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  • How to pair users? (Like Omegle.com)

    - by Carlos Dubus
    Hi, I'm trying to do an Omegle.com clone script, basically for learning purposes. I'm doing it in PHP/MySQL/AJAX. I'm having problems finding two users and connecting them. The purpose of omegle is connecting two users "randomly". What I'm doing right now is the following: When a user enters the website a session is assigned. There are 3 states for each session/user (Normal,Waiting,Chatting) At first the user has state Normal and a field "connected_to" = NULL If the users clicks the START button, a state of "Waiting" is assigned. Then it looks for another user with state Waiting, if doesn't find one then it keeps looping, waiting for the "connected_to" to change. The "connected_to" will change when other user click START and then find another user waiting and updates the sessions accordingly. Now this have several problems, like: A user only can be connected to one user at a time. In omegle you can open more than one chat simultaneously. I don't know if this is the best way. About the chat, each user is polling the events from the server with AJAX calls, I saw that omegle, instead of several HTTP requests each second (let's say), does ONE request and wait for an answer, that means that the PHP script is looping indefinitely until gets an answer.I did this using set_time_limit(30) each time the loop is started. Then when the Ajax call is done start over again. Is this approach correct? I will appreciate a LOT your answers, Thank you, Carlos

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  • Jquery Sorting by Letter

    - by Batfan
    I am using jquery to sort through a group of paragraph tags (kudos to Aaron Harun). It pulls the value "letter" (a letter) from the url string and displays only paragraphs that start with that letter. It hides all others and also consolidates the list so that there are no duplicates showing. See the code: var letter = '<?php echo(strlen($_GET['letter']) == 1) ? $_GET['letter'] : ''; ?>' function finish(){ var found_first = []; jQuery('p').each(function(){ if(jQuery(this).text().substr(0,1).toUpperCase() == letter){ if(found_first[jQuery(this).text()] != true){ jQuery(this).addClass('current-series'); found_first[jQuery(this).text()] = true; }else{ jQuery(this).hide(); } } else{ jQuery(this).hide();} }) } Been working with this all day and I have 2 Questions on this: Is there a way to get it to ignore the word 'The', if it's first? For example, if a paragraph starts with 'The Amazing', I would like it to show up on the 'A' page, not the 'T' page, like it currently is. Is there a way to have a single page for (all) numbers? For example, the url to the page would be something similar to domain.com/index.php?letter=0 and this would show only the paragraph tags that start with a number, any number. I can currently do this with single numbers but, I would like 1 page for all numbers.

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  • Upload file onto Server from the IPhone using ASIHTTPRequest

    - by Nick
    I've been trying to upload a file (login.zip) using the ASIHTTPRequest libraries from the IPhone onto the inbuilt Apache Server in Mac OS X Snow Leopard. My code is: NSString *urlAddress = [[[NSString alloc] initWithString:self.uploadField.text]autorelease]; NSURL *url = [NSURL URLWithString:urlAddress]; ASIFormDataRequest *request; NSArray *paths = NSSearchPathForDirectoriesInDomains(NSDocumentDirectory,NSUserDomainMask, YES); NSString *documentsDirectory = [paths objectAtIndex:0]; NSString *dataPath = [documentsDirectory stringByAppendingPathComponent:@"login.zip"]; NSData *data = [[[NSData alloc] initWithContentsOfFile:dataPath] autorelease]; request = [[[ASIFormDataRequest alloc] initWithURL:url] autorelease]; [request setPostValue:@"login.zip" forKey:@"file"]; [request setData:data forKey:@"file"]; [request setUploadProgressDelegate:uploadProgress]; [request setShowAccurateProgress:YES]; [request setDelegate:self]; [request startAsynchronous]; The php code is : <?php $target = "upload/"; $target = $target . basename( $_FILES['uploaded']['name']) ; $ok=1; if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) { echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded"; } else { echo "Sorry, there was a problem uploading your file."; } ?> I don't quite understand why the file is not uploading. If anyone could help me. I've stuck on this for 5 days straight. Thanks in advance Nik

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  • explode is not working to split string

    - by pmms
    we unable to split the string following code.please Help us. <?php $i=0; $myFile = "testFile.txt"; $fh = fopen($myFile, 'a') or die("can't open file"); $stringData = "no\t"; fwrite($fh, $stringData); $stringData = "username \t"; fwrite($fh, $stringData); $stringData ="password \t"; fwrite ($fh,$stringData); $newline ="\r\n"; fwrite ($fh,$newline); $stringData1 = "1\t"; fwrite($fh, $stringData1); $stringData1 = "srinivas \t"; fwrite($fh, $stringData1); $stringData1 ="malayappa \t"; fwrite ($fh,$stringData1); fclose($fh); ?> $fh = fopen("testFile.txt", "r"); $ while (!feof($fh)) { $line = fgets($fh); echo $line; } fclose($fh); $Beatles = array('pmm','malayappa','sreenivas','PHP'); for($i=0;$i<count($Beatles);$i++) { if($i==2) { echo $Beatles[$i-1]; echo $Beatles[$i-2]; } } $pass_ar=array(); $fh = fopen("testFile.txt", "r"); while (!feof($fh)) { $line = fgets($fh); echo $line; $t1=explode(" ",$line); print_r($t1); array_push($pass_ar,t1); } fclose($fh); Thanks & Regards pmms

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  • How to populate html combo box with mysql data

    - by user225269
    Please help, I'm having trouble loading mysql data on the combo box. The data that I'm loading is 1 column from a table. Here is my current code, and it crashed firefox for some reason: <td colspan=”2?>Religion</TD> <td> <select name="REL" onClick="submitCboSemester();"> <?php $query_disp="SELECT * FROM Religion ORDER BY RID"; $result_disp = mysql_query($query_disp, $conn); while($query_data = mysql_fetch_array($result_disp)) { ?> <option value="<? echo $query_data["RID"]; ?>"<?php if ($query_data["RID"]==$_POST['REL']) {?>selected<? } ?>><? echo $query_data["RELIGION"]; ?></option> <? } ?> </select> </td> The column is RELIGION and it ID is RID How do I populate the combo box with all the data in the column RELIGION

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  • How do I use curl to display a table on a page?

    - by user272899
    I want to use curl to retrieve a table from an external page. So far my code retrieves all data from the page. I have read that using preg_match or preg_replace is the way to go about it. This is my code so far: <?php $ch = curl_init() or die(curl_error()); curl_setopt($ch, CURLOPT_URL,"http://www.megaupload.com/?d=XE30L1GA&w=631&h=392"); curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); $data1=curl_exec($ch) or die(curl_error()); echo "<font color=black face=verdana size=3>".$data1."</font>"; echo curl_error($ch); curl_close($ch); ?> This is the data I want to retrieve from the page: <FORM method="POST" id="captchaform"> <INPUT type="hidden" name="captchacode" value="1a589e0a53c54f937eb8"> <INPUT type="hidden" name="megavar" value="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"> <TR> <TD>Enter this </TD> <TD width="100" align="center" height="40"><img src="http://wwwq32.megaupload.com/gencap.php?23243f17c2404dd4.gif" border="0" alt=""></TD> <TD> here:</TD> <TD><input type="text" name="captcha" id="captchafield" maxlength="4" style="border:solid 1px; border-color:#C4C4C4; background-color:#F9F9F9; width:50px; height:25px;"></TD> </TR> </FORM>

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  • Using a join with three tables when a field might be null

    - by John
    Hello, The code below works great. It combines data from two MySQL tables. I would like to modify it by pulling in some data from a third MySQL table called "comment." In the HTML table below, "title" is a field in the MySQL table "submission." Every "title" has a corresponding "submissionid" field. The field "submissionid" is also found in the "comment" MySQL table. In the HTML table below, I would like "countComments" to equal the number of times a field called "commentid" appears in the MySQL table "comment" for any given "submissionid," where the "submissionid" is the same in both the "submission" and "comment" tables, and where the "submissionid" corresponds to the "title" being used. Here's the catch: if there is no row in the MySQL table "comment" that corresponds with the "submissionid" being used for "table", I would like "countComments" to equal to zero. How can I do this? Thanks in advance, John $sqlStr = "SELECT s.loginid, s.title, s.url, s.displayurl, l.username FROM submission AS s, login AS l WHERE s.loginid = l.loginid ORDER BY s.datesubmitted DESC LIMIT 10"; $result = mysql_query($sqlStr); $arr = array(); echo "<table class=\"samplesrec\">"; while ($row = mysql_fetch_array($result)) { echo '<tr>'; echo '<td class="sitename1"><a href="http://www.'.$row["url"].'">'.$row["title"].'</a></td>'; echo '</tr>'; echo '<tr>'; echo '<td class="sitename2"><a href="http://www...com/sandbox/members/index.php?profile='.$row["username"].'">'.$row["username"].'</a><a href="http://www...com/sandbox/comments/index.php?submission='.$row["title"].'">'.$row["countComments"].'</a></td>'; echo '</tr>'; } echo "</table>";

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  • Open source real life license examples: yours or others

    - by donpal
    I'm aware of the usual list of open source licenses, so I'm not even going to list it here. What I'd like to ask is about your open source projects (whether out or planned for the future), and why you're planning to choose a certain license over the other. Basically say I went for X license because I wanted Y and that other license didn't provide it for me. I understand that the language itself can make a difference in the choice of license: interpreted languages like PHP vs. compiled languages like Java. I'm mostly interested in hearing about PHP projects, but of course additional insights are welcome. You may even have chosen that particular language for a licensing reason. Ideally I want to hear answers from people who were involved in the actual project (i.e. your own project), because that usually means you've put some thought into the license yourself and understand the implications of that license. But examples of existing projects that aren't your own are OK. Please just say why you think that license was good/bad for them. But first-hand experience is preferred. Looking forward to hearing some informative input.

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  • Preserving Tabs in POST Data

    - by byronh
    I need to preserve tab characters from a textarea through POST Data. It seems that there's no way to differentiate tabs from spaces in the $_POST array, and this is really frustrating me. I'm using a jQuery plugin from here to allow for tab and shift+tab usage within a textarea. http://teddevito.com/demos/textarea.html The JavaScript is using this as its TAB character: $.fn.tabby.defaults = {tabString : String.fromCharCode(9)}; For some reason, it shows an individual space instead of each tab character, so all my code formatting is lost: <textarea name="field0" rows="26" cols="123"><?php echo $_POST['field0']; ?></textarea> Neither does this. Apparently the tabs disappear before the data even reaches the str_replace function (the four spaces in the first double quotes is when I press TAB in my text editor). <textarea name="field0" rows="26" cols="123"><?php echo str_replace(" ", "\t", $_POST['field0']); ?></textarea> The reason I need tabs and not multiple spaces is because my application features and on-line code editor. Anyone have any ideas? I'm guessing the solution would involve modifying the data with javascript before it's sent through POST, but I haven't the slightest idea how to start.

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  • What's causing this permission's error and how can I work around it?

    - by Scott B
    Warning: move_uploaded_file(/home/site/public_html/wp-content/themes/mytheme/upgrader.zip) [function.move-uploaded-file]: failed to open stream: Permission denied in /home/site/public_html/wp-content/themes/mytheme/uploader.php on line 79 Warning: move_uploaded_file() [function.move-uploaded-file]: Unable to move '/tmp/phptempfile' to '/home/site/public_html/wp-content/themes/mytheme/upgrader.zip' in /home/site/public_html/wp-content/themes/mytheme/uploader.php on line 79 There was a problem. Sorry! Code is below for that line... // permission settings for newly created folders $chmod = 0755; // Ensures that the correct file was chosen $accepted_types = array('application/zip', 'application/x-zip-compressed', 'multipart/x-zip', 'application/s-compressed'); foreach($accepted_types as $mime_type) { if($mime_type == $type) { $okay = true; break; } } $okay = strtolower($name[1]) == 'zip' ? true: false; if(!$okay) { die("This upgrader requires a zip file. Please make sure your file is a valid zip file with a .zip extension"); } //mkdir($target); $saved_file_location = $target . $filename; //Next line is 79 if(move_uploaded_file($source, $saved_file_location)) { openZip($saved_file_location); } else { die("There was a problem. Sorry!"); }

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  • Combining Data from two MySQL tables.

    - by Nick
    I'm trying to combine data from two tables in MySQL with PHP. I want to select all the data (id, title, post_by, content and created_at) from the "posts" table. Then I would like to select the comment_id COUNT from the "comments" table IF the comment_id equals the posts id. Finally, I would like to echo/print something on this order: <? echo $row->title; ?> Posted by <? echo $row->post_by; ?> on <? echo $row->created_at; ?> CST <? echo $row->content; ?> <? echo $row->comment_id; ?> comments | <a href="comment.php?id=<? echo $row->id; ?>">view/post comments</a> I'm uncertain as to how to "combine" the data from two tables. I have tried numerous things and have spent several evenings and have had no luck. Any help would be greatly appreciated!

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  • Form validation

    - by kielie
    Hi guys, I need to create a form that has many of the same fields, that have to be inserted into a database, but the problem I have is that if a user only fills in one or two of the rows, the form will still submit the blank data of the empty fields along with the one or two fields the user has filled in. How can I check for the rows that have not been filled in and leave them out of the query? or check for those that have been filled in and add them to the query. . . The thank_you.php file will capture the $_POST variables and add them to the database. <form method="post" action="thank_you.php"> Name: <input type="text" size="28" name="name1" /> E-mail: <input type="text" size="28" name="email1" /> <br /> Name: <input type="text" size="28" name="name2" /> E-mail: <input type="text" size="28" name="email2" /> <br /> Name: <input type="text" size="28" name="name3" /> E-mail: <input type="text" size="28" name="email3" /> <br /> Name: <input type="text" size="28" name="name4" /> E-mail: <input type="text" size="28" name="email4" /> <input type="image" src="images/btn_s.jpg" /> </form> I am assuming that I could use javascript or jQuery to accomplish this, how would I go about doing this? Thanx in advance for the help.

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  • Help with an RSS Feed

    - by Pete Herbert Penito
    Hi Everyone! I've spent ages on this, all I'm trying to do is extract the "title" contents from an rss feed, everything else can be ignored. I've looked into simplepie, magpie and all that stuff, but I feel its kind of overkill for what I need to do. I realise there are google gadgets that are made that can do this, but I didn't want all the google logo stuff, and I wanted to personally make this. theres a whole bunch of unneeded tags thats coming in from the rss feed all I need is the title tag, it looks like this <title> My Title 3.0 </title> My server has PHP 5+ so I know I can use some of these simple xml functions which look promising. so far I've got <?php $blogfeed = file_get_contents("http://myblog.blogspot.com/feeds/posts/default?alt=rss"); echo $blogfeed; ?> And it gives me all the data, I was thinking of running through it with strpos and searching for <title> but is there any easier way to do this?? Thanks alot!

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  • Dealing with image upload on server

    - by user1073320
    I have got a the following problem: I have got multi-step form where in one step user upload image to server and then few steps further supplies other information, when this information is invalid no data should be commited - also the image should be deleted. I was thinking about PHP session, but I've read here PHP - Store Images in SESSION data? that it is inefficient way. Every time you proceed step in the form the image is reloaded (in the session) and as somebody mentioned "You will want it to only be as big as it needs to be and you need to delete it as soon as you don't need it because large pieces of information in the session will slow down the session startup." - here i got a question: will it slow down the stratup the session of user who upload file or sessions of all users? I have to mention that I'm looking for solution that doesn't rely on operating system scripts (cron or etc) - I have no permission to run such scripts. The perfect solution for me would be: saving image on disk (for example in some folder named after session id) then after the latest step of form move this image or delete depending on form validation. If user unexpectedly destroy the session (for example closing the browser) of course the folder with image should be deleted. In nutshell I need somethig like callback to event "destroying session".

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  • Toolbox/framework to construct lightweight public-facing web site

    - by aSteve
    I am aware of full-blown content management systems (CMS) such as SugarCRM and TikiWiki... where content is typically stored in a database... and edited through the same interface as it is published. While I like many of the features, the product is clearly aimed at enterprise-wide use rather than to be public-facing. What I'd like to establish are potential alternatives that fill the space between full-blown CMS and hand-coded bespoke site. I like the way that I can add modules to my CMS... allowing me to quickly introduce new functionality, and I'd like an analogous feature in a system for public web-content. Modules I know I'd like include moderated comments; web-form-to-email gateway; menus/tabs... in future, perhaps mapping or diaries or RSS integration - etc. Where my requirements differ from a CMS, I don't need (or want) most content to be editable through the main site... and, somehow, I do want to be able to preview how updates will be presented to the public rather than to make live changes. For these purposes, in contrast to those where a typical CMS would be ideal, presentation is of paramount importance - and trumps any desire to immediately disseminate information. I realise that this is a very high-level question... (suggestions of additional tags welcome) - I mentioned PHP only as - ideally - I'm looking for an open source solution and a PHP deployment is an easy option. What are my options?

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  • Parsing HTML: Call to a member function > children() on a non-object

    - by sm56d
    Hello all, I was just helped with this question but I can't get it to move to the next block of HTML. $html = file_get_html('http://music.banadir24.com/singer/aasha_abdoo/247.html'); $urls = $html->find('table[width=100%] table tr'); foreach($urls as $url){ $song_name = $url->children(2)->plaintext; $url = $url->children(6)->children(0)->href; } It returns the list of the names of the first album (Deesco) but it does not continue to the next album (The Best Of Aasha)? It just gives me this error: Notice: Trying to get property of non-object in C:\wamp\www\test3.php on line 26 Fatal error: Call to a member function children() on a non-object in C:\wamp\www\test3.php on line 28 Why is this and how can I get it to continue to the next table element? I appreciate any help on this! Please note: This is legal as the songs are not bound by copyright and they are available to download freely, its just I need to download a lot of them and I can't sit there clicking a button all day. Having said that, its taken me an hour to get this far.

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  • i read that for RESTful websites. it is not good to use $_SESSION. Why is it not good? how then do i

    - by keisimone
    I read that it is not good to use $_SESSION. http://www.recessframework.org/page/towards-restful-php-5-basic-tips I am creating a WEBSITE, not web service in PHP. and i am trying to make it more RESTful. at least in spirit. right now i am rewriting all the action to use Form tags POST and add in a hidden value called _method which would be "delete" for deleting action and "put" for updating action. however, i am not sure why it is recommended NOT to use $_SESSION. i would like to know why and what can i do to improve. To allow easy authorization checking, what i did was to after logging in the user, the username is stored in the $_SESSION. Everytime the user navigates to a page, the page would check if the username is stored inside $_SESSION and then based on the $_SESSION retrieves all the info including privileges from the database and then evaluates the authorization to access the page based on the info retrieved. Is the way I am implementing bad? not RESTful? how do i improve performance and security? Thank you.

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  • Database class is not correctly connecting to my database.

    - by blerh
    I'm just venturing into the world of OOP so forgive me if this is a n00bish question. This is what I have on index.php: $dbObj = new Database(); $rsObj = new RS($dbObj); This is the Database class: class Database { private $dbHost; private $dbUser; private $dbPasswd; private $dbName; private $sqlCount; function __construct() { $this->dbHost = 'localhost'; $this->dbUser = 'root'; $this->dbPasswd = ''; $this->dbName = 'whatever'; $this->sqlCount = 0; } function connect() { $this->link = mysql_connect($this->db_host, $this->db_user, $this->db_passwd); if(!$this->link) $this->error(mysql_error()); $this->selection = mysql_select_db($this->db_name, $this->link); if(!$this->selection) $this->error(mysql_error()); } } I've shortened it to just the connect() method to simplify things. This is the RS class: class RS { private $username; private $password; function __construct($dbObj) { // We need to get an account from the db $dbObj->connect(); } } As you can probably see, I need to access and use the database class in my RS class. But I get this error when I load the page: Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\includes\database.class.php on line 22 The thing is I have NO idea where it got the idea that it needs to use ODBC as a user... I've read up on doing this stuff and from what I can gather I am doing it correctly. Could anyone lend me a hand? Thank you.

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  • Getting list of Facebook friends with latest API

    - by Eric
    I'm using the most recent version of the Facebook SDK (which lets to connect to something called the 'graph API' though I'm not sure). I've adapted Facebook's example code to let me connect to Facebook and that works... but I can't get a list of my friends. $friends = $facebook->api('friends.get'); This produces the error message: "Fatal error: Uncaught OAuthException: (#803) Some of the aliases you requested do not exist: friends.get thrown in /mycode/facebook.php on line 543" No clue why that is or what that means. Can someone tell me the right syntax (for the latest Facebook API) to get a list of friends? (I tried "$friends = $facebook-api-friends_get();" and get a different error, "Fatal error: Call to a member function friends_get() on a non-object in /mycode/example.php on line 129".) I can confirm that BEFORE this point in my code, things are fine: I'm connected to Facebook with a valid session and I can get my info and dump it to the screen just... i.e. this code executes perfectly before the failed friends.get call: $session = $facebook->getSession(); if ($session) { $uid = $facebook->getUser(); $me = $facebook->api('/me'); } print_r($me);

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  • file cretaed using exec could not be accessed immediately after creation?

    - by Holicreature
    HI I'm using exec in php to execute a command and it will create a .png file in a temp folder.. After creating that i'm trying to open that file and read contents and process them,, but i end up file could not read error.. I think the time taken by the exec to execute and create a file is the cause for the issue.. but i dont know how to fix it? i tried sleep() but it makes my script to run slow <?php error_reporting(E_ALL); extension_loaded('ffmpeg') or die('Error in loading ffmpeg'); //db connection codes $max_width = 120; $max_height = 72; $path ="/path/"; $qry="select id, input_file, output_file from videos where thumbnail='' or thumbnail is null;"; $res=mysql_query($qry); $cnt = 1; while($row = mysql_fetch_array($res,MYSQL_ASSOC)) { $outfile = $row[output_file]; $imgname = $cnt.".png"; $srcfile = "/path/".$outfile; echo "####$srcfile####"; exec("ffmpeg -i ".$srcfile." -r 1 -ss 00:00:05 -f image2 -s 120x72 ".$path.$imgname); $nname = "./temp/".$imgname; echo "nname===== $nname"; $fileo = fopen($nname,"rb"); if($fileo) { $imgData = addslashes(file_get_contents($nname)); .. ... .... } else echo "Could not open<br><br>"; $cnt = $cnt + 1: } ?>

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  • $_SESSION v. $_COOKIE

    - by taeja87
    I learned about $_SESSION about several weeks ago when creating a login page. I can successfully login and use it with variables. Currently I am trying to understand $_SESSION and $_COOKIE. Please correct me if I am wrong, I can use $_SESSION when logging in and moving around pages. With $_COOKIE, it is used to remember when I last visit and preferences. Another thing involving cookies is that when websites use advertisements (for example: Google AdSense), they use the cookies to track when visitor click on a advertisement, right? I can use both ($_SESSION & $_COOKIE)? I read somewhere that you can store the session_id as value for the cookie. Also, I read about security which let to me finding this: What do I need to store in the php session when user logged in?. Is using session_regenerate_id good for when a user comes back to the site? And this: How to store a cookie with php involving uniqid. For those wanting to know about the login, I use email and password. That way the user can be able to change their username. I look forward to learning more about these two from anybody who would like to share their knowledge about it. If I asked too many question, you can just answer the one that you have more experience with. If you need more information, just ask since I might have forgotten to include something. Thank You. Found this: What risks should I be aware of before allowing advertisements being placed on my website?

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  • Error loading il8n console in CakePHP 1.3

    - by inkedmn
    I'm trying to generate the .po files for the site I'm working with. When I run this command from within the console directory: Warning: strtotime(): It is not safe to rely on the system's timezone settings. You are *required* to use the date.timezone setting or the date_default_timezone_set() function. In case you used any of those methods and you are still getting this warning, you most likely misspelled the timezone identifier. We selected 'America/Los_Angeles' for 'PDT/-7.0/DST' instead in /Users/bkelly/Development/[redacted]/www/trunk/about/trunk/cake/libs/cache.php on line 570 Warning: strtotime(): It is not safe to rely on the system's timezone settings. You are *required* to use the date.timezone setting or the date_default_timezone_set() function. In case you used any of those methods and you are still getting this warning, you most likely misspelled the timezone identifier. We selected 'America/Los_Angeles' for 'PDT/-7.0/DST' instead in /Users/bkelly/Development/[redacted]/www/trunk/about/trunk/cake/libs/cache.php on line 570 Error: Class Il8nShell could not be loaded. This also happens if I issue ./cake il8n help . The bake console seems to load fine (no errors, takes me to the expected menu). I'm running CakePHP 1.3.0 under Mac OSX 10.6 (using MAMP). Cake appears to be functioning normally otherwise. Thanks!

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  • How do I make a HTTP POST request to an url from within a c++ dll?

    - by tjorriemorrie
    Hi, there's an open-source application I need to use. It allows you to use a custom dll. Unfortunately, I can't code c++; i don't like it; it don't want to learn it. I do know PHP very well however, thus you can see that I'll rather do my logic within a PHP application. Thus I'm thinking about posting the data from c++/dll to a url on my localhost. (i have my local server set up, that's not the problem). I need to post a large amount of variables (thus a POST and not GET request required). The return value will only be one (int)variable, either 0, 1 or 2. So I need a c++ function that: 1) will post variables to an url. 2) Wait for, and receive the answer. The data type can be in xml, soap, json, whatever, doesn't matter. Is there anyone that can write a little c++ http function for me? pretty please? ;)

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  • Mysql Query is not working in edited jTable code, why?

    - by Furkan Kadioglu
    I'm using this example: www.jtable.org I've downloaded the jTable PHP version. I then edited the script. The jTable simple version is working, but my edited version isn't. I can create a list, but I can't add a row; this code is causing problems. However, PHP doesn't display any error messages. else if($_GET["action"] == "create") { //Insert record into database $result = mysql_query("INSERT INTO veriler(bolge, sehir, firma, adres, tel, web) VALUES('" . $_POST["bolge"] . "', '" . $_POST["sehir"] . "', '" . $_POST["firma"] . "', '" . $_POST["adres"] . "', '" . $_POST["tel"] . "', '" . $_POST["web"] . "'"); //Get last inserted record (to return to jTable) $result = mysql_query("SELECT * FROM veriler WHERE id = LAST_INSERT_ID();"); $row = mysql_fetch_array($result); //Return result to jTable $jTableResult = array(); $jTableResult['Result'] = "OK"; $jTableResult['Record'] = $row; print json_encode($jTableResult); } What is the problem?

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  • Linux apache developing configuration

    - by Jeffrey Vandenborne
    Recenly reinstalled my system, and came to a point where I need apache and php. I've been searching a long time, but I can't figure out how to configure apache the best way for a developer computer. The plan is simple, I want to install apache 2 + mysql server so I can develop some php website. I don't want to install lamp though, just the apache2, php5 and mysql. The problem that I've been looking an answer for is the permissions on the /var/www/ folder. I've tried making it my folder using the chown command, followed by a chmod -R 755 /var/www. Most things work then, but fwrite for example won't work, because I need to give write permissions to everyone, unless I change my global umask to 000 I'm not sure what I can do. In short: I want to install apache2, php5, mysql-server without using lamp, but configured in a way so I can open up netbeans, start a project with root in /var/www/, and run every single function without permission faults. Does anyone have experiences or workarounds to this? Extra: OS: Ubuntu 10.04 ARCH: x86_64

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