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frobak

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Posts posted by frobak

  1. Hi

     

    Im trying to write a very simple script that creates a php file that logs a download in a database.

     

    Basically, the user will upload some audio and I want to use fopen to automatically create a download counter. I get the file created ok, but my $sql, $result & $connection are stripped and not written, so the queries wont run.

     

    Is there anyway to escape this, so it writes the variables.

     

    Heres the code:

     

    <?php

     

    $file = "downloads/pbz99.php";

    $open = fopen($file, "w+");

    fwrite($open,"<?php include ('functions/db_fws_connect.inc.php');

    $sql = 'INSERT INTO `frobakco_fwsapps`.`downloadcount` (`downloadcountid`, `downloadid`) VALUES (NULL, '99')';

    $result = @mysql_query($sql,$connection) or die(mysql_error());?>

    <html><head><title>DJ Neil S audio download page</title></head><body><a href='../mixes/.$filename.mp3'></a></body></html>");

    ?>

     

    any ideas

  2. but heres the table, copied directly from mysql?  the columns there?

     

    customer_id  int(15)  No     

    company_name varchar(20) Yes  NULL 

    address_line1 varchar(20) Yes  NULL 

    address_line2 varchar(20) Yes  NULL 

    City varchar(20) Yes  NULL 

    postcode varchar(8) Yes  NULL 

    contact_name varchar(25) Yes  NULL 

    emailaddress varchar(25) Yes  NULL 

    webaddress varchar(25) Yes  NULL 

    phone varchar(20) Yes  NULL 

    fax varchar(20) Yes  NULL 

    mobile varchar(20) Yes  NULL

  3. CREATE TABLE customers(

    customer_id INT(15),

    company_name VARCHAR(20),

    address_line1 VARCHAR(20),

    address_line2 VARCHAR(20),

    City VARCHAR(20),

    postcode VARCHAR(8),

    contact_name VARCHAR(25),

    emailaddress VARCHAR(25),

    webaddress VARCHAR(25),

    phone VARCHAR(20),

    fax VARCHAR(20),

    mobile VARCHAR(20));

  4. Hi i have this simple php insert statement, but whatever i try it says:

     

    Unknown column 'company_name' in 'field list'

     

    all the column names are the same in my html documnt, this statement here and in my database table. Thsi is really simple and driving me mad

     

    <?

    if($_POST['add']) {

     

    include("includes/db_connect.inc.php");

    $sql = ("INSERT INTO customers

    (company_name, address_line1, address_line2, city, postcode, contact_name, emailaddress, webaddress, phone, fax, mobile) VALUES

    ('$_POST[company_name]', '$_POST[address_line1]', '$_POST[address_line2]', '$_POST[city]', '$_POST[postcode]', '$_POST[contact_name]',

    '$_POST[emailaddress]', '$_POST[webaddress]', '$_POST[phone]', '$_POST[fax]', '$_POST[mobile]')");

     

    $result = @mysql_query($sql,$connection) or die(mysql_error());

    }

     

    ?>

     

    any help much appreciated.

     

    cheers

  5. Thsi file open a dialog box to open the index page and not the mp3 file.

     

    ANy idea why this happen?

     

    <?php

    include '../functions/db_connect.inc.php';

     

    $sql = ("INSERT INTO downloadcount

      (downloadid) VALUES

      ('$_REQUEST[downloadreq]')");

     

    $result = @mysql_query($sql,$connection) or die(mysql_error());

     

    $filename = "milli.mp3";

    $myFile = "/httpdocs/djneils/mixes/milli.mp3";

     

    $mm_type="application/octet-stream";

     

    header("Cache-Control: public, must-revalidate");

    header("Pragma: hack"); // WTF? oh well, it works...

    header("Content-Type: " . $mm_type);

    header("Content-Length: " .(string)(filesize($myFile)) );

    header('Content-Disposition: attachment; filename="'.$filename.'"');

    header("Content-Transfer-Encoding: binary\n");

     

    readfile($myFile);

     

    ?>

  6. Ive tried this but it opens the index page and not the mp3?

     

    <?php

    include '../functions/db_connect.inc.php';

     

    $sql = ("INSERT INTO downloadcount

    (downloadid) VALUES

    ('$_REQUEST[downloadreq]')");

     

    $result = @mysql_query($sql,$connection) or die(mysql_error());

     

    $filename = "milli.mp3";

    $myFile = "/mixes/milli.mp3";

     

    $mm_type="application/octet-stream";

     

    header("Cache-Control: public, must-revalidate");

    header("Pragma: hack"); // WTF? oh well, it works...

    header("Content-Type: " . $mm_type);

    header("Content-Length: " .(string)(filesize($myFile)) );

    header('Content-Disposition: attachment; filename="'.$filename.'"');

    header("Content-Transfer-Encoding: binary\n");

     

    readfile($myFile);

     

    header('Location: ../index.php');

    ?>

     

     

     

  7. Hi

     

    Im making a real simple audio download page with a counter on each download, nothing fancy, just need the job done. So the basics are i want a user to click on a link, and i want it to update a database table, which works fine, but i want it to also open the audio file for download, which i cant get to work.

     

    The script im trying to run is as follows:

     

    <?php

    include '../functions/db_connect.inc.php';

     

    $sql = ("INSERT INTO downloadcount

    (downloadid) VALUES

    ('$_REQUEST[downloadreq]')");

     

    $result = @mysql_query($sql,$connection) or die(mysql_error());

     

    $handle = fopen("../mixes/milli.mp3", "r");  <----------this not working??????????

     

    header('Location: ../index.php');

    ?>

     

    Am i barking up the wrong tree? or can i get this to work? any help appreciated.

     

    cheers

     

    fro

     

     

  8. $sql = "UPDATE customers SET

    cust_firstname = '$_POST[cust_firstname]',

    cust_surname = '$_POST[cust_surname]',

    house_no = '$_POST[house_no]',

    streetname = '$_POST[streetname]',

    postcode = '$_POST[postcode]',

    emailaddress = '$_POST[emailaddress]',

    homephone = '$_POST[homephone]',

    mobphone = '$_POST[mobphone]',

    WHERE username = {$_SESSION['username']}";

     

    why am i being told i have an sql error near my where clause?

     

  9. session_start();

    if (isset($_POST['userid']) && isset($_POST['password'])) {

     

    $table_name = "customers";

     

    include '../functions/db_connect.inc.php';

     

    $userid = $_POST['userid'];

    $password = $_POST['password'];

     

      // check if the user id and password combination exist in database

      $sql = "SELECT * FROM customers WHERE userid= '$userid'";

     

      $result = mysql_query($sql)

                or die('Query failed. ' . mysql_error());

     

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

          // the user id and password match,

          // set the session

          $_SESSION['db_is_logged_in'] = true;

      $_SESSION[$userid] = $userid;

          // after login we move to the main page

          header('Location: ../pages/control_panel.php?$userid;');

          exit;

      } else {

          $errorMessage = 'Sorry, wrong user id / password';

      }

     

    }

     

    How do i get the posted value of userid to $_SESSION[$userid]?

     

    and then how do i call it in another script?

  10. Hi, im making a simple user control panel, the user logs in by entering a user id and passowrd, which is checked in a mysql database before displaying the control panel. from this control panel i want the user to be able to change user details, sign up for bits, register for stuff etc. When the user clicks on a link to update their user details for example, i want to use the user id originally posted with the login form.

     

    Is it possible to register a session variable from a posted field in a form, and then retreive and use throughout my control panel? Ive tried many different ways to pass variables but its not happening?

     

    Any ideas?

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