");

if (isset($_GET['submitted'])) {
  if (checkCreditCard ($_GET['CardNumber'], $_GET['CardType'], $errornumber, $errortext)) {
    $errortext = 'This card has a valid format'; // Default text if card is VALID
$result = mysql_query($sql)){
echo "$errortext"; // Echo/Show this text
     
}
}
else
{
 echo '<p align=center>';echo "$errortext"; // Display ERROR Type/Text
 echo "ERROR: ".mysql_error();
}

}
The code is to check details beign entered and to make sure thier correct and then print the correcrt message but i also want it to store the details into a table

i know i havent got the syntax correct was wondering is anyone can help ?

I need some help please.  I have a project that has a login script that is required before moving to a shopping cart and each use a session.  I have a login script that i know works separate of the cart and vice versa.  The problem is that when i access the cart i get the session errors and it quits working.  So i need help with getting the 2 combined so i can run the cart script while a user is logged in.  Here is the code i have.

This is the login script session:
<? ob_start(); session_start();include_once"config.php";

if(!isset($_SESSION['username']) || !isset($_SESSION['password'])){

header("Location: login.php");
}else{
$user_data = "".$_SESSION['username']."";
$fetch_users_data = mysql_fetch_object(mysql_query("SELECT * FROM `members` WHERE `username`='".$user_data."'"));
}
?>



This is the script session for the shopping cart:
<?php
require_once 'library/config.php';
require_once 'library/category-functions.php';
require_once 'library/product-functions.php';
require_once 'library/cart-functions.php';

$_SESSION['shop_return_url'] = $_SERVER['REQUEST_URI'];

$catId  = (isset($_GET['c']) && $_GET['c'] != '1') ? $_GET['c'] : 0;
$pdId   = (isset($_GET['p']) && $_GET['p'] != '') ? $_GET['p'] : 0;

require_once 'include/header.php';
?>  

and any help is very much appreciated.

I have created a button which when pressed should present the user with their details (whoever is logged in), here is the form code:

<form id="form1" name="form1" method="post" action="getdetails.php">
 
  <input type="submit" name="Get Details" value="Get Details" />
   </label>
   
  </p>
</form>

Here is the getdetails.php file

<?php
 mysql_connect("localhost","root","");
 mysql_select_db("test");
 $username = $_POST['textfield'];
  echo '</br>';
 $query = mysql_query("SELECT * FROM membersdetails WHERE name=`$username` ");

while($result = mysql_fetch_array($query)) {
//display
echo $result['firstname'];
echo $result['surname'];
}
?>

Its not workin at all I have attacthed the error i am getting

Any help please?

after the user has logged in, I would like to display their details by barcode id


Login.php

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



session_start();
// username and password sent from form
$barcodeID=$_POST['barcode'];



// To protect MySQL injection (more detail about MySQL injection)
$barcodeID = stripslashes($barcodeID);
$barcodeID = mysql_real_escape_string($barcodeID);


$sql="SELECT * FROM $tbl_name WHERE BarcodeID='$barcodeID'";


$result=mysql_query($sql);
     
$count=mysql_num_rows($result);

if($count > 0){
      $data = mysql_fetch_array ($result);
      
       $_SESSION["user_id"] = $data["BarcodeID"];
        $_SESSION["user_firstname"] = $data["Firstname"];
        $_SESSION["user_surname"] = $data["Surname"];
        $_SESSION["user_jobrole"] = $data["JobRole"];
        $_SESSION["user_manager"] = $data["Manager"];
        $_SESSION["user_priority"] = $data["Priority"];
        $_SESSION["user_datejoined"] = $data["DateJoined"];
        $_SESSION["user_times_loggged_in"] = $data["TimesLoggedOn"];

        if ($_SESSION["user_priority"] == '1')
        {
            header("Location: AdminSection.php");           
        }
        else
        {
            header("Location:LoggedIn.php");
      }      

      if ($_SESSION["user_times_loggged_in"] == '0')
      {
         
         header("Location:UsingTheSystem.html");
      }
}
         
?>



LoggedIn.php



I keep getting the error undefined index "barcode"?

<?php

$barcodeID = $_POST["barcode"];


include 'dbcon.php';

    $sql = "SELECT Firstname, Surname, JobRole, Manager"
        . " FROM users"
        . " WHERE BarcodeID = .'$barcodeID'" ;
    $rows = mysql_query($sql);
   
    echo $rows;
   
    ?>


Any help will be greatly appreciated


Thanks

Hey i have created the code below attempting to show a specific members details when their username is entered and the submit button is pressed. but when i enter the username all members details are shown. The code i have used is shown below, can anybody help me with the code to just show the single users details. I have attached an image aswell.

<?php
mysql_connect ("localhost","root","");
mysql_select_db ("test");

$sql = "select * from memberdetails";
       $result = mysql_query ($sql);

       while ($row = mysql_fetch_array($result))
              {
              $username= $row["username"];
              $firstname= $row["firstname"];
             $surname= $row["surname"];
              $dob= $row["dob"];
              $totalwins= $row["totalwins"];   
              $totalloses= $row["totalloses"];
              $email= $row["email"];
              $country= $row["country"];
              $info= $row["info"];

             

   echo "<b><u>Username:</b></u>  $username<br>";
   echo "<b><u>Firstname:</b></u> $firstname<br>";
   echo "<b><u>Surname: </b> </u> $surname<br>";
   echo "<b><u>Date of Birth:</b></u> $dob<br>";
   echo "<b><u>Total Chess Wins:</b></u>  $totalwins<br>";   
   echo "<b><u>Total Chess loses:</b></u> $totalloses<br>";
   echo "<b><u>Email Address: </b></u> $email<br>";
   echo "<b><u>Born in: </b></u> $country<br>";
   echo "<b><u>Other Details:</b></u>  $info<br>";
           }
?>

hi,

i'm new in php/mysql. i'm stored student marks values in following format in mysql db table.


id student_code Tamil English Maths Science Social
1                1        100      75       78      88       95
2                2         85      90        88     80      100

But i want to search and display the specific student marks in following format.

id:1

student_code:1

Tamil:100

English:75

Maths:78

Science:88

Social:92

Total:?

Avg:?

please give correct code for this format.