PHP - Crashed Os, How To Transfer Mysql Database

There is a "PHP ajax cascading dropdown using MySql" at codestips.com/php-ajax-cascading-dropdown-using-mysql/

I want to use this technique but with a XML or array file instead of mysql database, but my knowledge about mysql is very low. How I can modify this code to catch the categories and products from an array, instead of mysql database?
Code: [Select]
$connect=mysql_connect($server, $db_user, $db_pass)
      or die ("Mysql connecting error");
echo '<table align="center"><tr><td><center><form method="post" action="">Category:<select name="category" onChange="CategoryGrab('."'".'ajaxcalling.php?idCat='."'".'+this.value);">';
$result = mysql_db_query($database, "SELECT * FROM Categories");
$nr=0;
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
 $nr++;
 echo "<option value=".'"'.$row['ID'].'" >'.$row['Name']."</option>";
}
echo '</select>'."\n";
echo '<div id="details">Details:<select name="details" width="100" >';
$result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=1");
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option value=".$row['ID'].">".$row['Name']."</option>";
}
echo '</select></div>';
echo '</form></td></tr></table>';
mysql_close($connect);
ajaxcalling.php is
Code: [Select]
include("config.php");
$ID=$_REQUEST['idCat'];
$connect=mysql_connect($server, $db_user, $db_pass);
echo 'Details:<select name="details" width="100">';
$result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=".$ID);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "<option value=".$row['ID'].">".$row['Name']."</option>";
}
echo '</select>';
mysql_close($connect);

I'm trying to create a page that allows people to backup their database on a web page but I'm having trouble with the ending of it.

<?php

require("../include/config.php");

    $tables = array();
    $qTables = mysql_query("SHOW TABLES");
    while($row = mysql_fetch_row($qTables))
    {
      $tables[] = $row[0];
    }

   foreach($tables as $tab1)
  {
 
    $return.= "DROP TABLE IF EXISTS `".$tab1."`;";
    $row2 = mysql_fetch_row(mysql_query("SHOW CREATE TABLE `" . $tab1 . "`"));
    $return.= "\n\n".$row2[1].";\n\n";
   
   
    $result = mysql_query("SELECT * FROM ".$tab1) or die(mysql_error());
    $num_fields = mysql_num_fields($result);
   
       $return.= "INSERT INTO `".$tab1."`";
      $col = mysql_query('SELECT * FROM '.$tab1);
      $a = 0;
      $return.= " (";
      while ($a < mysql_num_fields($col)) {
         $meta = mysql_fetch_field($col, $a);

         $return.= "`" . "$meta->name";
         $a++;
         if ($a < mysql_num_fields($col)) {
         $return.= "`,";
         } else {
         $return.= "`";
         }
      }
      $return.= ")";
      $return.=" VALUES\n(";
   
      for ($i = 0; $i < $num_fields; $i++){
      while($row = mysql_fetch_row($result)){

        for($j=0; $j<$num_fields; $j++){
          if (isset($row[$j])) {
        $return.= "'".$row[$j]."'" ;
        } else {
        $return.= "''";
        }
       
          if ($j < ($num_fields-1)) {
        $return.= ",";
        }
       
      if (j < ($num_fields-1)){
      $return.= "),\n(";
      } else {
      $return.= ");\n";
      }
 
      }
      }       
      }
      }
 
      $handle = fopen("db-backup-".time()."-".(md5(implode(",",$tables))).".sql","w+");
      fwrite($handle,$return);

  ?>

At this part:

      if (j < ($num_fields-1)){
      $return.= "),\n(";
      } else {
      $return.= ");\n";
      }


If it has finished going through all of the values it will put ); at the end and if it hasn't it will put ), and then continue with the next one. The problem I'm having is it's only doing the ),

Can someone help me please?

Im trying to connect to a database from php. Heres the code:
<?php
  $dbc = mysqli_connect('192.168.0.122', 'boyyo', 'KiaNNa11', 'aliendatabase')
    or die('Error connecting to MySQL server.');
   
   
  $query = "INSERT INTO aliens_abduction (first_name, last_name, " .
    "when_it_happened, how_long, how_many, alien_description, " .
   "what_they_did, fang_spotted, other, email) " .
   "VALUES ('Sally', 'Jones', '3 days ago', '1 day', 'four', " .
   "green with six tentacles', 'We just talked and played with a dog', " .
   "'yes', 'I may have seen your dog. Contact me.', " .
   "'sally@gregs-list.net')";
   
   $result = mysqli_query($dbc, $query)
   or die('Error querying database.');
   ?>

I think i have a theory that my MySQL server location is wrong but i dont know. I use HostGator to do this and im using Phpmyadmin. But everytime i type in a form that i created it says Error querying database. Can someone tell me whats wrong with this code. Oh by the way im using head first into PHP and MySQL

Hello guys. Trying to connect php with mysql database and then display results on the screen. This is my code:

Code: [Select]
<?php

$dbhost = "localhost";
$dbuser = "username1";
$dbpass = "password1";
$db = "username1_myDB";

$connection = mysql_connect($dbhost, $dbuser, $dbpass)
or die ("Could not connect");
mysql_select_db($connection, $db);

$show = "SELECT Name, Description FROM people";
$result = mysql_query($show);


while($show = mysql_fetch_array($result)){
$field01 = $show[Name];
$field02 = $show[Description];
echo "id: $field01<br>";
echo "description: $field02<p>";
}


?>
However im getting this:
Warning: mysql_select_db() expects parameter 1 to be string, resource given in /home/pain33/public_html/index.php on line 20

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/pain33/public_html/index.php on line 26

Any ideas how to fix this? Thank you.

Hi,

I want to make a Shoutbox in Php without using a MySQL-database. This is the code I've got:
Code: [Select]
<?php
$nickname = strip_tags($_POST['nickname']);
$message = strip_tags($_POST['message']);
?>



<html>
<body>
<<<HERE<b><h3><center>Shout Box</center></h3></b>
<iframe src="chat.txt" width="700" height="250"></iframe>
<form method="post" name="form">
Nickname: <input type="text" name="nickname" value="<?php echo $nickname;?>"><br>
Message: <input type="text" name="message" value="<?php echo $message;?>">
<input type="submit" value="Submit">
</form>
<a href="#">Refresh</a>


<?php

if ($_POST['nickname'] && $_POST['message']){
$fp = fopen("chat.txt", "a");
$stuff = $nickname . ": " . $message . "n";
fputs($fp, $stuff);
} else {
echo "Please enter nickname and message.";
}

?>

</body>
</html>
Problem is: it isn't working. :-P
It does read things from chat.txt, but I cant get the input-box working (let people add stuff to chat.txt).
What do I do wrong?

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

Ok, I got someone to help me fix this but he had no idea what the error was...

I have 2 tables, one called points and the other called members.

In members i have got:
id
name

In points i have got:
id
memberid
promo

I have the following code:
Code: [Select]
<?php
$con = mysql_connect("localhost","slay2day_User","slay2day");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("slay2day_database",$con);
$sqlquery="SELECT Sum(points.promo) AS score, members.name, members.id = points.memberid Order By members.name ASC";
$result=mysql_query($sqlquery,$con);
while ($row = mysql_fetch_array($result))
{

//get data
$id = $row['id'];
$name = $row['name'];
$score = $row['score'];

echo "<b>Name:</b> $name<br />";
echo "<b>Points: </b> $score<br />" ;
echo "<b>Rank: </b>";
if  ($name == 'Kcroto1'):
  echo 'The Awesome Leader';
else:
if ($points >= '50'):
  echo 'General';
elseif ($points >= '20'):
  echo 'Captain!';
elseif ($points >= '10'):
  echo 'lieutenant';
elseif ($points >= '5'):
  echo 'Sergeant';
elseif ($points >= '2'):
  echo 'Corporal';
else:
  echo 'Recruit';
endif;
endif;
echo '<br /><br />';

}

?>
I am getting the following error when i do the query in mysql:
Code: [Select]
#1109 - Unknown table 'points' in field list
And when i open the webpage i get the following error:
Code: [Select]
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/slay2day/public_html/points/members.php on line 18
Please Help me?

Okay, I have been following this tutorial: http://www.freewebmasterhelp.com/tutorials/phpmysql/1 to achieve exactly what I wanted to get done. I also followed the way to have it formatted in tables and added extra columns that I needed, included an "Options" column which houses three links, including "edit" and "delete"

Now, I have everything working fine but I am stumped on how to get the "edit" and "delete" links to work for each individual entry that is listed. I have to have these features so the entries can be edited and deleted without having to physically go into the MySQL database to do it. The tutorial explains how to do it in Step 6, but I am confused. I'm not quite sure where to place the code for the links, which are generated automatically every time a new entry is inputted into the database.

Anybody available to help me out? Thanks!

Hi,

I wonder if you could help me try to find what i'm looking for, i have a problem with bogus users on my site.

I created a register page where the details are sent to the database, this is fine but someone is registering with the username: 1 and password: 1 multiple times.
I have about 50 of these now and i would like to know what to actually search for (how to word it) to find out how to stop this?

What would be the name of the script? i've looked for fake username script, multiple username/password prevention script, i'm just not getting it, sorry.

If any of you have any ideas i'd like to hear from you, many thanks in advance for that.

If you need anymore to go on please ask, once again thank you.

Can you upload videos to a mysql database? If so how would you go about doing it?

Hello,
I've been having trouble connecting to a MySQL database, I can't find the problem in the code, what am I doing wrong?

Getting the database file in the config file : require_once("db_connect.php");

db_connect.php :
<?php
$db = mysql_connect('83.172.155.14:3306', 'username', 'password') or die(mysql_error());
mysql_select_db('databasename', $db) or die(mysql_error());
?>

I need to connect to a PhpMyAdmin database.
I need this fixed asap since I'm doing this for someone and he wants the site done as quickly as possible.
P.S: The database used to work in php4 and now I need it to work on php5
Thanks in advance,

Hey guys,

I'm working a project that requires sessions be stored within the database, as the project I'm working on is on a shared host. But I'm having a problem with getting the data of a session in the database, the other fields like session_id, session_updated, session_created are working fine. I think I've got a bug in my code, but I just can't detect it (frustrating).

Database connection

class db extends mysqli {
private $host;
private $user;
private $pass;
private $db;

function __construct( $host='localhost', $user='user', $pass='pass', $db='website' ) {
$this -> host = $host;
$this -> user = $user;
$this -> pass = $pass;
$this -> db = $db;

parent::connect( $host, $user, $pass, $db );

if( mysqli_connect_error( ) ) {
die( 'Connection error ('.mysqli_connect_errno(  ).'): '.mysqli_connect_error(  ) );
}

}

function __destruct(  ) {
$this -> close(  );
}
}


Session handler

class sessionHandler {
private $database;
private $dirName;
private $sessTable;
private $fieldArray;

    function sessionHandler() {
        // save directory name of current script
$this -> database = new db;
        $this -> dirName = dirname(__file__);
        $this -> sessTable = 'sessions';
    }

    function open( $save_path, $session_name ) {
return TRUE;
    }

    function close() {
//close the session.
        if ( !empty( $this -> fieldarray ) ) {
            // perform garbage collection
            $result = $this->gc( ini_get ( 'session.gc_maxlifetime' ) );
            return $result;
        }
        return TRUE;
    }

    function read( $session_id ) {
$sql = "
SELECT *
FROM sessions
WHERE session_id=( '$session_id' )
LIMIT 1
";
$result = $this -> database -> query( $sql );
if( $result -> num_rows > 0  ) {
$data = $result -> fetch_array( MYSQLI_ASSOC );
$this -> fieldArray = $data;
$result -> close();
return $data;
}
return "";
    }

    function write( $session_id, $session_data ) {
//write session data to the database.
        if ( !empty( $this -> fieldArray ) ) {
            if ( $this -> fieldArray['session_id'] != $session_id ) {
                // user is starting a new session with previous data
                $this -> fieldArray = array();
            }
        }

$this -> fieldArray['session_id'] = $session_id;
$this -> fieldArray['session_data'] = $session_data;
$this -> fieldArray['session_updated'] = time();
$this -> fieldArray['session_created'] = time();

$session_id = $this -> database -> escape_string( $session_id );
$session_data = $this -> database -> escape_string( $session_data );
$session_updated = time();
$session_created = time();

$sql = "
INSERT INTO sessions ( session_id, session_data, session_updated, session_created )
VALUES ( '$session_id', '$session_data', '$session_updated', '$session_created' )
";
if( $this -> database -> query( $sql ) !== TRUE ) {
return FALSE;
}
return TRUE;
    }

    function destroy( $session_id ) {
$sql = "
DELETE FROM sessions
WHERE session_id=('$session_id')
";
if( $this -> database -> query( $sql ) !== TRUE ) {
return FALSE;
}
return TRUE;
    }

    function gc( $max_lifetime ) {
return TRUE;
    }

    function __destruct() {
//ensure session data is written out before classes are destroyed
//(see http://bugs.php.net/bug.php?id=33772 for details)
@session_write_close();
    }
}


The call

$session_class = new sessionHandler;
session_set_save_handler(
array( &$session_class, 'open' ),
array( &$session_class, 'close' ),
array( &$session_class, 'read' ),
array( &$session_class, 'write' ),
array( &$session_class, 'destroy' ),
array( &$session_class, 'gc' )
);
if( !session_start() ) {
exit();
}


Any help at all would be appreciated.

Kind Regards
Mike

Hi All,

Whenever I try to update any piece of PHP code to update a MySQL database, nothing happens. I have tried copying in some of the working codes of a website and tried the same, but no success.

I recently installed XAMPP. I am connecting using the correct user id and pass to the database. The scripts are not giving me any error, but just not connecting, that's all.

While making such a usage as noted below

<FORM name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >


I get the following error

Firefox can't find the file at /C:/xampp/htdocs/="<?php..  so on

Why does this happen?

I am pretty new to this, so please do help.

Thanks,
Satheesh P R

Hello all, first post here so i hope i'm doing this right and am putting this into the right place.
Im in the process of integrating a blog into my website, mostly it's set up however i'm currently working on the code to update the posts (theres only 2 parts a title and a comments[which is in actual fact the post content itself]). The code succsefully completes without any errors but for some reason it does not actually update the mysql database.. the code i'm using is as follows:

Code: [Select]
<?php
require_once('header.php');
include "../blog/blogconfig.php";

if(isset($_POST['submit'])){
$update="UPDATE eq_blogarticle SET title='".$_POST['title']."',comments='".$_POST['comment']."' WHERE artid='$aid'";
if(!mysql_query($update)){
echo mysql_error();
}else{
header("location:blog.php?action=listmsgs");
exit;
}
}
?>
<?php
// get value of aid that sent from address bar by blog.php?action=listmsgs
$aid=$_GET['aid'];

// Retrieve data from database
$sql="SELECT * FROM eq_blogarticle WHERE artid='$aid'";
$result=mysql_query($sql);

$row=mysql_fetch_array($result);
?>

<form name="form2" method="post" action="update.php">
<script type="text/javascript">var SITE_URL="<?php echo SITE_URL;?>";</script>
<script type="text/javascript" src="<?php echo SITE_URL;?>/includes/js/nicEdit.js"></script>
<script type="text/javascript">
//<![CDATA[
bkLib.onDomLoaded(function() { nicEditors.allTextAreas() });
//]]>
</script>
<body>
<table width="100%" border="0" cellspacing="1">
  <tr>
    <td colspan="2" class="temptitle">Equidisc Blog</td>
  </tr>
  <tr>
    <td width="74%" valign="top">
<table>
Edit Blog Post
<br>
<br>
Title:
<br>
<input name="title" type="text" class="input" id="title" value="<? echo $row['title']; ?>">
<br>
<br>
<span class="style1 style2 style3">Blog Post:</span>
<br>
<br>
<textarea name="comments" cols="55" rows="12" class="input" id="comments"><? echo stripslashes($row['comments']); ?></textarea>
<br>
<br>
<input name="aid" type="hidden" id="aid" value="<? echo $row['artid']; ?>">
<input type="submit" name="submit" value="Submit">
</form>
</table>
</td>
    <td width="26%" valign="top"><table width="100%" border="0" cellspacing="1">
      <tr>
        <td colspan="2"><img src="../blog/images/fb.gif" width="16" height="16" /> <strong>Blog Menu</strong></td>
        </tr>
       <tr>
        <td><a href="blog.php">Home</a></td>
       </tr>
      <tr>
        <td><a href="blog.php?action=newblogpost">New Post </a></td>
        </tr>
      <tr>
        <td><a href="blog.php?action=listmsgs">Manage Posts </a></td>
        </tr>
    </table></td>
  </tr>
</table>
</body>
<?php
require_once('footer.php');
?>
Explanation: header/footer.php obvious
../blog/blogconfig.php holds my mysql connection settings and connects to the sql, this is the same config as is used for creating the new posts which i have no issues with so i dont think the issue lays there. If i run the query on phpmyadmin with dummy data it works fine and updates the entry..
Any help would be very much appreciated as i'm at the end of my tether with this!.
Thanks in advance.
Jo

Hi there,
I have been creating a website which shows products of the companys (www.theadventurestartshere.org) and I have been trying to make some filters for the products using URL Parameters and recordsheet filters...

Can someone advise me on the easiest way to do this? As I have created a method to do it with, (check website) but it has to be entered manually and I was hoping there is an easier way?

Please Note I like to use drop down boxes for the filters but if there is a way to do the checkbox style you see on say Amazon then that would be brilliant...

I use Dreamweaver CS5, and the newest versions of PHP and MySql

Many Thanks,
Paul