PHP - Shoutbox In Php Without Mysql Database

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

Hello playERZ'

So, I have a dinky contact form that validates and uses:
Code: [Select]
<form name="infoget" action="<?=$_SERVER['PHP_SELF']?>" method="post">
to validate the form via an external validate.php, then generate the email.

Finally:

// Send the mail using PHPs mail() function
mail($to, $subject, $body);
header("Location:customize_search.php");


So, this takes the user to a more advanced form after they 'send' the first form.

1 - Is it possible to have the second form autofill with the POSTed data and NOT use MySQL or database whatever?
2 - If so, how??

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

Hi

On my xampp server's database I get the warning that my database is read-only when I try to insert something into a table. I think I may know what cause this, because I created an export/import app that exports/import important data to the database...

How can I remove the read-only from the database, because I am scared that the same thing might happen to my server's database? And is there a mysql query that test if the database is read-only?

Thanks in advance

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

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.

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

I'm trying to use php code that is stored in the sql database, but It doesn't seem to be executing the code.  when I see the page source, its there but the server is not executing the command how do I accomplish this.

Here is a simple code snippet to show what I am trying to do.

$result = mysql_query("select * from data");
$row = mysql_fetch_array($result);
echo $row['code'];

In the code field in data table this is whats there.
<?php echo "testing."; ?>

For some reason Im not able to connect to mysql database, when i fill in the form and select search, it just basically refreshes the page but does not come up with no error messages or any results from my database. any help is appreciated.

HTML Code

Code: [Select]

<table id="tb1">
<tr>
<td><p class="LOC">Location:</p></td>
<td><div id="LC">
<form action="insert.php" method="post">
<select multiple="multiple" size="5" style="width: 150px;" >
<option>Armley</option>
<option>Chapel Allerton</option>
<option>Harehills</option>
<option>Headingley</option>
<option>Hyde Park</option>
<option>Moortown</option>
<option>Roundhay</option>
</select>
</form>
</div>
</td>

<td><p class="PT">Property type:</p></td>
<td><div id="PS">
<form action="insert.php" method="post">
<select name="property type" style="width: 170px;">
<option value="none" selected="selected">Any</option>
<option value="Houses">Houses</option>
<option value="Flats / Apartments">Flats / Apartments</option>
</select>
</form>
</div>
</td><td>
<div id="ptype">
<form action="insert.php" method="post">
<input type="radio" class="styled" name="ptype" value="forsale"/> For Sale

<p class="increase">
<input type="radio" class="styled" name="ptype" value="forrent"/> To Rent
</p>
<p class="increase">
<input type="radio" class="styled" name="ptype" value="any"/> Any
</p>
</form>
</div>
</td>
</tr>

</table>

<div id="table2">
<table id="NBtable">
<tr>
<td><p class="NBS">Number of bedrooms:</p></td>
<td><div id="NB">
<form action="insert.php" method="post">
<select name="number of bedrooms">
<option value="none" selected="selected">No Min</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select> to

<select name="number of bedrooms">
<option value="none" selected="selected">No Max</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
</form>
</div>
</td>

<td><p class="PR">Price range:</p></td>
<td><div id="PR">
<form action="insert.php" method="post">
<select name="price range">
<option value="none" selected="selected">No Min</option>
<option value="50,000">50,000</option>
<option value="60,000">60,000</option>
<option value="70,000">70,000</option>
<option value="80,000">80,000</option>
<option value="90,000">90,000</option>
<option value="100,000">100,000</option>
<option value="110,000">110,000</option>
<option value="120,000">120,000</option>
<option value="130,000">130,000</option>
<option value="140,000">140,000</option>
<option value="150,000">150,000</option>
<option value="160,000">160,000</option>
<option value="170,000">170,000</option>
<option value="180,000">180,000</option>
<option value="190,000">190,000</option>
<option value="200,000">200,000</option>
<option value="210,000">210,000</option>
<option value="220,000">220,000</option>
<option value="230,000">230,000</option>
<option value="240,000">240,000</option>
<option value="250,000">250,000</option>
<option value="260,000">260,000</option>
<option value="270,000">270,000</option>
<option value="280,000">280,000</option>
<option value="290,000">290,000</option>
<option value="300,000">300,000</option>
<option value="310,000">310,000</option>
<option value="320,000">320,000</option>
<option value="330,000">330,000</option>
<option value="340,000">340,000</option>
<option value="350,000">350,000</option>
</select> to

<select name="price range">
<option value="none" selected="selected">No Max</option>
<option value="50,000">50,000</option>
<option value="60,000">60,000</option>
<option value="70,000">70,000</option>
<option value="80,000">80,000</option>
<option value="90,000">90,000</option>
<option value="100,000">100,000</option>
<option value="110,000">110,000</option>
<option value="120,000">120,000</option>
<option value="130,000">130,000</option>
<option value="140,000">140,000</option>
<option value="150,000">150,000</option>
<option value="160,000">160,000</option>
<option value="170,000">170,000</option>
<option value="180,000">180,000</option>
<option value="190,000">190,000</option>
<option value="200,000">200,000</option>
<option value="210,000">210,000</option>
<option value="220,000">220,000</option>
<option value="230,000">230,000</option>
<option value="240,000">240,000</option>
<option value="250,000">250,000</option>
<option value="260,000">260,000</option>
<option value="270,000">270,000</option>
<option value="280,000">280,000</option>
<option value="290,000">290,000</option>
<option value="300,000">300,000</option>
<option value="310,000">310,000</option>
<option value="320,000">320,000</option>
<option value="330,000">330,000</option>
<option value="340,000">340,000</option>
<option value="350,000">350,000</option>

</select>
</form>
</div>
</td>


</tr>

</table>

PHP Code
Code: [Select]

<?php
$server = "";      // Enter your MYSQL server name/address between quotes
$username = "";    // Your MYSQL username between quotes
$password = "";    // Your MYSQL password between quotes
$database = "";    // Your MYSQL database between quotes

$con = mysql_connect($server, $username, $password);       // Connect to the database
if(!$con) { die('Could not connect: ' . mysql_error()); }  // If connection failed, stop and display error

mysql_select_db($database, $con);  // Select database to use

$result = mysql_query("SELECT * FROM tablename");  // Query database

while($row = mysql_fetch_array($result)) {     // Loop through results
  echo "<b>" . $row['id'] . "</b><br>\n";      // Where 'id' is the column/field title in the database
  echo $row['location'] . "<br>\n";            // Where 'location' is the column/field title in the database
  echo $row['property_type'] . "<br>\n";       // as above
  echo $row['number_of_bedrooms'] . "<br>\n";  // ..
  echo $row['purchase_type'] . "<br>\n";       // ..
  echo $row['price_range'] . "<br>\n";         // ..
}

mysql_close($con);  // Close the connection to the database after results, not before.
?>

To connect to my database I an entering mysql database details just at the first top lines server,username,password and database. is thats correct?

Thank You for your help in adavance.

Hi,

I cant connect to my Mysql database.
I get this problem:
Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'esolarch_databas'@'localhost' (using password: YES) in /home7/esolarch/public_html/new/storescripts/connect_to_mysql.php on line 21
could not connect to mysql

Code: [Select]
<?php  
/*  
1: "die()" will exit the script and show an error statement if something goes wrong with the "connect" or "select" functions. 
2: A "mysql_connect()" error usually means your username/password are wrong  
3: A "mysql_select_db()" error usually means the database does not exist. 
*/ 
// Place db host name. Sometimes "localhost" but  
// sometimes looks like this: >>      ???mysql??.someserver.net 
$db_host = "localhost"; 
// Place the username for the MySQL database here 
$db_username = "esolarch_database";  
// Place the password for the MySQL database here 
$db_pass = "Password";  
// Place the name for the MySQL database here 
$db_name = "esolarch_admin2"; 

// Run the actual connection here  
mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db("$db_name") or die ("no database");              
?>

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

I looked into this everywhere,but the code does not make sense, is not properly documented and is outdated.


Ive been trying to achieve this for the past hour.

Anybody know how I can do this?

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've tried reading through some of the threads but couldnt understand some of them.

I've made a newsfeed script which works how i want it to. Now i want to add the function to delete a row from the database from an "admin panel" on the website.

So far i have this:
<?php
include("includes.php");

doConnect();

$get_news = "SELECT id, title, text, DATE_FORMAT(datetime, '%e %b %Y at %T') AS datetime FROM newsfeed ORDER BY datetime DESC";

$result= mysqli_query($mysqli, $get_news)
or die(mysqli_error($mysqli));

while ($row = mysqli_fetch_array($result)) {
echo '<strong><font size="3">'. $row['title'] .' </font></strong><br/><font size="3">'. $row['text'] .'</font><br/><font size="2">'. $row['datetime'] .'</font><br/><br/><a href="delnews.php?del_id=' .$row['id']. '">
<strong>DELETE</strong></a>';}
?>
then my delnews.php is:
<?php
include("includes.php");

doConnect();

$query = "DELETE FROM newsfeed WHERE id = "$_POST['id']""; 

$result = mysql_query($query); 

echo "The data has been deleted."; 

?>
I believe the problem is $_POST['id']. i've tried different things in there but none work. It displays the echo line but doesnt actually delete anything.

I am new to php so this may be a stupid mistake, but try and play nice!

Thanks

I know this is a very simple and probably stupid question - but what is a patch? i've tried searching online for an explanation but I just find articles on the 'best practice' and it doesn't break it down into what it actually is!   I've just been looking into new hosting and they offer a managed service, which includes database patching.   Could someone please enlighten me?
Edited by paddyfields, 10 June 2014 - 04:49 AM.