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  • Right way to handle friend-multi-select post on facebook, issues with session

    - by simple
    Do I need a infinite key with fbml facebook app that resides on fanpage? I am asking user to select user and posting it(facebook posts) to my server. On my server I want to get user_id and selected friends Id. Everything is fine with selected friend_ids, but have issues with getting id of a user. sometimes I can get it all fine, sometimes I am getting session expired exception, sometimes I get nothing. any ideas why this is happening?

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  • How can I join 3 tables with mysql & php?

    - by steven
    check out the page [url]http://www.mujak.com/test/test3.php[/url] It pulls the users Post,username,xbc/xlk tags etc which is perfect... BUT since I am pulling information from a MyBB bulletin board system, its quite different. When replying, people are are allowed to change the "Thread Subject" by simplying replying and changing it. I dont want it to SHOW the changed subject title, just the original title of all posts in that thread. By default it repies with "RE:thread title". They can easily edit this and it will show up in the "Subject" cell & people wont know which thread it was posted in because they changed their thread to when replying to the post. So I just want to keep the orginial thread title when they are replying. Make sense~?? Tables:mybb_users Fields:uid,username Tables:mybb_userfields Fields:ufid Tables:mybb_posts Fields:pid,tid,replyto,subject,ufid,username,uid,message Tables:mybb_threads Fields:tid,fid,subject,uid,username,lastpost,lastposter,lastposteruid I haev tried multiple queries with no success: $result = mysql_query(" SELECT * FROM mybb_users LEFT JOIN (mybb_posts, mybb_userfields, mybb_threads) ON ( mybb_userfields.ufid=mybb_posts.uid AND mybb_threads.tid=mybb_posts.tid AND mybb_users.uid=mybb_userfields.ufid ) WHERE mybb_posts.fid=42"); $result = mysql_query(" SELECT * FROM mybb_users LEFT JOIN (mybb_posts, mybb_userfields, mybb_threads) ON ( mybb_userfields.ufid=mybb_posts.uid AND mybb_threads.tid=mybb_posts.tid AND mybb_users.uid=mybb_posts.uid ) WHERE mybb_threads.fid=42"); $result = mysql_query(" SELECT * FROM mybb_posts LEFT JOIN (mybb_userfields, mybb_threads) ON ( mybb_userfields.ufid=mybb_posts.uid AND mybb_threads.tid=mybb_posts.tid ) WHERE mybb_posts.fid=42");

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  • Why won't the following PDO transaction won't work in PHP?

    - by jfizz
    I am using PHP version 5.4.4, and a MySQL database using InnoDB. I had been using PDO for awhile without utilizing transactions, and everything was working flawlessly. Then, I decided to try to implement transactions, and I keep getting Internal Server Error 500. The following code worked for me (doesn't contain transactions). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } Then I attempted to utilize transactions with the following code (which doesn't work). try { $DB = new PDO('mysql:host=localhost;dbname=database', 'root', 'root'); $DB->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); $dbh = $DB->beginTransaction(); $dbh->prepare("SELECT * FROM user WHERE username = :test"); $dbh->bindValue(':test', $test, PDO::PARAM_STR); $dbh->execute(); $dbh->commit(); } catch(Exception $e){ $dbh->rollback(); echo "an error has occured"; } When I run the previous code, I get an Internal Server Error 500. Any help would be greatly appreciated! Thanks!

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  • What is a good RPC model for building a AJAX web app using PHP & JS?

    - by user366152
    I'm new to writing AJAX applications. I plan on using jQuery on the client side while PHP on the server side. I want to use something like XML-RPC to simplify my effort in calling server-side code. Ideally, I wouldn't care whether the transport layer uses XML or JSON or a format more optimized for the wire. If I was writing a console app I'd use some tool to generate function stubs which I would then implement on the RPC server while the client would natively call into those stubs. This provides a clean separation. Is there something similar available in the AJAX world? While on this topic, how would I proceed with session management? I would want it to be as transparent as possible. For example, if I try to hit an RPC end-point which needs a valid session, it should reject the request if the client doesn't pass a valid session cookie. This would really ease my application development. I'd then have to simply handle the frontend using native JS functions. While on the backend, I can simply implement the RPC functions. BTW I dont wish to use Google Web Toolkit. My app wont be extremely heavy on AJAX.

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  • php weird bug where an array is not an array !

    - by iko
    I've been going mad trying to figure out why an array would not be an array in php. For a reason I can't understand I have a bug in a smarty class. The code is this : $compiled_tags = array(); for ($i = 0, $for_max = count($template_tags); $i < $for_max; $i++) { $this->_current_line_no += substr_count($text_blocks[$i], "\n"); // I tried array push instead to see // bug is here array_push($compiled_tags,$this->_compile_tag($template_tags[$i])); //$compiled_tags[] = $this->_compile_tag($template_tags[$i]); $this->_current_line_no += substr_count($template_tags[$i], "\n"); } the error message is Warning: array_push() expects parameter 1 to be array, integer given in .... OR before with [] Warning: Cannot use a scalar value as an array in .... I trying a var_debug on $compiled_tags and as soon I enter the for loop is not an array anymore but an integer. I tried renaming the variable, but same problem. I'm sure is something simple that I missed but I can't figure it out. Any help is (as always) welcomed !

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  • Are there any Open source Admin Panel based on PHP and jQuery?

    - by Pablo
    I've try many CMS flavors like MODx, Drupal, Joomla for the admin Panel but they can't seem to manage my existing data, like the users. Im planing on building a admin Control Panel for my site and i was wondering if there is something out there which i can start working with instead of starting from scratch. UPDATE: To do simple tasks like: Managing my users Managing my content( the site is an image sharing site, so in this case images) I was looking more into a graphical interface more then a system as i have lots of costume content and data like the users.

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  • Does a native php (5+) function exist that does the following in 1 line?

    - by Vinh
    function array_value_from_key($array,$key) { return !empty($array[$key]) ? $array[$key] : null; } The reason I ask is because I have a class function that returns an array. Instead of having to do $myArray = myClass::giveMeArray(); $myValue = $myArray[$myKey]; I'd like to do something along the lines of $myValue = array_value_from_key(myClass::giveMeArray(),$myKey); When an object is returned, you can chain the object such as $myValue = myClass::giveMeObject()->aValue; Voila, nice and clean.. not being able to find what seems to be a simple and trivial function is driving me crazy... PS.. one more example of how I'd like to use such a function if(arrayKeyVal(aClass::giveMeArray(),$myKey)) { do_something(); }

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  • Can anyone explain this impossible bit of PHP logic?

    - by user268208
    I'm attempting to debug a simple PHP script. Essentially, there's a variable which is defined with: $variable = ($_GET['variable'] == 'true') ? TRUE : FALSE; Then, in the view file, the following code is meant to display a box if $variable == TRUE: <? if ($variable == true) { ?> <p class="box">You have imported a new plan.</p> <? } ?> Now, even when that $variable, as shown by var_dump($variable); == FALSE, that HTML is printed between the if { } tags. To me, this defies logic. I simply can't figure out this problem out. Furthermore, this code works fine on many PHP4 and PHP5 installations except for one particular server running PHP5.2. Any possible suggestions? Leads? I'm pulling out my hair trying to figure this one out. Thank you.

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  • How to get values from SQL query made by php?

    - by Ole Jak
    So I made a query like this global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); in my DB there are filds ID, UID, SID, TIME (all INT type exept time) So I am triing to print query relult into form <form> <select class="multiselect" multiple="multiple" name="SIDs"> <?php global $connection; $query = "SELECT * FROM streams "; $streams_set = mysql_query($query, $connection); confirm_query($streams_set); $streams_count = mysql_num_rows($streams_set); for ($count=1; $count <= $streams_count; $count++) { echo "<option value=\"{$count}\""; echo ">{$count}</option>"; } ?> </select> <br/> <input type="submit" value="Submit Form"/> </form> How to print out as "option" "values" SID's from my sql query?

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  • How to write data option in jQuery.ajax() function when it include in a mysql_query?

    - by cj333
    I modify a php comment system. I want add it after every article witch are query from database. this is the php part <?php ... while($result = mysql_fetch_array($resultset)) { $article_title = $result['article_title']; ... ?> <form id="postform" class="postform"> <input type="hidden" name="title" id="title" value="<?=$article_title;?>" /> <input type="text" name="content" id="content" /> <input type="button" value="Submit" class="Submit" /> </form> ... <?php } ?> this is the ajax part. $.ajax({ type: "POST", url: "ajax_post.php", data: {title:$('#title').val(), content:$('#content').val() ajax_post.php echo $title; echo $content; How to modify the ajax data part that each article's comment can send each data to the ajax_post.php? thanks a lot.

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  • PHP - How do you secure a unique variable name?

    - by 102319141763223461745
    This function cropit, which I shamelessly stole off the internet, crops a 90x60 area from an existing image. In this code, when I use the function for more than one item (image) the one will display on top of the other (they come to occupy the same output space). I think this is because the function has the same (static) name ($dest) for the destination of the image when it's created (imagecopy). I tried, as you can see to include a second argument to the cropit function which would serve as the "name" of the $dest variable, but it didn't work. In the interest of full disclosure I have 22 hours of PHP experience (incidentally the same number of hours since the last I slept) and I am not that smart to begin with. Even if there's something else at work here entirely, seems to me that generally it must be useful to have a way to secure that a variable is always given a unique name. function cropit($srcimg, $dest) { $im = imagecreatefromjpeg($srcimg); $img_width = imagesx($im); $img_height = imagesy($im); $width = 90; $height = 60; $tlx = floor($img_width / 2) - floor ($width / 2); $tly = floor($img_height / 2) - floor ($height / 2); if ($tlx < 0) { $tlx = 0; } if ($tly < 0) { $tly = 0; } if (($img_width - $tlx) < $width) { $width = $img_width - $tlx; } if (($img_height - $tly) < $height) { $height = $img_height - $tly; } $dest = imagecreatetruecolor ($width, $height); imagecopy($dest, $im, 0, 0, $tlx, $tly, $width, $height); imagejpeg($dest); imagedestroy($dest); } $img = "imagefolder\imageone.jpg"; $img2 = "imagefolder\imagetwo.jpg"; cropit($img, $i1); cropit($img2, $i2); ?

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  • Can I check if e-mail address is valid?

    - by simple
    How can I implement following logic? User registers with an e-mail address If provided e-mail address is a valid email address Then user account get's activated or if it is a fake email then user account is not activated I doubt that I can catch the - "Delivery failed reply message", right? anyhow how would you suggest to implement the above logic? PS. I will have to find a way no matter what, client wants it =)

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  • PHP class extends not working why and is this how to correctly extend a class?

    - by Matthew
    Hi so I'm trying to understand how inherteince works in PHP using object oriented programming. The main class is Computer, the class that is inheriting is Mouse. I'm extedning the Computer class with the mouse class. I use __construct in each class, when I istinate the class I use the pc type first and if it has mouse after. For some reason computer returns null? why is this? class Computer { protected $type = 'null'; public function __construct($type) { $this->type = $type; } public function computertype() { $this->type = strtoupper($this->type); return $this->type; } } class Mouse extends Computer { protected $hasmouse = 'null'; public function __construct($hasmouse){ $this->hasmouse = $hasmouse; } public function computermouse() { if($this->hasmouse == 'Y') { return 'This Computer has a mouse'; } } } $pc = new Computer('PC', 'Y'); echo $pc->computertype; echo $pc->computermouse;

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  • PHP, better to set the variable before if or use if/else?

    - by DssTrainer
    So a simple one that I just never could find a straight answer on. What is better (performance or otherwise): $var = false; If ($a == $b) { $var = true; } or If ($a == $b) { $var = true; } else { $var = false; } I've heard arguments for both ways. I find the first cleaner to ensure I have it set, and a little less code too. The pro being that you may only need to set it once without conditional. But the con being that if the argument is true, it gets set twice. I am assuming the second way is probably best practice

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  • ZF2 namespace and file system

    - by user1918648
    I have standard Application module in ZF2. It's configured by default, I didn't change anything. I just added some stuff: module/ Application/ src/ Application/ Entity/ Product/ **Product.php** Controller/ **IndexController.php** Product.php namespace Application\Entity; class Product { } IndexController.php namespace Application\Controller; use Zend\Mvc\Controller\AbstractActionController; use Zend\View\Model\ViewModel; use Application\Entity\Product; class IndexController extends AbstractActionController { public function indexAction() { $product = new Product(); } } and I get following error: Fatal error: Class 'Application\Entity\Product' not found in \module\Application\src\Application\Controller\IndexController.php on line 20 I use the same namespace, but it doesn't see it. Why? P.S: If I will change Product.php to be the following: namespace Application\Entity\Product; class Product { } then in the IndexController.php the following code will be working: namespace Application\Controller; use Zend\Mvc\Controller\AbstractActionController; use Zend\View\Model\ViewModel; use Application\Entity\Product\Product; class IndexController extends AbstractActionController { public function indexAction() { $product = new Product(); } }

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  • How can I use Amazon's API in PHP to search for books?

    - by TerranRich
    I'm working on a Facebook app for book sharing, reviewing, and recommendations. I've scoured the web, searched Google using every search phrase I could think of, but I could not find any tutorials on how to access the Amazon.com API for book information. I signed up for an AWS account, but even the tutorials on their website didn't help me one bit. They're all geared toward using cloud computing for file storage and processing, but that's not what I want. I just want to access their API to search info on books. Kind of like how http://openlibrary.org/ does it, where it's a simple URL call to get information on a book (but their databases aren't nearly as populated as Amazon's). Why is it so hard to find the information I need on Amazon's AWS site? If anybody could help, I would greatly appreciate it.

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  • How to check successful copy() function execution in php?

    - by OM The Eternity
    How to check successful copy() function execution in php? I am using the following code: <? function full_copy( $source, $target ) { if ( is_dir( $source ) ) { @mkdir( $target ); $d = dir( $source ); while ( FALSE !== ( $entry = $d->read() ) ) { if ( $entry == '.' || $entry == '..' ) { continue; } $Entry = $source . '/' . $entry; if ( is_dir( $Entry ) ) { full_copy( $Entry, $target . '/' . $entry ); continue; } copy( $Entry, $target . '/' . $entry ); } $d->close(); }else { copy( $source, $target ); } } $source ='.'; $destination = '/html/parth/'; full_copy($source, $destination); ?> I do not get anything in my parth folder. Why? I am using Windows, Script is executed on fedora system (its my server)...

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  • object-oriented question

    - by user522962
    I am attempting to put all my database connections in 1 php file, rather than in each of my individual php pages. I have the following: //conn.php: <?php class conn { var $username = "name"; var $password = "password"; var $server = "localhost"; var $port = "3306"; var $databasename = "db"; var $tablename = "tablename"; var $connection; public function getConnected() { $this->connection = mysqli_connect( $this->server, $this->username, $this->password, $this->databasename, $this->port ); } } ?> // file.php: <?php require_once("conn.php"); class myClass{ public function con() { $conn = new conn(); $conn->getConnected(); } public function myF() { $stmt = mysqli_prepare($conn->connection, "SELECT * FROM $conn->tablename"); mysqli_stmt_execute($stmt); } } ?> I then call this as follows: $myNew = new myClass(); $myNew-con(); $myNew-myF(); When I call this, I get the following error: Undefined property: myClass::$connection What am I doing wrong?

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  • Why does this sql statement keep saying it is a boolean and not a parameter? (php/Mysql)

    - by ggfan
    In this statement, I am trying to see if there if the latest posting in the database that has the exact same title, price, city, state, detail. If there is, then it would say to the user that the exact post has been already made; if not then insert the posting into the dbc. (This is one type of check so that users can't accidentally post twice. This may not be the best check, but this statement error is annoying me, so I want it to work :)) Why won't this sql work? I think it's not letting the title=$title and not getting the value in the $title... ERROR: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in postad.php on line 365 //there is a form that users fill out that has title, price, city, etc <form> blah blah </form> //if users click submit, then does all the checks and if all okay, insert to dbc if (isset($_POST['submit'])) { // Grab the pposting data from the POST and gets rid of any funny stuff $title = mysqli_real_escape_string($dbc, trim($_POST['title'])); $price = mysqli_real_escape_string($dbc, trim($_POST['price'])); $city = mysqli_real_escape_string($dbc, trim($_POST['city'])); $state = mysqli_real_escape_string($dbc, trim($_POST['state'])); $detail = mysqli_real_escape_string($dbc, trim($_POST['detail'])); if (!is_numeric($price) && !empty($price)) { echo "<p class='error'>The price can only be numbers. No special characters, etc</p>"; } //Error problem...won't let me set title=$title, detail=$detail, etc. //this statement after all the checks so that none of the variables are empty $query="Select * FROM posting WHERE user_id={$_SESSION['user_id']} AND title=$title AND price=$price AND city=$city AND state=$state AND detail=$detail"; $data = mysqli_query($dbc, $query); if(mysqli_num_rows($data)==1) { echo "You already posted this ad. Most likely caused by refreshing too many times."; } }

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  • PHP "header (location)" isnide IFRAME, to load in _top location?

    - by Spoonk
    Hi Webmasters. I have a simple form which is inside IFRAME. When user click on SUBMIT, it redirects to a specific page on my server. The function I use for the redirect is header ('Location: mypage2.html'); exit (); But I want the new page to open in _top location, not inside the same IFRAME that I use. How can I tell the browser to open the new page in _top not inside the IFRAME? Thanks in advance.

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