PHP - How To Share A Page Content Into My Facebook Page

Hi, I have a page whihc gets Daily motion URL from another page and also location of Image. I have Face book share button on this page and want to show my inage while trying to share on FB.   unfortunatley face book picks something wrong , and I use Open Graph Debug library then temporarily this picture issue gets resolved but it re-occures again if another URL is selected for sharing purpose.   Code written in  header of my page:   <meta property="og:title" content="Greate Web SIte"/>

<meta property ="og:url" content="http://www.MySIte.co...howMovie.php"/>
<meta property ="og:type" content="website"/>

<meta property="og:image:width" content="200" />
<meta property="og:image:height" content="200" />
<?php
    $selectedImage=$_GET['selectedImg'];  // this param is passed from qry string whihc is sent by a previos page, I want to change this param every time parent page sends a request of new image.     $ImageAddress='http://www.mySIte.com/'.$selectedImage; //creating URL of image , if i paste this URL in browser , image is fetched successfully.     echo '<meta property="og:image" content="'.$ImageAddress.'"/>';
?>                


<meta property="og:site_name" content="my Heading"/>
<meta property="og:description" content="Bla Bla bla."/>       Following is the java script that I have pasted under body tag: <div id="fb-root"></div>
<script>(function(d, s, id) {
  var js, fjs = d.getElementsByTagName(s)[0];
  if (d.getElementById(id)) return;
  js = d.createElement(s); js.id = id;
  js.src = "//connect.facebook.net/en_US/sdk.js#xfbml=1&version=v2.0";
  fjs.parentNode.insertBefore(js, fjs);
}(document, 'script', 'facebook-jssdk'));</script>


following is button for FB sha   echo '<div class="fb-post" data-href="'.$pageURL.'" data-width="500"></div>';  // URL that is passed as qry stirng are passed here.     kindly do let me know that how can I display the exact image on every attelp to share something.
                  

About a year ago I wrote a mod_rewrite in at sites htaccess file.  It was to call a page that queried the database based on a unite code that was assigned to a member.  If the member was assigned a code that was FL1001, the URL www.marketingteammates.com/FL1001 would bring up a page that quaried that member and displayed their database content.   In the users table of the site, it quaried the field webPageID and the information was called.  It called a page webpage.php that quaried the info.  The URL in the address bar still displayed www.marketingteammates.com/FL1001.  Here are the lines of code in the htaccess file.

RewriteEngine on
RewriteCond %{REQUEST_FILENAME} !-f
RewriteCond %{REQUEST_FILENAME} !-d
RewriteRule ^([A-Za-z])([A-Za-z])([0-9]+)$ /webpage.php?webPageID=$1$2$3 [NC,L]

I now have a second page completely different from the webpage.php.  It is a page storeNumber.php and would query a field storeNumberID.  This field could contain any kind of string such as Store#101, or mystore101, or #101 is my store.   I tried using the same rewite rule just replaceing the webpage.php and variable with storeNumber.php and $storeNumberID.   This doesn't work.  It also leads me to wonder if there can even be two different pages in the mod_rewrite.  How would it diferentiate one from the other.   Thanks for any help in advance.   Mike

Hi guys,

I was wondering if it is possible to like a page on facebook through a checkbox, instead of using the FB like function/script. In other words; people fill out a form and can check the box if they like the page, such as:
Code: [Select]
<?php
if (isset($_POST['form'])) {
    
if (isset($_POST['facebook_checkbox'])) {
//like page
    
    }
    
    else{
    //don't like
    
    }
}
?>

<form action="<? $_SERVER['PHP_SELF'];?>" method="post">
<input type="checkbox" id="facebook_checkbox" name="facebook_checkbox" />
<input type="submit" name="form" />
Is there any possibility to do that?
Thanks in advance.

I am trying to allow a user to proceed to another page but only after a user has clicked a Facebook like button and a check has been done to check if the currently logged in user has liked the specific Facebook Page.

How can I go about this?

Thanks in advance 

Hello every one

i need your help please about this

i would like to add this tabs to my facebook fan page i got App allowing me to put php coding or html


but i need when some one add his info and click on submit i receive it at my email
so can any 1 here help me or tell me what shall i do ?

i newbie and hope to get your help guys please

I have been trying to replicate the facebook "share" functionality , well actually i was in need of this functionality even before i knew that fb already had it..

I learnt a few things that this can be achieved by using the "Curl"  and I also hear that this setting may have been switched in some servers so if anyone can refer me to other options that would always fetch me the - Title- Description- Images for sure I' d be thankful 

Regards

Quesion: Show each movie in the database on its own page, and give the user links in a "page 1, Page 2, Page 3" - type navigation system. Hint: Use LIMIT to control which movie is on which page. I have provided 3 files: 1st: configure DB, 2nd: insert data, 3rd: my code for the question. I would appreciate the help. I am a noob by the way.

First set up everything for DB:

<?php
//connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//create the main database if it doesn't already exist
$query = 'CREATE DATABASE IF NOT EXISTS moviesite';
mysql_query($query, $db) or die(mysql_error($db));

//make sure our recently created database is the active one
mysql_select_db('moviesite', $db) or die(mysql_error($db));

//create the movie table
$query = 'CREATE TABLE movie (
        movie_id        INTEGER UNSIGNED  NOT NULL AUTO_INCREMENT,
        movie_name      VARCHAR(255)      NOT NULL,
        movie_type      TINYINT           NOT NULL DEFAULT 0,
        movie_year      SMALLINT UNSIGNED NOT NULL DEFAULT 0,
        movie_leadactor INTEGER UNSIGNED  NOT NULL DEFAULT 0,
        movie_director  INTEGER UNSIGNED  NOT NULL DEFAULT 0,

        PRIMARY KEY (movie_id),
        KEY movie_type (movie_type, movie_year)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die (mysql_error($db));

//create the movietype table
$query = 'CREATE TABLE movietype (
        movietype_id    TINYINT UNSIGNED NOT NULL AUTO_INCREMENT,
        movietype_label VARCHAR(100)     NOT NULL,
        PRIMARY KEY (movietype_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

//create the people table
$query = 'CREATE TABLE people (
        people_id         INTEGER UNSIGNED    NOT NULL AUTO_INCREMENT,
        people_fullname   VARCHAR(255)        NOT NULL,
        people_isactor    TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,
        people_isdirector TINYINT(1) UNSIGNED NOT NULL DEFAULT 0,

        PRIMARY KEY (people_id)
    )
    ENGINE=MyISAM';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Movie database successfully created!';
?>



********************************************************************
***********************************************************************

second file to load info into DB:



<?php
// connect to MySQL
$db = mysql_connect('localhost', 'root', '000') or
    die ('Unable to connect. Check your connection parameters.');

//make sure you're using the correct database
mysql_select_db('moviesite', $db) or die(mysql_error($db));

// insert data into the movie table
$query = 'INSERT INTO movie
        (movie_id, movie_name, movie_type, movie_year, movie_leadactor,
        movie_director)
    VALUES
        (1, "Bruce Almighty", 5, 2003, 1, 2),
        (2, "Office Space", 5, 1999, 5, 6),
        (3, "Grand Canyon", 2, 1991, 4, 3)';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the movietype table
$query = 'INSERT INTO movietype
        (movietype_id, movietype_label)
    VALUES
        (1,"Sci Fi"),
        (2, "Drama"),
        (3, "Adventure"),
        (4, "War"),
        (5, "Comedy"),
        (6, "Horror"),
        (7, "Action"),
        (8, "Kids")';
mysql_query($query, $db) or die(mysql_error($db));

// insert data into the people table
$query  = 'INSERT INTO people
        (people_id, people_fullname, people_isactor, people_isdirector)
    VALUES
        (1, "Jim Carrey", 1, 0),
        (2, "Tom Shadyac", 0, 1),
        (3, "Lawrence Kasdan", 0, 1),
        (4, "Kevin Kline", 1, 0),
        (5, "Ron Livingston", 1, 0),
        (6, "Mike Judge", 0, 1)';
mysql_query($query, $db) or die(mysql_error($db));

echo 'Data inserted successfully!';
?>



**************************************************************
****************************************************************

MY CODE FOR THE QUESTION:


<?php
$db = mysql_connect('localhost', 'root', '000') or
die ('Unable to connect. Check your connection parameters.');

mysql_select_db('moviesite', $db) or die(mysql_error($db));

//get our starting point for the query from the URL
if (isset($_GET['offset'])) {
   $offset = $_GET['offset'];
} else {
   $offset = 0;
}

//get the movie
$query  = 'SELECT
   movie_name, movie_year
FROM
   movie
ORDER BY
   movie_name
LIMIT ' . $offset . ' , 1';
$result = mysql_query($query, $db) or die(mysql_error($db));
$row = mysql_fetch_assoc($result);
?>
<html>
<head>
<title><?php echo $row['movie_name']; ?></title>
</head>
<body>
<table border = "1">
<tr>
<th>Movie Name</th>
<th>Year</th>
</tr><tr>
<td><?php echo $row['movie_name']; ?></td>
<td><?php echo $row['movie_year']; ?></td>
</tr>
</table>
<p>
<a href="page.php?offset=0">Page 1</a>,
<a href="page.php?offset=1">Page 2</a>,
<a href="page.php?offset=2">Page 3</a>
</p>
</body>
</html>

Some code from my pages ,   Page1 ( Redirecting page )

<html>
<title>login_redirect.</title>
body>

<form name="redirect" action="http://mysite/page2.php" method="post">
<input type="hidden" name="mac" value="$(mac)">
</form>

<script language="JavaScript">
<!--
document.redirect.submit();
//-->
</script>
</body>
</html>

Page 2 ( select product )

<?php


session_start();

ini_set('display_errors',1);
error_reporting(E_ALL); 

include '../lib/config.php';
include '../lib/opendb.php';

// get user mac adres from redirect post page1
$_SESSION['macid'] = $_POST['mac'];
// set $macid for other use ( maybe not needed, am learning )
$macid = $_SESSION['macid'];

// echo $macid does show mac adress, so variable is not empty here


if (!empty($_POST["submit"]))

{
$product_choice = $_POST['accounttype'];
$query= "SELECT AccountIndex, AccountCost, AccountName FROM AccountTypes WHERE AccountIndex='$product_choice'";
$result = mysql_query($query) or die('Query failed. ' . mysql_error()); 
while($row = mysql_fetch_array($result))
{
$_SESSION['AccountIndex'] = $row['AccountIndex'];
$_SESSION['AccountCost'] = $row['AccountCost'];
$_SESSION['AccountName'] = $row['AccountName'];
}


header('Location: page3.php');

}

// did leave out the other/html/form stuff here

Page 3 ( show Session variables )

<?php


ini_set('display_errors',1);
error_reporting(E_ALL); 

session_start();

print_r($_SESSION);

?>

Now, on page 3 i do see the right session varables, only the "macid" is empty.   why ?    

Currently, in my files, I echo content through either a function called error() or a funtion called info().

Each function has 3 require statements for the header, body and footer files and that is all that is contained in the function.

What I want to do is, if error() is called, is kind of leave the script and exit it. Surprisingly I have used exit; at the end of the functions to stop any more of the script being executed.

Is this the best way of doing it?

Hello.  I am attempting to display content from a database to a web page.  I have SELECT * etc to do this, plus the menu that should be displayed has links to pages.  This would display text to the page depending on the link, for this to work I use the id of the page ($Get['id']).  Following advice (not code), I have attempted the following:

 

<!DOCTYPE html>
<html>

  <?php include 'includes/headsection.php'; ?>

  <body>

    <?php

      // query database
      if (!empty($_GET['id']) && (intval($_GET['id']) == $_GET['id'])) {
      	$pdo_statement = $conn->prepare("SELECT * FROM pages WHERE id=:id ORDER BY id ASC");
      	$pdo_statement->execute(['id' => $_GET['id']]);
      	$result = $pdo_statement->fetchAll();
      }

    ?>

    <!-- topMenu -->
    <table id="topMenu">

      <tr>
        <td><h1 class="siteName">Site name</h1></td>

        <td>
          <?php
          if(!empty($result)) {
            foreach($result as $row) {
          ?>
          <a href="index.php?id=<?php echo $row['id']; ?>"><?php echo $row['menuheader'];  ?></a>

          <?php }
          }
        ?>

        </td>

      </tr>

    </table>

This issue is that the top menu (home, etc), does not appear, and there are no errors to suggest what I have done wrong.  Any help will be appreciated. I include a picture of what I mean.

There should be Home | other | etc...

 

I have a php page in which javascript called like this
<script language="JavaScript" type="text/javascript" src="http://www.test.com/test.php"></script>

the output of the page shows a well detail table , but when i use file_get_content method to get that all detail in a varaible its just took above javascript like $a=<script language="JavaScript" type="text/javascript" src="http://www.test.com/test.php"></script>
when i right click on page to view source it show just <script language="JavaScript" type="text/javascript" src="http://www.test.com/test.php"></script>.
But on actual page there is a html table which detail i have to store in table,
Can any one tell me method.

Good day everyone!

I know this is possible but I do not know how to do it.

I have a single page, this page has no dynamic content.

I have a set of links. What I need is to have content placed on the page based on the link the user clicks.

Example: I have pictures of animals. When the user click a Zebra, the Zebra info is placed on the page.

So the page does not change, meaning the content on the page changes but not the page it self.

The idea is instead of creating different html pages for each animal profile, the profile is generated to the page from a php include file.

Is is this possible?

Can anyone, some one please point me to how I can get this going.


Thanks everyone!


IC

What I'm trying to do is get a php file with content to load into my basic page template based on the page id, however it's not loading anything, however there are no errors coming up. I've looked everywhere I could find, and tried everything I could think of. I really could use some help on this. Below is the page template code and the code I'm trying to use to get the content to load up.

page_templ.php
Code: [Select]
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title></title>
<meta http-equiv="Content-Type" content="text/html;charset=utf-8" >
<link rel="stylesheet" type="text/css" href="templates/content/style.css" >
<!--This is the css used for IE because we need to correct some things for it.-->
<!--[if IE]>
  <style type="text/css">
    @import 'templates/content/styleIE.css';
  </style>
<![endif]-->
</head>
  <body>
    <?php
    //We need to load our language files with our common text varibles
      require("content/langeng.php");
    ?>
   
<div id="header">
      <?
      //Lets show the header. I use html where I can so smartphones load as little advanced code as possible
        require_once("header.php");
      ?>
</div>

<div id="menu">
      <?//Now our menu.
        require_once("menu.php");
      ?>
</div>

<div id="mbody">
      <?//Now the primary site content.
        //Load up our postvars.php file
        require_once("content/postvars.php");
      ?>
</div>
  </body>
</html>

postvars.php
Code: [Select]
<?php
$id  =  $_POST["id"];
if ( $id == "home" ) {
include 'main_c.php';
}
elseif ( $id == "guides" ) {
include  'guides_c.php';
}
elseif ( $id == "trips" ) {
include  'trips_c.php';
}
?>

Most of the websites I have built have not been large and not dynamic in nature, but I need to make one now with a boatload of content on it. and I’m needing a little resource help where I can research something. I’m sure all developers do this, and I know I need a massive database or 2 and the PHP knowledge I already have, but I’m just looking for a little guidance about how to do this the easiest way. It may actually sound like a stupid question and I might be over-thinking a bit. So…

Lets say I have 500 articles that I want to make available to a site visitor and give them the option to choose what they want to look at through nav menus and submenus. I want the same page to display to the user, regardless of what article is chosen.  how does PHP play a part in that? Would I be right by saying the following?

If the home page had this content on it:

This website is a course in PHP coding. <a href="courseTemplate.php" target="_blank">Click here</a> to view course #1.

and I wanted all the course content to be displayed through the use of the “courseTemplate.php” file, how simple is that? I would assume that these types of things would be the results and the techniques to accomplish the goal, right?

=> a resulting URL that looks like this:

www.site.com/courseTemplate.php?id=1

=> storing all the text of the course material in one single field of a DB record. NOT storing the layout-oriented code (HTML, etc.) and echoing it out with the rest of the text.

=> making use of the GET() function somehow to pull the course’s text content out of the database.

Can someone show me a website that demonstrates this? I don’t think this is very difficult, and I’m sure there are web resources available that show how this is done. I’m almost sure that most news agency websites do this, and I know for a fact that most forum software has this template technique written into it, regardless if SEO is included or not. Sorry for the basic nature of this question. I know some of my previous posts have been such where I’ve asked much more difficult questions than this. I have seen youtube tutorials on issues similar to this, but nothing pulled up by google really shows this in a very simplistic nature. thanks.

I'm doing a cURL post to a server and it returns "true No_Error Data was accepted 10356546 3770 false 0 "


Form --> Method=Post --> submit.php (this is the page that's doing the cURL post and the server returns data that appears on this page).


How do I take this and put it into a string?