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  • Cannot Call WordPress Plugin Files Under wp-content

    - by Volomike
    I have a client who has many blog customers. Each of these WordPress blogs calls a plugin that provides a product link. The way that link is composed looks like this: {website}/wp-content/plugins/prodx/product?id=432320 This works fine on all blogs except two. On those two, when you try to call the URL, you get a 404. So, I disabled all plugins except prodx and reverted the theme to default (Kubrick), thinking perhaps a plugin intercept with add_action() API was doing this, such as intercepting URLs and redirecting them. However, this did not help. So, I upgraded the WordPress to the latest version. Again, didn't fix. So, I checked permissions, comparing with a blog that worked just fine. Again, didn't fix. So I replaced the .htaccess, using one from a working blog. Again, didn't fix. So I replaced all the files using some from a working blog that was identical to this one, and then restored the wp-config.php file back so that it talked to the right blog database. Again, didn't fix. Again I checked permissions meticulously, comparing to a perfectly working blog. Again, didn't fix. So, I created a test.php that looks like so: <?php print_r($_GET); echo "hello world"; I then copied it into another plugin folder and used my browser to get to it -- again, 404. So I copied it into the root of wp-content/plugins and tried to call it there -- again, 404. So I copied it into wp-content -- again, 404. Last, I copied it into the root of the WordPress blog website, and this time, it worked! Doesn't make sense. I started to think that perhaps something was going on with /etc/httpd/conf/httpd.conf for this customer, but the only thing I saw different in their for this customer was the IP address was different than the customer's blog that worked. Each customer gets their own IP in this environment my client has built. My client sysop is baffled too. What do you think is going on? Is there something wrong in the WP database for this customer? Is there something wrong in httpd.conf?

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  • What would you do to make this code more "over-engineered"? [closed]

    - by Mez
    A friend and I got bored, and, long story short, decided to make an over-engineered FizzBuzz in PHP <?php interface INumber { public function go(); public function setNumber($i); } class FBNumber implements INumber { private $value; private $fizz; private $buzz; public function __construct($fizz = 3 , $buzz = 5) { $this->setFizz($fizz); $this->setBuzz($buzz); } public function setNumber($i) { if(is_int($i)) { $this->value = $i; } } private function setFizz($i) { if(is_int($i)) { $this->fizz = $i; } } private function setBuzz($i) { if(is_int($i)) { $this->buzz = $i; } } private function isFizz() { return ($this->value % $this->fizz == 0); } private function isBuzz() { return ($this->value % $this->buzz == 0); } private function isNeither() { return (!$this->isBuzz() AND !$this->isFizz()); } private function isFizzBuzz() { return ($this->isFizz() OR $this->isBuzz()); } private function fizz() { if ($this->isFizz()) { return "Fizz"; } } private function buzz() { if ($this->isBuzz()) { return "Buzz"; } } private function number() { if ($this->isNeither()) { return $this->value; } } public function go() { return $this->fizz() . $this->buzz() . $this->number(); } } class FizzBuzz { private $limit; private $number_class; private $numbers = array(); function __construct(INumber $number_class, $limit = 100) { $this->number_class = $number_class; $this->limit = $limit; } private function collectNumbers() { for ($i=1; $i <= $this->limit; $i++) { $n = clone($this->number_class); $n->setNumber($i); $this->numbers[$i] = $n->go(); unset($n); } } private function printNumbers() { $return = ''; foreach($this->numbers as $number){ $return .= $number . "\n"; } return $return; } public function go() { $this->collectNumbers(); return $this->printNumbers(); } } $fb = new FizzBuzz(new FBNumber()); echo $fb->go(); In theory, what could we/would you do to make it even more "over-engineered"?

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  • PHP - Cannot modify header information...

    - by Scott W.
    Hi, I am going crazy with this error: Cannot modify header information - headers already sent by... Please note that I know about the gazillion results on google and on stack overflow. My problem is the way I've constructed my pages. To keep html separate from php, I use include files. So, for example, my pages look something like this: <?php require_once('web.config.php'); ?> <!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>Login</title> <link rel="shortcut icon" href="images/favicon.gif"/> <link rel="shortcut icon" href="images/favicon.ico"/> <link rel="stylesheet" type="text/css" href="<?php echo SITE_STYLE; ?>"/> </head> <body> <div id="page_effect" style="display:none;"> <?php require_once('./controls/login/login.control.php'); ?> </div> </body> </html> So, by the time my php file is included, the header is already sent. Part of the include file looks like this: // redirect to destination if($user_redirect != 'default') { $destination_url = $row['DestinationUrl']; header('Location:'.$user_redirect); } elseif($user_redirect == 'default' && isset($_GET['ReturnURL'])) { $destination_url = $_GET['ReturnURL']; header('Location:'.$destination_url); } else { header('Location:'.SITE_URL.'login.php'); } But I can't figure out how to work around this. I can't have the header redirect before the output so having output buffering on is the only thing I can do. Naturally it works fine that way - but having to rely on that just stinks. It would be nice if PHP had an alternative way to redirect or had additional parameters to tell it to clear the buffer.

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  • Using jQuery's ajax get request with parameters, returning page content

    - by Stevie Jenowski
    Thank you for looking at my question, as I appreciate your time. Okay, so I'm trying to use jQuery's get function to call my php script which ultimately returns a variable which is a built template of the main content of my page minus my static header/footer, for which I would like to replace the "old" page content without the page reloading. Can anyone point me in the right direction as to where I'm going wrong here? My code is as follows... jQuery: function getData(time, type) { $.ajax({ type: "GET", url: "getData.php", data: "time=" + time + "&type=" + type, success: function(data){ $('#pageContent').fadeOut("slow"); $('#pageContent').html(data); $('#pageContent').fadeIn("slow"); } }); return false; } getData.php(paraphrased): .... $time = empty($_GET['time']) ? '' : $_GET['time']; $type = empty($_GET['type']) ? '' : $_GET['type']; echo getData($time, $type); function getData($time, $type) ...... ..... $tmpl = new Template(); $tmpl->time= $time; $tmpl->type = $type; $builtpage = $tmpl->build('index.tmpl'); return $builtpage; ..... ...... jQuery function call: <a href="#" onclick="getData("<?php print $tmpl->time; ?>", "Orange")">Orange</a> <a href="#" onclick="getData("<?php print $tmpl->time; ?>", "Apple")">Apple</a> <a href="#" onclick="getData("<?php print $tmpl->time; ?>", "Banana")">Banana</a> When I click any link, the ajax seems to run fine, and the page does refuse to reload, but the content still remains unchanged... Anyone happen to know what's up?

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  • PHP miniwebsever file download

    - by snikolov
    $httpsock = @socket_create_listen("9090"); if (!$httpsock) { print "Socket creation failed!\n"; exit; } while (1) { $client = socket_accept($httpsock); $input = trim(socket_read ($client, 4096)); $input = explode(" ", $input); $input = $input[1]; $fileinfo = pathinfo($input); switch ($fileinfo['extension']) { default: $mime = "text/html"; } if ($input == "/") { $input = "index.html"; } $input = ".$input"; if (file_exists($input) && is_readable($input)) { echo "Serving $input\n"; $contents = file_get_contents($input); $output = "HTTP/1.0 200 OK\r\nServer: APatchyServer\r\nConnection: close\r\nContent-Type: $mime\r\n\r\n$contents"; } else { //$contents = "The file you requested doesn't exist. Sorry!"; //$output = "HTTP/1.0 404 OBJECT NOT FOUND\r\nServer: BabyHTTP\r\nConnection: close\r\nContent-Type: text/html\r\n\r\n$contents"; function openfile() { $filename = "a.pl"; $file = fopen($filename, 'r'); $filesize = filesize($filename); $buffer = fread($file, $filesize); $array = array("Output"=$buffer,"filesize"=$filesize,"filename"=$filename); return $array; } $send = openfile(); $file = $send['filename']; $filesize = $send['filesize']; $output = 'HTTP/1.0 200 OK\r\n'; $output .= "Content-type: application/octet-stream\r\n"; $output .= 'Content-Disposition: attachment; filename="'.$file.'"\r\n'; $output .= "Content-Length:$filesize\r\n"; $output .= "Accept-Ranges: bytes\r\n"; $output .= "Cache-Control: private\n\n"; $output .= $send['Output']; $output .= "Content-Transfer-Encoding: binary"; $output .= "Connection: Keep-Alive\r\n"; } socket_write($client, $output); socket_close ($client); } socket_close ($httpsock); Hello, I am snikolov i am creating a miniwebserver with php and i would like to know how i can send the client a file to download with his browser such as firefox or internet explore i am sending a file to the user to download via sockets, but the cleint is not getting the filename and the information to download can you please help me here,if i declare the file again i get this error in my server Fatal error: Cannot redeclare openfile() (previously declared in C:\User s\fsfdsf\sfdsfsdf\httpd.php:31) in C:\Users\hfghfgh\hfghg\httpd.php on li ne 29, if its possible, i would like to know if the webserver can show much banwdidth the user request via sockets, perl has the same option as php but its more hardcore than php i dont understand much about perl, i even saw that a miniwebserver can show much the client user pulls from the server would it be possible that you can assist me with this coding, i much aprreciate it thank you guys.

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  • PHP Strange error "0:". Possible Object syntax problem?

    - by Michael Stone
    I'm trying to transfer a site from one hosting company to another. I was using 000webhost until the client had hosting and a domain. Now that the client has their domain and hosting, using FatCow.com, I can not for the life of me, debug my PHP code. I'm not getting any errors. I have a successful DB connection. If you procedurally display data, it works, but when I try to use my original objects, something is breaking and just returns "0:". I have all errors on. On old server where the site worked: PHP Version 5.2.11 MySQL Version: 5.0.81 On new server where I get the "0:": PHP Version 5.2.12 MySQL Version 5.0.32 I've setup a test page to test just the output of the variables with the DB connection. Below is my code: <?php error_reporting(E_ALL); ini_set('display_errors', '1'); try { $link = mysql_connect('connectionstring', 'username', 'password'); if (!$link) { die('Could not connect: ' . mysql_error()); } else{ $db = mysql_select_db('a8210422_lit'); } if($db){ include_once('admin/classes/clsPOW.php'); include_once('admin/classes/clsProviders.php'); $pow = new POW(); $prov = new Providers(); $new = $pow->getNew(); $newAr = $new->val(); $get = $prov->getAll($newAr['providerId']); $getAr = $get->val(); $img = $getAr['image']; $name = $getAr['provider']; $desc = $getAr['description']; $zip = $getAr['zip']; $web = $getAr['link']; if($zip==0){ $zip = "Unavailable"; } print $img; print $name; print $desc; print $zip; print $web; } else{ print 'fail!'; } } //catch exception catch(Exception $e) { echo 'Message: ' .$e->getMessage(); } ?>

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  • Python and displaying HTML

    - by Tyler Seymour
    I've gotten pretty comfortable with Python and now I'm looking to make a rudimentary web application. I was somewhat scared of Django and the other Python frameworks so I went caveman on it and decided to generate the HTML myself using another Python script. Maybe this is how you do it anyways - but I'm just figuring this stuff out. I'm really looking for a tip-off on, well, what to do next. My Python script PRINTS the HTML (is this even correct? I need it to be on a webpage!), but now what? Thanks for your continued support during my learning process. One day I will post answers! -Tyler Here's my code: from SearchPhone import SearchPhone phones = ["Iphone 3", "Iphone 4", "Iphone 5","Galaxy s3", "Galaxy s2", "LG Lucid", "LG Esteem", "HTC One S", "Droid 4", "Droid RAZR MAXX", "HTC EVO", "Galaxy Nexus", "LG Optimus 2", "LG Ignite", "Galaxy Note", "HTC Amaze", "HTC Rezound", "HTC Vivid", "HTC Rhyme", "Motorola Photon", "Motorola Milestone", "myTouch slide", "HTC Status", "Droid 3", "HTC Evo 3d", "HTC Wildfire", "LG Optimus 3d", "HTC ThunderBolt", "Incredible 2", "Kyocera Echo", "Galaxy S 4g", "HTC Inspire", "LG Optimus 2x", "Samsung Gem", "HTC Evo Shift", "Nexus S", "LG Axis", "Droid 2", "G2", "Droid x", "Droid Incredible" ] print """<!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>table of phones</title> </head> <body> </body> </html> """ #table print '<table width="100%" border="1">' for x in phones: y = SearchPhone(x) print "\t<tr>" print "\t\t<td>" + str(y[0]) + "</td>" print "\t\t<td>" + str(y[1]) + "</td>" print "\t\t<td>" + str(y[2]) + "</td>" print "\t\t<td>" + str(y[3]) + "</td>" print "\t\t<td>" + str(y[4]) + "</td>" print "\t</tr>" print "</table>

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  • Converting my lightweight MySQL DB wrapper into MySQLi. Pesky Problems

    - by Chaplin
    Here is the original code: http://pastebin.com/DNxtmApY. I'm not that interested in prepared statements at the moment, I just want this wrapper updating to MySQLi so once MySQL becomes depreciated I haven't got to update a billion websites. Here is my attempt at converting to MySQLi. <? $database_host = "127.0.0.1"; $database_user = "user"; $database_pass = "pass"; $database_name = "name"; $db = new database($database_host, $database_user, $database_pass, $database_name); class database { var $link, $result; function database($host, $user, $pass, $db) { $this->link = mysqli_connect($host, $user, $pass, $db) or $this->error(); mysqli_select_db($db, $this->link) or $this->error(); } function query($query) { $this->result = mysqli_query($query, $this->link) or $this->error(); $this->_query_count++; return $this->result; } function countRows($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_num_rows($result); } function fetch($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result); } function fetch_num($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_NUM); } function fetch_assoc($result = "") { if ( empty( $result ) ) $result = $this->result; return mysqli_fetch_array($result, mysqli_ASSOC); } function escape($str) { return mysqli_real_escape_string($str); } function error() { if ( $_GET["debug"] == 1 ){ die(mysqi_error()); } else { echo "Error in db code"; } } } function sanitize($data) { //apply stripslashes if magic_quotes_gpc is enabled if(get_magic_quotes_gpc()) $data = stripslashes($data); // a mysqli connection is required before using this function $data = trim(mysqli_real_escape_string($data)); return $data; } However it chucks all sorts of errors: Warning: mysql_query(): Access denied for user 'www-data'@'localhost' (using password: NO) in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_query(): A link to the server could not be established in /home/count/Workspace/lib/classes/user.php on line 7 Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /home/count/Workspace/lib/classes/user.php on line 8 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, object given in /home/count/Workspace/lib/classes/database.php on line 31

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  • How to send variable to php with ajax?

    - by Dee1983
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"> <head> <script type="text/javascript"> function load(thediv, thefile) { if (window.XMLHttpRequest) { xmlhttp = new XMLHttpRequest(); } else { xmlhttp = new ActiveXObject('Microsoft.XMLHTTP)'); } xmlhttp.onreadystatechange = function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById(thediv) .innerHTML = xmlhttp.responseText; } } xmlhttp.open('GET', thefile, true); xmlhttp.send(); } </script> </head> <body> <?php //connection to db and mysql query $result = mysql_query($query) or die(mysql_error()); $options=""; while ($row=mysql_fetch_array($result)) { $id=$row["idProducts"]; $thing=$row["country"]; $options.="<OPTION VALUE=\"$id\">".$thing.'</option>'; } mysql_close(); ?> <SELECT id="countrySearch" NAME=countrySearch onchange="load('divtest', 'step2.search.php')";> <OPTION VALUE=0>Choose <?=$options?> </SELECT> <div id="divtest"> test </div> </body> step2.search.php consists of: <?php echo "I want it to store the users selection as a variable for php to use"; ?> The problem I have is I want to store what the user selects from the drop down box and use it in php to do a mysql query using the variable from the user select form to form the WHERE part of the mysql statement. Then use ajax to put new data in "divtest". How can I store the user selection into a variable then send it to be used in step2.search.php?

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  • Mootools 1.2.4 delegation not working in IE8...?

    - by michael
    Hey there everybody-- So I have a listbox next to a form. When the user clicks an option in the select box, I make a request for the related data, returned in a JSON object, which gets put into the form elements. When the form is saved, the request goes thru and the listbox is rebuilt with the updated data. Since it's being rebuilt I'm trying to use delegation on the listbox's parent div for the onchange code. The trouble I'm having is with IE8 (big shock) not firing the delegated event. I have the following HTML: <div id="listwrapper" class="span-10 append-1 last"> <select id="list" name="list" size="20"> <option value="86">Adrian Franklin</option> <option value="16">Adrian McCorvey</option> <option value="196">Virginia Thomas</option> </select> </div> and the following script to go with it: window.addEvent('domready', function() { var jsonreq = new Request.JSON(); $('listwrapper').addEvent('change:relay(select)', function(e) { alert('this doesn't fire in IE8'); e.stop(); var status= $('statuswrapper').empty().addClass('ajax-loading'); jsonreq.options.url = 'de_getformdata.php'; jsonreq.options.method = 'post'; jsonreq.options.data = {'getlist':'<?php echo $getlist ?>','pkey':$('list').value}; jsonreq.onSuccess = function(rObj, rTxt) { status.removeClass('ajax-loading'); for (key in rObj) { status.set('html','You are currently editing '+rObj['cname']); if ($chk($(key))) $(key).value = rObj[key]; } $('lalsoaccomp-yes').set('checked',(($('naccompkey').value > 0)?'true':'false')); $('lalsoaccomp-no').set('checked',(($('naccompkey').value > 0)?'false':'true')); } jsonreq.send(); }); }); (I took out a bit of unrelated stuff). So this all works as expected in firefox, but IE8 refuses to fire the delegated change event on the select element. If I attach the change function directly to the select, then it works just fine. Am I missing something? Does IE8 just not like the :relay? Sidenote: I'm very new to mootools and javascripting, etc, so if there's something that can be improved code-wise, please let me know too.. Thanks!

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  • How to use Facebook graph API to retrieve fan photos uploaded to wall of fan page?

    - by Joe
    I am creating an external photo gallery using PHP and the Facebook graph API. It pulls thumbnails as well as the large image from albums on our Facebook Fan Page. Everything works perfect, except I'm only able to retrieve photos that an ADMIN posts to our page. (graph.facebook.com/myalbumid/photos) Is there a way to use graph api to load publicy uploaded photos from fans? I want to retrieve the pictures from the "Photos from" album, but trying to get the ID for the graph query is not like other albums... it looks like this: http://www.facebook.com/media/set/?set=o.116860675007039 Another note: The only way i've come close to retreiving this data is by using the "feed" option.. ie: graph.facebook.com/pageid/feed EDIT: This is about as far as I could get- it works, but has certain issues stated below. Maybe someone could expand on this, or provide a better solution. (Using FB PHP SDK) <?php require_once ('config.php'); // get all photos for album $photos = $facebook->api("/YourID/tagged"); $maxitem =10; $count = 0; foreach($photos['data'] as $photo) { if ($photo['type'] == "photo"): echo "<img src='{$photo['picture']}' />", "<br />"; endif; $count+= 1; if ($count >= "$maxitem") break; } ?> Issues with this: 1) The fact that I don't know a method for graph querying specific "types" of Tags, I had to run a conditional statement to display photos. 2) You cannot effectively use the "?limit=#" with this, because as I said the "tagged" query contains all types (photo, video, and status). So if you are going for a photo gallery and wish to avoid running an entire query by using ?limit, you will lose images. 3) The only content that shows up in the "tagged" query is from people that are not Admins of the page. This isn't the end of the world, but I don't understand why Facebook wouldn't allow yourself to be shown in this data as long as you posted it "as yourself" and not as the page.

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  • PHP Parse Error unexpected '{'

    - by Laxmidi
    Hi, I'm getting a "Parse error: syntax error, unexpected '{' in line 2". And I don't see the problem. <?php class pointLocation {     var $pointOnVertex = true; // Check if the point sits exactly on one of the vertices     function pointLocation() {     }                   function pointInPolygon($point, $polygon, $pointOnVertex = true) {         $this->pointOnVertex = $pointOnVertex;                  // Transform string coordinates into arrays with x and y values         $point = $this->pointStringToCoordinates($point);         $vertices = array();          foreach ($polygon as $vertex) {             $vertices[] = $this->pointStringToCoordinates($vertex);          }                  // Check if the point sits exactly on a vertex         if ($this->pointOnVertex == true and $this->pointOnVertex($point, $vertices) == true) {             return "vertex";         }                  // Check if the point is inside the polygon or on the boundary         $intersections = 0;          $vertices_count = count($vertices);              for ($i=1; $i < $vertices_count; $i++) {             $vertex1 = $vertices[$i-1];              $vertex2 = $vertices[$i];             if ($vertex1['y'] == $vertex2['y'] and $vertex1['y'] == $point['y'] and $point['x'] > min($vertex1['x'], $vertex2['x']) and $point['x'] < max($vertex1['x'], $vertex2['x'])) { // Check if point is on an horizontal polygon boundary                 return "boundary";             }             if ($point['y'] > min($vertex1['y'], $vertex2['y']) and $point['y'] <= max($vertex1['y'], $vertex2['y']) and $point['x'] <= max($vertex1['x'], $vertex2['x']) and $vertex1['y'] != $vertex2['y']) {                  $xinters = ($point['y'] - $vertex1['y']) * ($vertex2['x'] - $vertex1['x']) / ($vertex2['y'] - $vertex1['y']) + $vertex1['x'];                  if ($xinters == $point['x']) { // Check if point is on the polygon boundary (other than horizontal)                     return "boundary";                 }                 if ($vertex1['x'] == $vertex2['x'] || $point['x'] <= $xinters) {                     $intersections++;                  }             }          }          // If the number of edges we passed through is even, then it's in the polygon.          if ($intersections % 2 != 0) {             return "inside";         } else {             return "outside";         }     }               function pointOnVertex($point, $vertices) {         foreach($vertices as $vertex) {             if ($point == $vertex) {                 return true;             }         }          }                   function pointStringToCoordinates($pointString) {         $coordinates = explode(" ", $pointString);         return array("x" => $coordinates[0], "y" => $coordinates[1]);     }           } $pointLocation = new pointLocation(); $points = array("30 19", "0 0", "10 0", "30 20", "11 0", "0 11", "0 10", "30 22", "20 20"); $polygon = array("10 0", "20 0", "30 10", "30 20", "20 30", "10 30", "0 20", "0 10", "10 0"); foreach($points as $key => $point) { echo "$key ($point) is " . $pointLocation->pointInPolygon($point, $polygon) . "<br>"; } ?> Does anyone see the problem? Thanks, -Laxmidi

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  • newbie problems with codeigniter

    - by Patrick
    hi, i'm trying to learn codeigniter (following a book) but don't understand why the web page comes out empty. my controller is class Welcome extends Controller { function Welcome() { parent::Controller(); } function index() { $data['title'] = "Welcome to Claudia's Kids"; $data['navlist'] = $this->MCats->getCategoriesNav(); $data['mainf'] = $this->MProducts->getMainFeature(); $skip = $data['mainf']['id']; $data['sidef'] = $this->MProducts->getRandomProducts(3, $skip); $data['main'] = "home"; $this->load->vars($data); $this->load->view('template'); } the view is: <--doctype declaration etc etc.. --> </head> <body> <div id="wrapper"> <div id="header"> <?php $this->load->view('header');?> </div> <div id='nav'> <?php $this->load->view('navigation');?> </div> <div id="main"> <?php $this->load->view($main);?> </div> <div id="footer"> <?php $this->load->view('footer');?> </div> </div> </body> </html> Now I know the model is passing back the right variables, but the page appears completely blank. I would expect at least to see an error, or the basic html structure, but the page is just empty. Moreover, the controller doesn't work even if I modify it as follows: function index() { echo "hello."; } What am I doing wrong? Everything was working until I made some changes to the model - but even if I delete all those new changes, the page is still blank.. i'm really confused! thanks, P.

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  • dynamically bound elements don't subscribe to CSS rules

    - by user961627
    I'm using ajax to dynamically populate a menu. The problem is the dynamically created elements do not follow the CSS rules, although they're created with the right classname. This is surprising. Here's my HTML: <div id='menu1'> <ul class='menu'> <?php $sql = /// /* some sql string, which works */ $result = mysql_query($sql); while($row = mysql_fetch_array( $result )) { $prod_id = $row['product_id']; $prod_name = $row['product_name']; echo "<li data-title='$prod_id' class='menu1_item'>".$prod_name."</li> "; } ?> </ul> </div> <div id='menu2'> <ul class='menu'> </ul> </div> <div id='menu3'> <ul class='menu'> </ul> </div> Here's my jquery: $(".menu1_item").click( function() { $.ajax({ type: "POST", data: { m: '2', id: $(this).attr('data-title') }, url: "fetch_designs.php", success: function(msg){ if (msg != ''){ $("#menu2").html(msg).show(); $(".menu2_item").bind('click',function() { $.ajax({ type: "POST", data: { m: '3', id: $(this).attr('data-title') }, url: "fetch_designs.php", success: function(msg){ if (msg != ''){ $("#menu3").html(msg).show(); } } }); //end menu2 click }); //end if } //end success } }); //end menu1 click }); My CSS is as follows: ul.menu li { cursor:pointer; list-style: none; } #menu1, #menu2, #menu3 { margin:50px; position:relative; display:block; float:left; } .menu1-item .menu2-item .menu3-item { padding:4px; background-color:lightgray; color:black; cursor:pointer; }

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  • tablednd post issue help please

    - by netrise
    Hi plz i got a terrible headache my script is very simple Why i can’t get $_POST['table-2'] after submiting update button, i want to get ID numbers sorted # index.php <head> <script src="jquery.js" type="text/javascript"></script><br /> <script src="jquery.tablednd.js" type="text/javascript"></script><br /> <script src="jqueryTableDnDArticle.js" type="text/javascript"></script><br /> </head> <body> <form method='POST' action=index.php> <table id="table-2" cellspacing="0" cellpadding="2"> <tr id="a"><td>1</td><td>One</td><td><input type="text" name="one" value="one"/></td></tr> <tr id="b"><td>2</td><td>Two</td><td><input type="text" name="two" value="two"/></td></tr> <tr id="c"><td>3</td><td>Three</td><td><input type="text" name="three" value="three"/></td></tr> <tr id="d"><td>4</td><td>Four</td><td><input type="text" name="four" value="four"/></td></tr> <tr id="e"><td>5</td><td>Five</td><td><input type="text" name="five" value="five"/></td></tr> </table> <input type="submit" name="update" value="Update"> </form> <?php $result[] = $_POST['table-2']; foreach($result as $value) { echo "$value<br/>"; } ?> </body> # jqueryTableDnDArticle.js …………. $(“#table-2?).tableDnD({ onDragClass: “myDragClass”, onDrop: function(table, row) { var rows = table.tBodies[0].rows; var debugStr = “Row dropped was “+row.id+”. New order: “; for (var i=0; i<rows.length; i++) { debugStr += rows[i].id+" "; } //$("#debugArea").html(debugStr); $.ajax({ type: "POST", url: "index.php", data: $.tableDnD.serialize(), success: function(html){ alert("Success"); } }); }, onDragStart: function(table, row) { $("#debugArea").html("Started dragging row "+row.id); } });

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  • MySQL error problem

    - by comma
    I keep getting the error listed below but it only says line 1 what does this mean and how do I fix it? Here is the error I keep getting? You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1 here is the code. if (isset($_POST['info_submitted'])) { $user_id = '5'; $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT learned_skills.*, users_skills.* FROM learned_skills INNER JOIN users_skills ON learned_skills.id = users_skills.skill_id WHERE user_id='$user_id'"); $skill = $_POST['skill']; $experience = $_POST['experience']; $year = $_POST['year']; if (mysqli_num_rows($dbc) == 0) { $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $query1 = mysqli_query($mysqli,"INSERT INTO learned_skills (skill, experience, year) VALUES ('" . $skill . "', '" . $experience . "', '" . $year . "')"); } if (!mysqli_query($mysqli, $query1)) { print mysqli_error($mysqli); return; } $mysqli = mysqli_connect("localhost", "root", "", "sitename"); $dbc = mysqli_query($mysqli,"SELECT id FROM learned_skills WHERE id='" . $skill . "' AND experience='" . $experience . "' AND year='" . $year . "'"); if (!$dbc) { print mysqli_error($mysqli); } else { while($row = mysqli_fetch_array($dbc)){ $id = $row["id"]; } } $query2 = "INSERT INTO users_skills (skill_id, user_id, date_created) VALUES ('$id', '$user_id', NOW())"; if (!mysqli_query($mysqli, $query2)) { print mysqli_error($mysqli); return; } if ($dbc == TRUE) { $dbc = mysqli_query($mysqli,"UPDATE learned_skills JOIN users_skills ON (users_skills.skill_id = learned_skills.id) SET skill = '$skill', experience = '$experience', year = '$year'"); echo '<p class="changes-saved">Your changes have been saved!</p>'; } if (!$dbc) { print mysqli_error($mysqli); return; } }

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  • How to tag photos in facebook-api?

    - by Camillo
    Hey, I wanted to ask if/how is it possible to tag a photo using the FB API (Graph or REST). I've managed to create an album and also to upload a photo in it, but I stuck on tagging. I've got the permissions and the correct session key. My code until now: try { $uid = $facebook->getUser(); $me = $facebook->api('/me'); $token = $session['access_token'];//here I get the token from the $session array $album_id = $album[0]; //upload photo $file= 'images/hand.jpg'; $args = array( 'message' => 'Photo from application', ); $args[basename($file)] = '@' . realpath($file); $ch = curl_init(); $url = 'https://graph.facebook.com/'.$album_id.'/photos?access_token='.$token; curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, false); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $args); $data = curl_exec($ch); //returns the id of the photo you just uploaded print_r(json_decode($data,true)); $search = array('{"id":', "}"); $delete = array("", ""); // picture id call with $picture $picture = str_replace($search, $delete, $data); //here should be the photos.addTag, but i don't know how to solve this //above code works, below i don't know what is the error / what's missing $json = 'https://api.facebook.com/method/photos.addTag?pid='.urlencode($picture).'&tag_text=Test&x=50&y=50&access_token='.urlencode($token); $ch = curl_init(); $url = $json; curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, false); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_POST, true); curl_exec($ch); } catch(FacebookApiException $e){ echo "Error:" . print_r($e, true); } I really searched a long time, if you know something that might help me, please post it here :) Thanks for all your help, Camillo

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  • get renamed file names of multiple upload form [js array] in codeigniter

    - by artmania
    Hi friends, I use codeigniter. I have a multiple image upload form. The code below is working well for uploading, but I also need to save file names to database. How can I get the names in here? I spent hours & hours :/ but could not sort it :/ Appreciate helps!!! uploadform.php echo form_open_multipart('gallery/upload'); <input type="file" name="photo" size="50" /> <input type="file" name="thumb" size="50" /> <input type="submit" value="Upload" /> </form> I have a controller between form view and model load model (of course : )) but didnt post here because of no need. gallery_model.php function multiple_upload($upload_dir = 'uploads/', $config = array()) { /* Upload */ $CI =& get_instance(); $files = array(); if(empty($config)) { $config['upload_path'] = realpath($upload_dir); $config['allowed_types'] = 'gif|jpg|jpeg|jpe|png'; $config['max_size'] = '2048'; } $CI->load->library('upload', $config); $errors = FALSE; foreach($_FILES as $key => $value) { if( ! empty($value['name'])) { if( ! $CI->upload->do_upload($key)) { $data['upload_message'] = $CI->upload->display_errors(ERR_OPEN, ERR_CLOSE); // ERR_OPEN and ERR_CLOSE are error delimiters defined in a config file $CI->load->vars($data); $errors = TRUE; } else { // Build a file array from all uploaded files $files[] = $CI->upload->data(); } } } // There was errors, we have to delete the uploaded files if($errors) { foreach($files as $key => $file) { @unlink($file['full_path']); } } elseif(empty($files) AND empty($data['upload_message'])) { $CI->lang->load('upload'); $data['upload_message'] = ERR_OPEN.$CI->lang->line('upload_no_file_selected').ERR_CLOSE; $CI->load->vars($data); } else { return $files; } /* ------------------------------- Insert to database */ // problem is here, i need file names to add db. // if there is already same names file at the folder, it rename file itself. so in such case, I need renamed file name :/ } }

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  • PHP 5.3: Late static binding doesn't work for properties when defined in parent class while missing in child class

    - by DavidPesta
    Take a look at this example, and notice the outputs indicated. <?php class Mommy { protected static $_data = "Mommy Data"; public static function init( $data ) { static::$_data = $data; } public static function showData() { echo static::$_data . "<br>"; } } class Brother extends Mommy { } class Sister extends Mommy { } Brother::init( "Brother Data" ); Sister::init( "Sister Data" ); Brother::showData(); // Outputs: Sister Data Sister::showData(); // Outputs: Sister Data ?> My understanding was that using the static keyword would refer to the child class, but apparently it magically applies to the parent class whenever it is missing from the child class. (This is kind of a dangerous behavior for PHP, more on that explained below.) I have the following two things in mind for why I want to do this: I don't want the redundancy of defining all of the properties in all of the child classes. I want properties to be defined as defaults in the parent class and I want the child class definition to be able to override these properties wherever needed. The child class needs to exclude properties whenever the defaults are intended, which is why I don't define the properties in the child classes in the above example. However, if we are wanting to override a property at runtime (via the init method), it will override it for the parent class! From that point forward, child classes initialized earlier (as in the case of Brother) unexpectedly change on you. Apparently this is a result of child classes not having their own copy of the static property whenever it isn't explicitly defined inside of the child class--but instead of throwing an error it switches behavior of static to access the parent. Therefore, is there some way that the parent class could dynamically create a property that belongs to the child class without it appearing inside of the child class definition? That way the child class could have its own copy of the static property and the static keyword can refer to it properly, and it can be written to take into account parent property defaults. Or is there some other solution, good, bad, or ugly?

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  • expand div with focus-jquery

    - by Joel
    Hi guys, I'm revisiting this after a few weeks, because I could never get it to work before, and hoping to now. Please look at this website-notice the newsletter signup form at the top right. http://www.rattletree.com I am wanting it to look exactly this way for now, but when the user clicks in the box to enter their email address, the containing div will expand (or simply appear) above the email field to also include a "name" and "city" field. I'm using jquery liberally in the sight, so that is at my disposal. This form is in the header so any id info, etc can't be withing the body tag... This is what I have so far: <div class="outeremailcontainer"> <div id="emailcontainer"> <?php include('verify.php'); ?> <form action="index_success.php" method="post" id="sendEmail" class="email"> <h3 class="register2">Newsletter Signup:</h3> <ul class="forms email"> <li class="email"><label for="emailFrom">Email: </label> <input type="text" name="emailFrom" class="info" id="emailFrom" value="<?= $_POST['emailFrom']; ?>" /> <?php if(isset($emailFromError)) echo '<span class="error">'.$emailFromError.'</span>'; ?> </li> <li class="buttons email"> <button type="submit" id="submit">Send</button> <input type="hidden" name="submitted" id="submitted" value="true" /> </li> </ul> </form> <div class="clearing"> </div> </div> css: p.emailbox{ text-align:center; margin:0; } p.emailbox:first-letter { font-size: 120%; font-weight: bold; } .outeremailcontainer { height:60px; width: 275px; background-image:url(/images/feather_email2.jpg); /*background-color:#fff;*/ text-align:center; /* margin:-50px 281px 0 auto ; */ float:right; position:relative; z-index:1; } form.email{ position:relative; } #emailcontainer { margin:0; padding: 0 auto; z-index:1000; display:block; position:relative; } Thanks for any help! Joel

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  • How to change the JSON output format and how to support chinese character?

    - by sky
    Currently I using the following code to get my JSON output from MySQL. <?php $session = mysql_connect('localhost','name','pass'); mysql_select_db('dbname', $session); $result= mysql_query('SELECT message FROM posts', $session); $somethings = array(); while ($row = mysql_fetch_assoc($result)) { $somethings[] = $row; } ?> <script type="text/javascript"> var somethings= <?php echo json_encode($somethings); ?>; </script> And the output is: <script type="text/javascript"> var somethings= [{"message":"Welcome to Yo~ :)"},{"message":"Try iPhone post!"},{"message":"????"}]; </script> Here is the question, how can I change my output into format like : <script type="text/javascript"> userAge = new Array('21','36','20'), userMid = new Array('liuple','anhu','jacksen'); </script> Which I'll be using later with following code : var html = ' <table class="map-overlay"> <tr> <td class="user">' + '<a class="username" href="/' + **userMid[index]** + '" target="_blank"><img alt="" src="' + getAvatar(signImgList[index], '72x72') + '"></a><br> <a class="username" href="/' + **userMid[index]** + '" target="_blank">' + userNameList[index] + '</a><br> <span class="info">' + **userSex[index]** + ' ' + **userAge[index]** + '?<br> ' + cityList[index] + '</span>' + '</td> <td class="content">' + picString + somethings[index] + '<br> <span class="time">' + timeList[index] + picTips + '</span></td> </tr> </table> '; Thanks for helping and reading!

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  • Inserting checkbox values

    - by rabeea
    hey i have registration form that has checkboxes along with other fields. i cant insert the selected checkbox values into the data base. i have made one field in the database for storing all checked values. this is the code for checkbox part in the form: Websites, IT and Software Writing and Content <pre><input type="checkbox" name="expertise[]" value="Design and Media"> Design and Media <input type="checkbox" name="expertise[]" value="Data entry and Admin"> Data entry and Admin </pre> <pre><input type="checkbox" name="expertise[]" value="Engineering and Skills"> Engineering and Science <input type="checkbox" name="expertise[]" value="Seles and Marketing"> Sales and Marketing </pre> <pre><input type="checkbox" name="expertise[]" value="Business and Accounting"> Business and Accounting <input type="checkbox" name="expertise[]" value="Others"> Others </pre> and this is the corresponding php code for inserting data $checkusername=mysql_query("SELECT * FROM freelancer WHERE fusername='{$_POST['username']}'"); if (mysql_num_rows($checkusername)==1) { echo "username already exist"; } else { $query = "insert into freelancer(ffname,flname,fgender,femail,fusername,fpwd,fphone,fadd,facc,facc_name,fbank_details,fcity,fcountry,fexpertise,fprofile,fskills,fhourly_rate,fresume) values ('".$_POST['first_name']."','".$_POST['last_name']."','".$_POST['gender']."','".$_POST['email']."','".$_POST['username']."','".$_POST['password']."','".$_POST['phone']."','".$_POST['address']."','".$_POST['acc_num']."','".$_POST['acc_name']."','".$_POST['bank']."','".$_POST['city']."','".$_POST['country']."','".implode(',',$_POST['expertise'])."','".$_POST['profile']."','".$_POST['skills']."','".$_POST['rate']."','".$_POST['resume']."')"; $result = ($query) or die (mysql_error()); this code inserts data for all fields but the checkbox value field remains empty???

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  • PHP: Displaying Dom object and Creating xml file

    - by pavun_cool
    <?php $books = array(); $books [] = array( 'title' => 'PHP Hacks', 'author' => 'Jack Herrington', 'publisher' => "O'Reilly" ); $books [] = array( 'title' => 'Podcasting Hacks', 'author' => 'Jack Herrington', 'publisher' => "O'Reilly" ); $doc = new DOMDocument(); $doc->formatOutput = true; $r = $doc->createElement( "books" ); $doc->appendChild( $r ); foreach( $books as $book ) { $b = $doc->createElement( "book" ); $author = $doc->createElement( "author" ); $author->appendChild( $doc->createTextNode( $book['author'] ) ); #$author->appendChild( $doc->createTextNode( 'pavunkumar')); $new = $doc->createElement("Developer"); $a=$doc->createTextNode('I am developer ' ); $new->appendChild($a); $b->appendChild( $author ); $b->appendChild($new); $b->appendChild($new); $title = $doc->createElement( "title" ); $title->appendChild( $doc->createTextNode( $book['title'] ) ); $b->appendChild( $title ); $publisher = $doc->createElement( "publisher" ); $publisher->appendChild( $doc->createTextNode( $book['publisher'] ) ); $b->appendChild( $publisher ); $r->appendChild( $b ); } echo $doc->SaveXml() ; ?> When I run this code in command line. I am getting following things <?xml version="1.0"?> <books> <book> <author>Jack Herrington</author> <Developer>I am developer </Developer> <title>PHP Hacks</title> <publisher>O'Reilly</publisher> </book> <book> <author>Jack Herrington</author> <Developer>I am developer </Developer> <title>Podcasting Hacks</title> <publisher>O'Reilly</publisher> </book> </books> When I run the code in web browser it gives me following things Jack Herrington I am developer O'Reilly Jack Herrington I am developer O'Reilly I want to above output to be like command line output. And one more things is that instead of displaying , how could I create a xml file using $doc Dom object.

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  • How can I create a Searchstring for a Google AJAX Search API?

    - by elmaso
    Hello, i have this code to get the search resutls from the api: querygoogle.php: <?php session_start(); // Here's the Google AJAX Search API url for curl. It uses Google Search's site:www.yourdomain.com syntax to search in a specific site. I used $_SERVER['HTTP_HOST'] to find my domain automatically. Change $_POST['searchquery'] to your posted search query $url = 'http://ajax.googleapis.com/ajax/services/search/web?rsz=large&v=1.0&start=20&q=' . urlencode('' . $_POST['searchquery']); // use fopen and fread to pull Google's search results $handle = fopen($url, 'rb'); $body = ''; while (!feof($handle)) { $body .= fread($handle, 8192); } fclose($handle); // now $body is the JSON encoded results. We need to decode them. $json = json_decode($body); // now $json is an object of Google's search results and we need to iterate through it. foreach($json->responseData->results as $searchresult) { if($searchresult->GsearchResultClass == 'GwebSearch') { $formattedresults .= ' <div class="searchresult"> <h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3> <p class="resultdesc">' . $searchresult->content . '</p> <p class="resulturl">' . $searchresult->visibleUrl . '</p> </div>'; } } $_SESSION['googleresults'] = $formattedresults; header('Location: ' . $_SERVER['HTTP_REFERER']); exit; ?> search.php <?php session_start(); ?> <form method="post" action="querygoogle.php"> <label for="searchquery"><span class="caption">Search this site</span> <input type="text" size="20" maxlength="255" title="Enter your keywords and click the search button" name="searchquery" /></label> <input type="submit" value="Search" /> </form> <?php if(!empty($_SESSION['googleresults'])) { echo $_SESSION['googleresults']; unset($_SESSION['googleresults']); } ?> but with this code, I cant add a searchstring.. how can i add a search string like search.php?search=keyword ? thanks

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  • PHP: Remove Simple Session with Get-Method

    - by elmaso
    Hello, I want to Remove the Sessions from this php code, actually if someone searches i get this url search.php?searchquery=test but if I reload the page, the results are cleaned. how can I remove the Sessions to get the Results still, if someone reloads the page? this are the codes: search.php <?php session_start(); ?> <form method="get" action="querygoogle.php"> <label for="searchquery"><span class="caption">Search this site</span> <input type="text" size="20" maxlength="255" title="Enter your keywords and click the search button" name="searchquery" /></label> <input type="submit" value="Search" /> </form> <?php if(!empty($_SESSION['googleresults'])) { echo $_SESSION['googleresults']; unset($_SESSION['googleresults']); } ?> querygoogle.php <?php session_start(); $url = 'http://www.example.com'; $handle = fopen($url, 'rb'); $body = ''; while (!feof($handle)) { $body .= fread($handle, 8192); } fclose($handle); $json = json_decode($body); foreach($json->responseData->results as $searchresult) { if($searchresult->GsearchResultClass == 'GwebSearch') { $formattedresults .= ' <div class="searchresult"> <h3><a href="' . $searchresult->unescapedUrl . '">' . $searchresult->titleNoFormatting . '</a></h3> <p class="resultdesc">' . $searchresult->content . '</p> <p class="resulturl">' . $searchresult->visibleUrl . '</p> </div>'; } } $_SESSION['googleresults'] = $formattedresults; header("Location: search.php?searchquery=" . $_GET['searchquery']); exit; ?> thank you for your help!!

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