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PHP - Sync Facebook With Mysql Database/php Problem.
Hey there I'm trying to gather a users Facebook info and use that to create a user account for them on my database.
My code works perfectly for the first person who links his Facebook account with my site and adds all the user info to the database. After that with any other users it will display the logged in Facebook users name with their profile pic. The problem arises when it tries to add this person to the mysql database. It returns the error "An error has occurred and we are unable to sync your Facebook account." Before the <head> Code: [Select] <?php define('FACEBOOK_APP_ID', 'imagine_app_id_here'); define('FACEBOOK_SECRET', 'imagine_secret_here'); function get_facebook_cookie($app_id, $application_secret) { $args = array(); parse_str(trim($_COOKIE['fbs_' . $app_id], '"'), $args); ksort($args); $payload = ''; foreach ($args as $key => $value) { if ($key != 'sig') { $payload .= $key . '=' . $value; } } if (md5($payload . $application_secret) != $args['sig']) { return null; } return $args; } $cookie = get_facebook_cookie(FACEBOOK_APP_ID, FACEBOOK_SECRET); ?> In the <body> Code: [Select] <?php if ($cookie) { //###cookie is set, user is logged in $user = json_decode(file_get_contents('https://graph.facebook.com/'.$cookie['uid'])); define('INCLUDE_CHECK',true); require("connect.php"); $fbookid = $user->{'id'}; $fbookname = $user->{'name'}; $query = mysql_query("SELECT * FROM tz_members WHERE facebook_id='$fbookid'"); $numrows = mysql_num_rows($query); if ($numrows == 1){ echo "<font color='#FF0000'>You are logged in through Facebook.</font>"; } else { mysql_query("INSERT INTO tz_members(facebook_id,usr,regIP,dt) VALUES( '".$fbookid."', '".$fbookname."', '".$_SERVER['REMOTE_ADDR']."', NOW() )"); $query = mysql_query("SELECT * FROM tz_members WHERE facebook_id='$fbookid'"); $numrows = mysql_num_rows($query); if ($numrows == 1){ echo "<font color='#FF0000'>Your Facebook account has been synced with our user database.</font>"; } else { echo "<font color='#FF0000'>An error has occured and we are unable to sync your Facebook account.</font>"; } mysql_close(); } echo '<img src="http://graph.facebook.com/'.$user->{'id'}.'/picture" alt="'.$user->{'name'}.'"/>'; echo $user->{'name'}; echo '<fb:login-button perms="email,user_birthday" onlogin="window.location.reload(true);" autologoutlink="true"></fb:login-button>'; } else { //###user is not logged in, display the Facebook login button echo '<fb:login-button perms="email,user_birthday" autologoutlink="true"></fb:login-button>'; } ?> Javascript: Code: [Select] <div id="fb-root"></div> <script src="http://connect.facebook.net/en_US/all.js"></script> <script> FB.init({appId: '<?= FACEBOOK_APP_ID ?>', status: true, cookie: true, xfbml: true}); FB.Event.subscribe('auth.login', function(response) { window.location.reload(); }); </script> Any help would be greatly appreciated! Similar TutorialsHi. I have a database named 'Forums' with a table named 'forumlist'. The fields in the table are 'id', 'name', 'link', and 'keys'. I am trying to search the database in either the 'name' or 'keys' field and display the matches. I am getting this error: Parse error: syntax error, unexpected T_STRING, expecting ']' in C:\xampp\htdocs\search3.php on line 51 (The define fields) <html> <head> <title>My Forum Search</title> <style type="text/css"> table { background-color: #CCC } th { width: 150px; text-align: left; } </style> </head> <body> <h1>Forum Search</h1> <form method="post" action="search3.php"> <input type="hidden" name="submitted" value="true" /> <label>Search Category: <select name="category"> <option value="keys">Keyword</option> <option value="name">Name</option> </select> </label> <label><input type="text" name="criteria" /></label> <input type="submit" /> </form> <?php if (isset($_POST['submitted])) { DEFINE ('DB_USER', 'root'); DEFINE ('DB_PSWD', '******'); DEFINE ('DB_HOST', 'localhost'); DEFINE ('DB_NAME', 'Forums'); $dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME); $category = $_POST['category']; $criteria = $_POST['criteria']; $query = "SELECT * FROM forumlist WHERE $category LIKE '$criteria'"; $result = mysqli_query($dbcon, $query) or die ('Error retrieving data') echo "<table>"; echo "<tr> <th>Name</th><th>Link</th> </tr>"; while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { echo "<tr><td>"; echo $row['name']; echo "</td><td>"; echo $row['link']; echo "</td></tr>"; } echo "</table>"; } // end of main if state ?> </body> </html> Thanks in advance I'm having problems posting date from my MYSQL database. The date in my database is this "2011-01-08 02:53:14" but the it only echo's "01.01.70" Could someone help me figure out why? Here is my code: Code: [Select] <?php $servername='localhost'; $dbusername='root'; $dbpassword=''; $dbname='store'; connecttodb($servername,$dbname,$dbusername,$dbpassword); function connecttodb($servername,$dbname,$dbuser,$dbpassword) { global $link; $link=mysql_connect ("$servername","$dbuser","$dbpassword"); if(!$link){die("Could not connect to MySQL");} mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error()); } // Get all the data from the "example" table $result = mysql_query("SELECT * FROM henvendelser WHERE status = 'Ubehandlet' ORDER by id desc ") or die(mysql_error()); echo "<table cellspacing='12px' cellpaddomg='5px' align='center'>"; echo "<tr> <th>ID</th> <th> Opprettet </th> <th>Navn</th> <th>Telefon</th> <th>Emne</th> </tr>"; // keeps getting the next row until there are no more to get while($row = mysql_fetch_array($result)) { $postTime = $row['date']; $thisTime = time(); $timeDiff = $thisTime-$postTime; if($timeDiff <= 604800) { // Less than 7 days[60*60*24*7] $color = '#D1180A'; } else if($timeDiff > 604800 && $timeDiff <= 1209600) { // Greater than 7 days[60*60*24*7], less than 14 days[60*60*24*14] $color = '#D1B30A'; } else if($timeDiff > 1209600) { // Greater than 14 days[60*60*24*14] $color = '#08C90F'; } echo '<tr style="color: '.$color.';">'; echo '<td>'. $row['id'] .'</td>'; echo '<td>'. date('d.m.y', $postTime) .'</td>'; echo '<td><a href="detaljer.php?view='. $row['id'] .'">'. $row['Navn'] .'</a></td>'; echo '<td>'. $row['Telefon'] .'</td>'; echo '<td>'. $row['Emne'] .'</td>'; echo '</tr>'; } echo '</table>'; ?> Okay, I'll make this short and sweet, and all side comments are irrelevant. I'm doing this for my own happies So using Facebook.com's API you can extract all of a specific user's friends in an array. Well I'm making a array check to see who on that users friend list was removed or deleted them that week using array_dif(); in this method: 1) User allows the application 2) Application catches the user's friends_id's in an array and stores them // this is where i'm having trouble, i'll come back to it 3) User then waits a few days and notices their friend count is no longer 400, it's now 399! 4) User logs into my application, and checks the previously uploaded array against their current array of friends. Okay now to the part i'm struggling with is, what would you do? Should i store them each uniquely in a database? Because some users have 4,000+ friends and that gets quite strenuous. OR! Should i save them into a txt file unique to that person's user_id. I kind of like the 2nd option, however it's less ( to my newbish mind ) customizable versus the sql_queries. Once again, i'd like to reiterate, i don't care that this is again Facebooks terms of service, it's for my own personal gain and just enjoy taking on projects in my spare time that force my brain to work hard. That's all, hope you excuse the typos and poor grammar. Typed this pretty quickly, any questions - feel free to ask! Dear all friends, I don't know this forum already have kindly of this question or not. I'm coding my won php and mysql. but I have no idea with two wanted idea as below: 1. when I post on my website. It will automaticall post in my page in facebook 2. when friend in facebook comment on that item it will also show that comment on my facebook page and also my website too. I'm please to getting your idea event from general idea and technical idea. thanks for your answer. All of your answer and idea very important to me. Best Regards Steve. I want a php mysql script that works like a facebook social networking . A user can add,update and delete other users to their circle in 3 privacy categories (e.g. Private, Limited, Public). So that a user can define what level of detail is visible based on the privacy categories, to other users. 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 $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. I have this code curl_setopt($ch, CURLOPT_POST, 0); curl_setopt($ch, CURLOPT_URL, $link); $wall = curl_exec($ch); but the $link is a facebook wall page that takes a couple of seconds to load fully. and you have to scroll down the page before more loads or you have to click older posts when i run the script the $wall doesnt have the full page. Is there a way that i can delay the $link from saving the file so that it reads more of the page?? This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=309715.0 My game on facebook is being displayed in the facebook application iframe. When users register in firefox or google chrome everything is fine and to a certain extent in IE as well. However, when logging into the game, through the browsers the game loads fine except when loaded in IE.... Basically i am using a session to state that the game is currently in facebook mode: Code: [Select] <?php session_start(); $_SESSION['isface'] = 1; include("game.php"); ?> All is good until it hits this code: Code: [Select] <html> <frameset cols="130" noresize border=0> <frame src="nav.php" name="nav"> <frame src="hq.php" name="main"> </frameset> </html> The frames then output error message: Code: [Select] <?php if ($isface !="1") { if (sha1($info[4]) != $pass) { mysql_close(); echo $isface; die("<br><br><center><body bgcolor='#000000'><b><font color='#FFFFFF' face='Verdana' size='2pt'>{$isface} Your username cannot be found or password doesnt match</font></b></center></body></html>"); } } ?> essentially the game is using cookies for authentication on the main site, but since this is facebook this is not needed so it SHOULD skip this check and proceed with using the facebook details as the login. this is only happening in IE... i dont get it. any ideas? Hi i have a facebook login social plugin made but my ? is why do i get symbols in my url link ?. I login with the plugin to facebook but when page returns me back to my site i have symbols at the end of url. Like this http://www.games-flash.co.uk/#_=_ Ps sry if i have made this post in wrong section. This topic has been moved to Other Libraries and Frameworks. http://www.phpfreaks.com/forums/index.php?topic=347238.0 I'm updating my database with a set of data from another database. When there is an item deleted from the source db, how would I make sure it's deleted from the db copy given that I have to run the update in a batch and can't run it when the item is deleted? Would I truncate the db at the beginning of each update, then pull all the of data over? I am working with a site that needs some kind of calendar solution. Each user will have their own events calendar. The plan is that your events calendar could sync with google, ical, BB calendar when done. Should I build the calendar in PHP, then worry about syncing with these other apps? Or should I use come kind of 3rd party calendar to power the one I have? Or...does anyone know of any applications that help you sync calendars if I were to create one myself? I looked into google cals, but they have a limit on the amount of calendars per day you can create. Any ideas as to how I should go about this? All answers appreciated! Commands out of sync; you can't run this command now Query: DELETE FROM case_ref WHERE user_id='3';DELETE FROM updates WHERE user_id='3' php]<?php // Check if the form has been submitted: if (isset($_POST['delete'])) { if ($_POST['sure'] == 'Yes') { // Delete the record. // Make the query: $query = "DELETE FROM case_ref WHERE user_id='$id';"; $query .= "DELETE FROM updates WHERE user_id='$id'"; $r = mysqli_multi_query ($dbc, $query); if (mysqli_affected_rows($dbc) >= 1) { // If it ran OK. echo '<h3>All cases & updates associated with user have been deleted</h3>'; } // Make the query: $q = "DELETE FROM users WHERE user_id='$id' LIMIT 1"; $r = $r = mysqli_query ($dbc, $q); if (mysqli_affected_rows($dbc) == 1) { // If it ran OK. echo '<h3>User has been deleted</h3>'; } else { // If the query did not run OK. $errors[] = '<p class="error">The user could not be deleted due to a system error.</p>'; // Public message. $errors[] = '<p>' . mysqli_error($dbc) . '<br />Query: ' . $query . '</p>'; // Debugging message. } } else { // No confirmation of deletion. $errors[] = '<p class="error">The user has NOT been deleted.</p>'; } }else{ //show the form echo '<form id="form1" name="form1" method="post" action="delete_user.php?user_id=' . $id . '"> <table width="100%" border="1" cellpadding="5" cellspacing="5"> <tr> <td width="30%">Do you wish to delete this user?</td> <td width="19%"><input type="radio" name="sure" value="Yes" /> Yes </td> <td width="17%"><input type="radio" name="sure" value="No" checked="checked" /> No</td> <td width="34%"> <input type="submit" name="button" id="button" value="Submit" /> <input name="delete" type="hidden" id="delete" value="1" /></td> </tr> </table>'; } // Print any error messages, if they exist: if (!empty($errors)) { foreach ($errors as $msg) { echo "$msg\n"; } } mysqli_close($dbc); ?>[/php] Hi, I want to create a simple online buying site. I have an admin site with login where admin can add items to the database table, and change other fields such as price and description of product and I want to make a public html page to display the updates made in my admin page. I appreciate it if someone give me some suggestions on how to get started. My admin php scripts is working fine (so SQL queries for UPDATE, DELETE, INSERT) but I don't know how to 'sync' it with the an nice html table of products the user can checkbox off to purchase on the public website. Let me know, thanks. We are building a site for users to post articles on certain topics and we want to show their number of followers on various social media platforms. Apart from them having the means to add it manually, there must be a method of syncing with their account, that they punch in, so that it dynamically updates our database with their number of followers. How is it done? How would I go about doing the following: I have a csv like this Quote "Division","Section","Group","Product Code","Description","Description + Secondary Description" "Division 1","Section 1","Group 1","BMSLPL25","Test Name","Test Description" "Division 1","Section 1","Group 2","BMSLPL26","Test Name 2","Test Description 2" "Division 2","Section 2","Group 2","BMSLPL27","Test Name 3","Test Description 3" I have a database structured like this Quote Divisions --- id name parent_id Groups --- id name division_id Products --- id code description secondary_description Section is a sub division. What is the best way to get the information from CSV into this database? Should I have another table and store the CSV data as is and then query that to make the other tables. Any help much appreciated. 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? |
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