PHP - Load More From Database

hello everyone,
I am about to start coding my pages to display results from a database but before i do i want to know information about the following :

Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable?

Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page?

Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two .
Thanks in advance.

I have a very tricky php problem here. At least for me.
On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered.

So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out.

Description of what I'm trying to do:
I'm generating the the boxes with a call to my database for the list of checkboxes.
Then I'm building an array with all the possible items that could be checked to compare against.
 
Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient"

       Code: [Select]
<div class="text">Job Type&nbsp;(<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div>
<div class="field">

    <?PHP

    $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database.");
    $db=mysql_select_db ("database", $connection) or die (mysql_error());
    $query = "SELECT type FROM typelist ORDER BY type ASC";
    $sql_result = mysql_query($query, $connection) or die (mysql_error());
    $i=1;
    echo '<table valign="top"><tr><td>';
    while ($row = mysql_fetch_array($sql_result)) {
    $type = $row["type"];
        if ($i > 1 && $i % 26 == 0)
        echo '</td><td>';
        else if ($i)
        echo '';
        ++$i;
       
        $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)");
       
        $dbTypes = explode(',',$row['type']);
       
        foreach ($aTypes as $type){
       
        if(in_array($aType,$dbTypes)){
       
      echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>";
       
        }else
       
        {
       
            echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>";  
       
               
        }

        }

    }
      echo '</td></tr></table>';
   
    ?>

</div>

after cloasing connection of database i still got the values form database.

Code: [Select]
<?php
session_start();
/* 
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
require_once '../database/db_connecting.php';
$dbname="sahansevena";//set database name
 $con=  setConnections();//make connections use implemented methode in db_connectiong.php
 mysql_select_db($dbname, $con);
//update the time and date of the admin table
 $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4";
 //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time
 $link= mysql_query($update_time);
 // mysql_select_db($dbname, $link);
//$con=mysql_connect('localhost', 'root','ijts');

       $result="select * from admin where username='a'";

        $result=mysql_query($result);
mysql_close($con);
//here i just check after closing data baseconnection whether i do get reselts but i do, why?
echo "after the cnnection was closed";
if(!$result){
echo "cont fetch data";

}else{

   $row= mysql_fetch_array($result);
   echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4];
}
//  echo "<html>";
//echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>";
//      echo "<thead>";
//           echo"<tr>";
//                   echo "<th>";
//                   echo ID;
//                   echo"</th>";
//                   echo" <th>";echo Username; echo"</th>";
//                    echo"<th>";echo Password; echo"</th>";
//                    echo"<th>";echo Last_logged_date; echo "</th>";
//                   echo "<th>";echo Last_logged_time;  echo "</th>";
//               echo" </tr>";
//           echo" </thead>";
//           echo" <tbody>";
//while($row= mysql_fetch_array($result,MYSQL_BOTH)){
//             echo "<tr>";
//              echo "<td>";
//              echo $row[0];
//               echo "</td>";
//                   echo "<td>";
//              echo $row[1];
//              echo "</td>";
//                  echo "<td>";
//              echo $row[2];
//               echo "</td>";
//                   echo "<td>";
//                echo $row[3];
//              echo "</td>";
//                  echo "<td>";
//              echo $row[4];
//               echo "</td>";
//            echo "</tr>";
//     }
//         echo" </tbody>";
//         echo "</table>";
//         echo "</html>";
     session_destroy();
session_commit();

echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin'];
?>

session is destroyed but database connection is not closed.

thanks

At the moment  I am creating a search function for my website.
The approach I have in mind is a pseudo-PHP database. To give an example:

A HTML form will submit the results to a PHP file.

HTML FORM - Colour: Black
PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");}
HTML FORM - Price: <$50
PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");}

The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP.

I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.   my form page test.php

<?PHP
require_once("./include/membersite_config.php");

if(!$fgmembersite->CheckLogin())
{
    $fgmembersite->RedirectToURL("login.php");
    exit;
}

?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?>  "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?>  "><br><br>

<button>Submit</button>

my editform.php

<?php

$con = mysqli_connect("localhost","root","user","pass");

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");



header("Location: test.php");

?>

Hi guys,

I was just wondering if anyone could help me. I've got a My_SQL database containing articles, a summary for the article and a date. I have a basic CMS system set-up, but I want to create a script that when users sign up to a mail list it forwards the summary and dates of the articles database. If that makes sense? But I only want it to forward the most recent 5 rows. I'm pretty new to PHP and I've been mostly following tutorials thus far, but this is quite specific.

Thanks in advance!

hello
I want query from one table and insert in another table on another domain .
each database on one domain name.
for example http://www.site.com $con1
and http://www.site1.com $con.

can anyone help me?
my code is :
<?php



$dbuser1 = "insert in this database";
$dbpass1 = "insert in this database";
$dbhost1 = "localhost";
$dbname1 = "insert in this database";

// Connecting, selecting database

$con1 = mysql_connect($dbhost1, $dbuser1, $dbpass1)
   or die('Could not connect: ' . mysql_error());



mysql_select_db($dbname1) or die('Could not select database');


$dbuser = "query from this database";
$dbpass = "query from this database";
$dbhost = "localhost";
$dbname = "query from this database";

// Connecting, selecting database

$con = mysql_connect($dbhost, $dbuser, $dbpass)
   or die('Could not connect: ' . mysql_error());

mysql_select_db($dbname) or die('Could not select database');

//query from database
$query = mysql_query("SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1");
    while($row=mysql_fetch_array($query)){
$result=$row[0];
$text=$row[1]."</br>Size:(".$row[4].")";
$alias=$row[2];
$link = '<a target="_blank" href='.$row[3].'>Download</a>';
echo $result;
}  
//insert into  database
mysql_query("SET NAMES 'utf8'", $con1); 
$query3= " INSERT INTO `jos_content`
 (`id`,
  `title`,
   `alias`,
    `) VALUES 
                        (NULL, 
                        '".$result."', 
                        '".$alias."', 
                       '')";
if (!mysql_query($query3,$con1))
  {
  die('Error: text add' . mysql_error());
  }


            mysql_close($con);
                        mysql_close($con1);
             


?>

the second database found on the cloud

 

i try to get JSON data but how to insert and update  them to another online database with the same table

my php script to return json data 

<?php        

 include_once('db.php');

        $users = array();

        $users_data = $db -> prepare('SELECT id, username FROM  users');

        $users_data -> execute();

        while($fetched = $users_data->fetch()) {

            $users[$fetchedt['id']] = array (

                'id' => $fetched['id'],

                'username' => $fetched['name']

            );

        }

        echo json_encode($leaders);

 

 

 

 

i get

 

{"1":{"id":1,"username":"jeremia"},"2":{"id":2,"username":"Ernest"}}

 

Edited March 24 by mahenda

<?php

$URL1 = file_get_contents('http://www.yahoo.com');
$URL2 = file_get_contents('http://www.google.com');
?>

How would I make PHP load URL1 and URL2 at the same time..

I would add more than 2 urls also, Im tyring to test something with my website, load up to 50 links at once, how would I do this?

Thanks

Hello frnds,
Im new in ur forum....I want to know that my one page is taking too much time to load....in that page no images exists..but on that page around 6000 records are loding from database....I m confuse that may be such a large records it is taking too much time...so will u help me out that what can i do so the page can be loaded quickly with these records....
this is urgent for me.....pls pls pls....give me solution soon
Thnx a lot in advance

hello,

i have a page that parses a string from espn and i have a style sheet that makes it refresh every 2 minutes.  the thing that is buggin me is that when i load the page it loades with the last cached data so i have to wait 2 minutes for the info to update.

is there any way to force the current content on load, and keep my refresh every 2 minutes still?

thanks

I'm pretty new to php.  But the below script should get news articles from google news rss.  It is failing to load the simplexml and returns false.  Any idea why?

Code: [Select]
$url = "http://news.google.com/news?geo=90210";
if ($xml = @simplexml_load_file($url))
{
for ($i=0; $i<4; $i++)
{
$item= $xml->channel->item[$i];

$link= '<a href = "'.(string) $item->link.'", target= "_blank">'.(string) substr($item->title, 0, 50).'...</a><br />';
$news[]= $link;
}
}

Hi, so....i have these 2 pages, but they don't show up in browser....can't find the error so will ask your help guys

index.php
<?php
session_start();
if(!isset($_SESSION["manager"])){
header("Location: admin_login.php");
exit();
}
//Be sure to check that manager SESSION value is in database
$managerID = preg_replace('#[^0-9]#i',",$_SESSION["id"]); // Filter everything but numbers and letters
$manager = preg_replace('#[^A-Za-z0-9]#i',",$_SESSION["manager"]); //Filter everything but numbers and letters
$password = preg_replace('#[^A-Za-z0-9]#i',",$_SESSION["password"]); //Filter everything but numbers and letters
//Run MySQL query to be sure that this person is an admin and that their password session var equals the database
//Connect to the MySQL Database
include("../storescripts/connect_to_mysql.php");
$sql = mysql_query("SELECT * FROM admin WHERE id='$managerID AND username='$manager' AND password='$password' LIMIT 1"); //search user
$existsCount = mysql_num_rows($sql); //count the row numbers
if ($existsCount == 0){ //evaluate the count
header("Location: ../index.php");
exit();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>MyOnlineStore | Admin Area</title>
<link rel="stylesheet" href="../style/style.css" type="text/css" media="screen"/>
</head>

<body>
<div align="center" id="mainWrapper">
  <?php include_once("../template_header.php");?>
  <p><br  />
    <br />
  </p>
  <p>&nbsp;</p>
  <p><br />
  </p>
  <div id="pageContent">
    <div id="admin" align="left">
      <h3>Hello what you want to do?</h3>
      <p><a href="#">Manage Inventory</a><br />
      <a href="#">Manage afggagdfag</a></p>
</div>
</div>
<?php include_once("../template_footer.php");?>
</div>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p>&nbsp;</p>
</body>
</html>


admin_login.php
<?php
session_start();
if(!isset($_SESSION["manager"])){
header("Location: admin_login.php");
exit();
}
//Be sure to check that manager SESSION value is in database
$managerID = preg_replace('#[^0-9]#i',",$_SESSION["id"]); // Filter everything but numbers and letters
$manager = preg_replace('#[^A-Za-z0-9]#i',",$_SESSION["manager"]); //Filter everything but numbers and letters
$password = preg_replace('#[^A-Za-z0-9]#i',",$_SESSION["password"]); //Filter everything but numbers and letters
//Run MySQL query to be sure that this person is an admin and that their password session var equals the database
//Connect to the MySQL Database
include("../storescripts/connect_to_mysql.php");
$sql = mysql_query("SELECT * FROM admin WHERE id='$managerID AND username='$manager' AND password='$password' LIMIT 1"); //search user
$existsCount = mysql_num_rows($sql); //count the row numbers
if ($existsCount == 0){ //evaluate the count
header("Location: ../index.php");
exit();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>MyOnlineStore | Admin Area</title>
<link rel="stylesheet" href="../style/style.css" type="text/css" media="screen"/>
</head>

<body>
<div align="center" id="mainWrapper">
  <?php include_once("../template_header.php");?>
  <p><br  />
    <br />
  </p>
  <p>&nbsp;</p>
  <p><br />
  </p>
  <div id="pageContent">
    <div id="admin" align="left">
      <h3>Hello what you want to do?</h3>
      <p><a href="#">Manage Inventory</a><br />
      <a href="#">Manage afggagdfag</a></p>
</div>
</div>
<?php include_once("../template_footer.php");?>
</div>
<p>&nbsp;</p>
<p>&nbsp;</p>
<p>&nbsp;</p>
</body>
</html>

I have some FLV files above web root. I want to feed my flv player a php file and an ID. I need that php file to return an flv file basically. So if someone has firebug or something they wont see a video path, but instead see something like:

fileserve.php?id=4679

I dont care if a logged in user go directly to that link and it prompts them with a download or whatever. I just want to avoid having a non logged in user do that.

Hi....

I'm trying to dive into imagick.
But I cant seem to load it.

My web host has Imagic 6.3.X installed but from phpinfo() it does not show, and my scripts bring fatal error.
Code: [Select]
Fatal error: Class 'Imagick' not found in...I have made my own php.ini and it lodes fine with .htaccess.

Code: [Select]
<IfModule mod_suphp.c>
  suPHP_ConfigPath /home/2/f/fesan/
</IfModule>
In php.ini i have added:
Code: [Select]
;My Extesions:
[Imagick]
extension=imagick.so
Does anyone know how to load imagick correctly to php?
I'm on a linux server.