PHP - Help With Php/mysql Database Function

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

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

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 all,   I haven't worked with PHP and MySQL very much, but have been working with HTML and JavaScript for years. This time, I want to build a simple website that allows users to query a database of coins. I just want to make sure that I approach this in the right direction. I'd like to describe what I'm trying to do and hopefully someone can prevent me from becoming a failure.   For the database, each coin has an ID and row, the first column of which is the date, and the following columns attributes of that particular coin. Several entries can exist with the same date.   Anyway, there will basically be three types of pages:   1. A query page from which users will click a date or enter a term in a search box.   2. A query result page displaying a list of search results, like '1909' or 'Lincoln'. Each result will be linked to an individual page for that coin.   3. An individual page. This template page will have an html table whose cells are each defined by the values for that coin in the database.      So for instance, the user searches or clicks on '1909' on the query page and sees 10 records listed in the database on the result page. The user can then click one of those 10 records and view a new page that displays more information on that particular coin.    

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.

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");              
?>

I'm having problems updating my database, I have 4 fields i want to change. I checked all the { on the page, that's not the problem, I tried to echo information from the database and it displayed my information so that's not the problem, i tried yelling at my computer, that didn't work, i tried to input data into the database with the insert function it worked but is not practical in my situation. I'm probably going to face palm when i find out whats wrong, help please 

btw, the $_SESSION['usr'] was set in another page and works.


Code: [Select]

<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Edit Info</title>
   
    <link rel="stylesheet" type="text/css" href="demo.css" media="screen" />
   
</head>

<body>

<div id="main">


<div class="container">
<font size="5" face="sans-serif">Change Settings <?php echo "{$_SESSION['usr']}"; ?></font>
<form action="" method="POST">
<table cellpadding="3" cellspacinf="4" border="0">

<tr>
<td>Name</td>
<td><input type="text" name="name" /></td>
</tr>
<tr>
<td>Age</td>
<td><input type="text" name="age" /></td>
</tr>
<tr>
<td>Gender</td>
<td><input type="text" name="mf" /></td>
</tr>
<tr>
<td>Location</td>
<td><input type="text" name="loc" /></td>
</tr>
<tr>
<td><input type="submit" name="submit" value="submit" /></td>
</tr>
</table>
</form>



<?php
if ($_POST['submit']){
define('INCLUDE_CHECK',true);
require 'connect.php';

$usr = $_SESSION['usr'];

$sql = 
mysql_query("UPDATE members 
SET name='{$_POST['name']}', age='{$_POST['age']}, mf='{$_POST['mf']}', loc='{$_POST['loc']}' 
WHERE usr='{$_SESSION['usr']}'");

if($sql){
echo 'Changes Saved!';

}else{
echo 'Error';
} 
}

?>
</div>
</div>
</body>
</html>

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.

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

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?

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

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);

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

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