PHP - Making Php Check A Mysql Value And Then Acting Upon It

When I move threads on my forum, I want there to be a old thread in the section it was moved from saying "Moved: new link". To do this, I'm inserting a new thread in the database with the value moved equal to the new location, so when threads are being displayed users can see the thread in the new section.

Is there a quick function in MySQL that will allow me to create a duplicate row in MySQL, and then just run a quick update statement?

I don't want to have to insert a whole new thread like this (unless it's the only way):


//insert new thread
mysql_query("INSERT INTO `threads` VALUES (null, '$id','$title', '$content', '". getUsername() ."', NOW(), '". qfc() ."', NOW(), '". getUsername() ."', '','{$_SERVER['REMOTE_ADDR']}', '0', '0', '0', '', '0')");

MySQL Version: 5.5.20 (<-- This is my WAMPServer version to test my program, my hosting has 5.1 though)
PHP Version: 5.3.10 (My Hosting says its PHP 5 but not sure what version, i know its not the newest)

What im trying to do is have my site have a page like "www.mysite.com/data.php?user=Nicholas&country=USA"

Thing is, i dont know what to add in my data.php source code to make it work like that. What im trying to achieve is for my users to sign up for my page but I DO NOT want to add my MySQL user and pass in the page where it can be cracked and hacked. When the user goes to the page, a script will automatically put the user and country and then the page will automatically add the user and his country to the MySQL table on my database.

Im also making a program that will work like this and that will have a higher chances of being hacked if i add my MySQL info into the program in order to add data.

Can someone please provide me of a sample source i can try? I will not only be adding user and country but i want to test this out.

Thanks!

Hi Everyone.

I get the following data from a result from database search

<?php echo $rsjobs['CompanyURL'];?>  It prints out the url but when I try to turn it into a clickable link it does not work

Here is the coding I have so far.


<a href="<?php echo $rsjobs['CompanyURL'];?>"><?php echo $rsjobs['CompanyURL'];?></a>


 What am I doing wrong ? please help.

I am trying to check if email already in DB. This is what I have so far and does not work.
Code: [Select]
if($email){
// checking user input
$check_email=mysql_query("SELECT * FROM userlist WHERE email ='".$email."' ");
$get_email=mysql_fetch_array($check_email);
$existsmail=$get_email['email'];

  if($email == $existsmail){
$resultado = "Email address already exists!";
}
  if($email != $existsmail){
    $resultado = "Your Account was created successfully! You may now login.";
$date=getdate();
$insert=mysql_query("INSERT INTO userlist VALUES ('', '$name', '$phone', '$fax', '$company_name', '$billing_address', '$billing_city', '$billing_state', '$billing_zip_code', '$email', AES_ENCRYPT('$mysecretkey''password') ,now())");
  }else{
    $resultado = "";
}
 echo $resultado;
 }
Thanks in advance.

Hi guys,

I need your help. I am checking on a database as I want to see if I have the same value in the url and in the database.


Code: [Select]
<?php
session_start();
    define('DB_HOST', 'localhost');
    define('DB_USER', 'mydbuser');
    define('DB_PASSWORD', 'mydbpass');
    define('DB_DATABASE', 'mydbtable');
       
    $errmsg_arr = array();
    $errflag = false;

    $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
    if(!$link) {
  die('Failed to connect to server: ' . mysql_error());
    }

    $db = mysql_select_db(DB_DATABASE);
    if(!$db) {
  die("Unable to select database");
    }

   function clean($var){

return mysql_real_escape_string(strip_tags($var));
    }
  
    $username = clean($_GET['user']);
    $password = clean($_GET['pass']);
    $test = clean($_GET['test']);
    $public = clean($_GET['public']);
   
   
   if (isset($_GET['user']) && (isset($_GET['pass']))) {
    if($username == '' || $password == '') {
  $errmsg_arr[] = 'username or password are missing';
  $errflag = true;
    }
} elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) {
    if($username == '' || $test == '' || $public == '') {
  $errmsg_arr[] = 'user or others are missing';
  $errflag = true;
    }
  }

    if($errflag) {
  $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
  echo implode('<br />',$errmsg_arr);
   }
   else {
  $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'";
  $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error());


if ($username && $password) {
  if(mysql_num_rows($result) > 0) {
     $qrytable1="SELECT images, id, test, links, Public FROM user_list WHERE username='$username'";
    $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());

    while ($row = mysql_fetch_array($result1)) {
        echo "<p id='test'>";
        echo $row['test'] . "</p>";
        echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' .  $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>';
     }
   } else {
    echo "user not found";
  }
} elseif($username && $test && $public) {
    $qry="SELECT * FROM members WHERE username='$username'";
    $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error());
    
    if(mysql_num_rows($result1) > 0) {     
      $qrytable1="SELECT Public FROM user_list WHERE username='$username' && test='$test'";
      $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error());
      
      if(mysql_num_rows($result2) > 0) {
        $row = mysql_fetch_row($result2);
        mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'");
        echo "update!";
      } else {
        echo "already updated!";
      }
    } else {
      echo "user not found";
    }
  }
} 
?>
This is the function I use to check the value in the database:

Code: [Select]
if (mysql_affected_rows($result2) > 0) {
         mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'");
         echo "you have update it!";
       } else if (mysql_affected_rows($result2) < 0) {
         echo "it is not on the database";
       } else {
         echo "you have already updated!";
       }

When i input the different value in a url bar while the records are not the same as the value in the url and in the database, i can't get passed and I am keep getting "you have already updated!!" when the value in a database are different than I have input in a url.

Do you know how i can get pass it when I have input the different value in the url while it is not the same in the database?

Any advice would be much appreicated.

Thanks,
Mark