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  • GameKit server disconnect makes session invalid

    - by Markus
    Hi everyone, I'm trying to build a 4 player multiplayer game using GameKit. I tried many things and I'm currently using client/server mode, meaning that I have 1 session in GKSessionModeServer and 3 other sessions connecting to this one in GKSessionModeClient. My goal is, that when a user leaves the game or gets disconnected for whatever reason, that the others can continue their game. So far this works if any of the Clients leave the game, but if the Sevrer leaves the game, send and receive calls are not reaching the other peers anymore. I tried the same with 4 peers in GKSessionModePeer with the same result (where the player who accepts connections cannot quit). Does anyone had better success in making this work? Any help is highly appreciated. Thanks!

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  • Prevent PHP sesison hijack, are these good ideas?

    - by matthew Rhodes
    I'm doing a simple shopping cart for a small site. I plan to store cart items as well as logged in user_id in session variables. to make things a little more secure, I thought I'd do this: sha1() the user_id before storing it in the session. Also sha1() and store the http_user_agent var with some salt, and check this along with the user_id. I know there is more one can do, but I thought this at least helps quite a bit right? and is easy for me to implement.

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  • Approaches to timing out sessions on a web app using AJAX autorefreshes

    - by Braintapper
    I'm writing a web application that autorefreshes data with an AJAX call at set intervals. Because it's doing that, server side user sessions never time out, since the last activity is refreshed with every ajax call. Are there good client side rules I could implement to time out the user? I.e. should I track mouse movements in the browser, etc., or should I point the AJAX calls to URLs that don't refresh the session? I like that my AJAX calls hit a session-enabled URL, because I can also validate that the user is logged in, etc. Any thoughts in terms of whether I should even bother timing out the users?

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  • deny direct access to a php file by typing the link in the url

    - by aeonsleo
    hi, I am using php session for a basic login without encryption for my site. I want to prevent a user from directly accessing a php page by typing the url when he/she is not signed in. But this is not happening. I am using session_start(), initializing session variables and aslo unsetting and destroying sesssion during logout. Also if I type the link in a different browser the page is getting displayed. I am not very well versed with php , only a beginner. I googled for such problem and found few alternatives as keeping all files in a seperate folder from the web root, using .htaccess etc. Can someone explain in simple terms what could be a good solution.thanks in advance.

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  • Removing DOM event handlers in long-running browser session

    - by Chris Beck
    I have a browser interface with a ul#contacts list on the left and div#contact property panel (email, phone) on the right. Click a contact in the list and my app makes an XHR request to get the contact property HTML fragment and update div#contact.innerHTML. Each contact fragment has an "Edit Contact" link. With JS, I progressively upgrade that link with an event listener that performs an XHR request to replace the static property panel with an in-place edit form. This can happen many times during a single browser session. How should I clean up my "Edit Contact" event listener? Do I need to remove it manually before the form overwrites the property panel? Or is the event listener cleaned up automatically when the contents of div#contact (and the node that I'm listening on) is overwritten? FWIW, I still consider IE6 to be part of my target market.

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  • How and when to log account access login with PHP?

    - by Nazgulled
    I want to implement a basic login system in some PHP app where no cookies will be involved. I mean, the user closes the browser and the login expires, it will remain active during the browser session (or if the user explicitly logs out) otherwise. I want to log all this activity and I'm thinking that every time the user refreshes the page, opens a different link or logs out, I record that time as the last access made by that user, overwriting the previous access log. But my problem is when and how should I insert another record into the database instead of overwriting the last one? Should I just define a timeout and if the last access was made above that timeout, another log should be inserted into the database? Should the session expire too after that timeout? Or is there a better way? Ideally, I would like to log the "log out action" when the browser was closed, but I don't think there's a way to detect that is there? Suggestions?

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  • Make user object available to all Controllers in Zend?

    - by Sled
    Hey guys, I'm using Zend_Auth to identify a user in my application. This creates a session with the userobject. My question is how do I make this object available to every Controller and action, so I don't have to pull it out of the session every time I need data from this object? I'm guessing this should be done in bootstrap.php or index.php but I don't really know how to makte it available to every controller.. so any code examples would be appreciated! Thanks!

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  • PHP Captcha without session

    - by Anton N
    Ok, here is an issue: in the project i'm working on, we can't rely on server-side sessions for any functionality. The problem is that common captcha solutions from preventing robotic submits require session to store the string to match captcha against. The question is - is there any way to solve the problem without using sessions? What comes to my mind - is serving hidden form field, containing some hash, along with captcha input field, so that server then can match these two values together. But how can we make this method secure, so that it couldn't be used to break captcha easily.

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  • In grails how to insert additional parameters (from session) in all url's

    - by HeDinges
    I would like to add an additional parameter in my url, the use case is the following: When user do their login they also specify a 'company' name and from that moment on, all urls should map to: /$company/$controller/$action/$id The main idea is to have the current company name available in all url's, have it bookmarkable, and not to have to pass the company name everywhere as a request parameter. Also, once users are logged in it is acceptable to have the chosen company name in session scope. What is the right way of inserting this parameter in all our urls? I tried to modify my UrlMappings mapping, but I didn't found a way to insert the company name. Thanks,

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  • Sharing session (or cookie) using Grails acegi plugin

    - by firnnauriel
    Is it possible for two different Grails project, also having different domains, to share a session/cookie? Let's say I have 2 sites: www.mycompany.com, and, www.othercompany.com. Assume that both sites are having same domains, and same database and records too. What I want to know is if this code: authenticateService.userDomain() or even the authenticateService.isLoggedIn() will behave and return exactly the same object/result whether it is called in either of the site. Basically, what we need is a solution for sharing/identifying logged in user between two different sites. Need more details on how to implement this using acegi 0.5.2 and grails 1.2.1. Hoping for any leads on this. Thank you.

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  • How do I stack Plack authentication handlers?

    - by Schwern
    I would like to have my Plack app try several different means of authorizing the user. Specifically, check if the user is already authorized via a session cookie, then check for Digest authentication and then fall back to Basic. I figured I could just enable a bunch of Auth handlers in the order I wanted them to be checked (Session, Digest, Basic). Unfortunately, the way that Plack::Middleware::Auth::Digest and Plack::Middleware::Auth::Basic are written they both return 401 if digest or basic auth doesn't exist, respectively. How is this normally dealt with in Plack?

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  • Hudson build fails when a user logs out of RDP session

    - by sjohnston
    We are using Hudson to build mixed C++/Java projects with an Ant script. It is running in Tomcat 6, on a Win XP virtual machine. I have noticed recently that when a user logs off the machine (from a remote desktop session), builds that are currently running tend to suddenly fail without an error message. Has anyone encountered anything similar or have an idea what might be causing this effect? I can post additional information about our setup if needed, I'm just not sure what's relevant in this case.

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  • session set for some Actions in Zend framework

    - by user202127
    I'm working on a website , in CV part users have some articles that only logged in users can download them.I want to make changes to the log in Action or preDispatch() to set session for guess users to download the articles, can some one tell me how it can be or give me some reference links. here is my preDispatch(): public function preDispatch() { $userInfo=$this->_auth->getStorage()->read(); $identity= $this->_auth->getIdentity(); if(!$this->_auth->hasIdentity()) { return $this->_helper->redirector('login','login'); } if(!isset($userInfo["member_id"]) || strlen($userInfo["member_id"])==0) { return $this->_helper->redirector('forbidden','login'); } $this->_accessType=2; }

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  • Passing PHP session variable to AJAX URL

    - by user547794
    Hello, I am trying to pass a session variable into an AJAX loaded page. Here is the code I am using: jQuery(document).ready(function(){ $("#userdetail").click(function() { $.ajax({ url: "userdetail.php?id=<?php $_SESSION['uid']?>", success: function(msg){ $("#results").html(msg); } }); }); }); This is the HTML URL I had working, not sure how to get this into the AJAX call: userdetail.php?id=<?php $_SESSION['uid']?> I should also mention that if I manually pass in the userID it works fine url: "userdetail.php?id=1",

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  • Rails - Dynamic cookie domains using Rack

    - by Tim B.
    I'm fairly new to Rails and Rack, but this guy had a seemingly straightforward write-up about using Rack to implement dynamic session domain middleware. The code looks good to and I've implemented it here on my local machine, but I'm still not able to transcend top level domains on a single login. Here's the middleware code: class SetCookieDomain def initialize(app, default_domain) @app = app @default_domain = default_domain end def call(env) host = env["HTTP_HOST"].split(':').first env["rack.session.options"][:domain] = custom_domain?(host) ? ".#{host}" : "#{@default_domain}" @app.call(env) end def custom_domain?(host) domain = @default_domain.sub(/^\./, '') host !~ Regexp.new("#{domain}$", Regexp::IGNORECASE) end end And then in environment.db: config.load_paths += %W(#{RAILS_ROOT}/app/middlewares) Lastly in production.db (and development.db): config.middleware.use "SetCookieDomain", ".example.org" Any help is greatly appreciated. EDIT: I'm running Rails 2.3.3 and Rack 1.0

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  • Displaying objects based on if a user is logged in or not

    - by MaxMackie
    I'm learning about PHP sessions for user authentication on my website. I know how to restrict the viewing of a complete page using sessions (simply check if the 'uid' session variable is set and if it is, show content, if not redirect to an error). However I'm trying to figure out the best way to selectively show and hide different objects (div, text, images) based on if a user is logged in or not. Is it as simple as checking for the 'uid' session variable and displaying based on if it set or not? Is there a more efficient way of doing this id there are a lot of conditional elements on a page?

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  • Warning: session_start(): Cannot send session cache limiter - headers already sent

    - by shyam
    I am receiving this warning in a page while I try starting the session. <?php session_start(); ?> <!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> <title>Video</title> </head> -this is the part of the code. There are no characters (not even space) before the first line. This page is reached after logging in from another page, through redirection. Any help? (PHP version is 5.2)

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  • What magical thing could be killing my Drupal session and anywhere from 15-45 minutes of activity?

    - by jini
    I am using a standard Drupal install hosted on a LAMP stack. My settings.php has the following set: ini_set('session.gc_probability', 1); ini_set('session.gc_divisor', 100); ini_set('session.gc_maxlifetime', 200000); ini_set('session.cookie_lifetime', 2000000); my php.ini file has: session.gc_probability = 1 session.gc_divisor = 1000 session.gc_maxlifetime = 1440 Also I have checked that the safe mode is off so that my settings.php file is able to override main php.ini variables. Also since the person can get log out at 15 minutes, it is making me wonder whether php.ini has anything to do with it anyways. I have combed through my code and it seems to work fine on my local host however on server it is having issues. Where else can i possibly check?????

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  • PHP Session Array Value keeps showing as "Array"

    - by Nerathas
    Hello, When sending data from a form to a second page, the value of the session is always with the name "Array" insteed of the expected number. The data should get displayed in a table, but insteed of example 1, 2, 3 , 4 i get : Array, Array, Array. (A 2-Dimensional Table is used) Is the following code below a proper way to "call" upon the stored values on the 2nd page from the array ? $test1 = $_SESSION["table"][0]; $test2 = $_SESSION["table"][1]; $test3 = $_SESSION["table"][2]; $test4 = $_SESSION["table"][3]; $test5 = $_SESSION["table"][4]; What exactly is this, and how can i fix this? Is it some sort of override that needs to happen? Best Regards.

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  • Pass a variable from javascript to php in the same session OnClickFunction

    - by MickyScion
    I was seeing through stackoverflow the solutions for this kind of problems but any of them are executing the code of javascript in the same session...please i want some help with this...i have this in my session <script> function show_alert() { var ProdAntes = document.getElementById("productoseleccionado").value; var CantAntes = document.getElementById("cantidadantes").value; var PrecAntes = document.getElementById("precioantes").value; var FecAntes = document.getElementById("fechaantes").value; var ProdAhora = document.getElementById("SoyaProductorProduccionProducto").value; var CantAhora = document.getElementById("SoyaProductorProduccionCantidadtm").value; var PrecAhora = document.getElementById("SoyaProductorProduccionPreciodolar").value; var FecAhora = document.getElementById("select_date").value; } </script> and in my html stuff i have this <?php echo $this->Form->create('SoyaProductorProduccion');?> <fieldset> <?php echo $this->Form->hidden('id', array('value' => $this->data['SoyaProductorProduccion']['id'])); echo $this->Form->input('operacion', array('type' => 'hidden', 'value'=>'Produccion')); //-------------------------------------------------------------- $productoseleccionado = $this->data['SoyaProductorProduccion']['producto']; echo $this->Form->input('productoseleccionado', array('type' => 'hidden','style'=>'width:500px; height:30px;','id' => 'productoseleccionado' , 'value' => $productoseleccionado)); echo $this->Form->input('producto', array( 'options' => array( $productoseleccionado => $productoseleccionado, 'Torta solvente de soya' => 'Torta solvente de soya', 'Torta solvente de girasol' => 'Torta solvente de girasol', 'Harina integral de soya' => 'Harina integral de soya', 'Harina de girasol' => 'Harina de girasol', 'Cascarilla de soya' => 'Cascarilla de soya', 'Cascarilla de girasol' => 'Cascarilla de girasol', 'Aceite de soya refinado' => 'Aceite de soya refinado', 'Aceite de soya crudo' => 'Aceite de soya crudo', 'Aceite de girasol refinado' => 'Aceite de girasol refinado', 'Aceite de girasol crudo' => 'Aceite de girasol crudo', ),'label'=>'Tipo de Producto' )); foreach ($soyacambiodolares as $soyacambiodolar): $dolar=$soyacambiodolar['SoyaCambioDolar']['cambio']; endforeach; echo $this->Form->input('cambio', array('type' => 'hidden','value' => $dolar)); //----------------------------------------------------------------------------- $cantidadantes = $this->data['SoyaProductorProduccion']['cantidadtm']; echo $this->Form->input('cantidadantes', array('type' => 'hidden','style'=>'width:500px; height:30px;', 'value' => $cantidadantes,'id' => 'cantidadantes')); echo $this->Form->input('cantidadtm', array('label' => 'Cantidad en tonelada(s) métrica(s) del producto (TM)','style'=>'width:500px; height:30px;')); //----------------------------------------------------------------------------- $precioantes = $this->data['SoyaProductorProduccion']['preciodolar']; echo $this->Form->input('precioantes', array('type' => 'hidden','style'=>'width:500px; height:30px;', 'value' => $precioantes,'id' => 'precioantes')); echo $this->Form->input('preciodolar', array('label' => 'Precio en Dolares Americanos por tonelada métrica (TM / $us)','style'=>'width:500px; height:30px;')); //----------------------------------------------------------------------------- ?> <table style="width: 600px"> <tr> <td > <?php //---------------------------------------------------------------- $fechaantes = $this->data['SoyaProductorProduccion']['fecharegistro']; echo $this->Form->input('fechaantes', array('type' => 'hidden','style'=>'width:500px; height:30px;', 'value' => $fechaantes, 'id' => 'fechaantes')); //---------------------------------------------------------------- echo $this->Form->input("fecharegistro", array( 'label' => '<strong>Periodo al que corresponde la declaración</strong>', 'type' => 'text', 'style' => 'width: 110px', 'class' => 'fl tal vat w300p', 'error' => false , 'id' => 'select_date')); ?> <?php echo $this->Html->div('datepicker_img w100p fl pl460p pa', $this->Html->image('datepicker_calendar_icon.gif'),array('id' => 'datepicker_img')); ?> <?php echo $this->Html->div('datepicker fl pl460p pa', ' ' ,array('id' => 'datepicker')); ?> </td> </tr> </table> <?php echo $this->Form->submit('Modificar Existencia', array('class' => 'form-submit', 'title' => 'Presione aqui para agregar datos', 'onclick' => 'return show_alert();')); ?> </fieldset> <?php echo $this->Form->end(); ?> my function is ok but i want these: when i click the submit button i want to compare wich field had been changed, and i want to create a chain of detailed changes like "change in the field 1, change in the fiel 2.--" and so on...and this has to be saved in my database so i have to pass to a variable in my php before saving...thanks!

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  • Redirecting a page when session expires using asp.net mvc

    - by Naidu
    In my web.config file i have the following code: <system.web> <assemblies> <authentication mode="Forms"> <forms loginUrl="/Account/Login" slidingExpiration="true" timeout="1" /> </authentication> <sessionState timeout="1"></sessionState> </assemblies> </system.web> And I have main page Project and in that there will sub pages. I have given the [Authorize] attribute for each view index method. After the session complete when we select any view then the page inside the project main page will be redirecting. But I want the whole page to be redirected. Any Help is appreciated.

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  • Help me with Php session vs Header redirect?

    - by python
    I have the following pages: *page1.php <?php if (isset($_GET['link'])) { session_start(); $_session['myvariable'] = 'Hello World'; header('Location: http://' . $_SERVER['SERVER_NAME'] . dirname($_SERVER['REQUEST_URI']) . '/page2.php'); exit; } ?> <a href="<?php print $_SERVER['REQUEST_URI'] . '?link=yes';?>">Click Here</a> *page2.php <?php print 'Here is page two, and my session variable: '; session_start(); print $_session['myvariable']; //This line could not output. exit; ?> When I try output $_session['myvariable'] I did not get the result hello world message. I could not find out the solution to fix it .?

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  • Multiple sessions or one?

    - by user1314285
    I am using a security token for a form, the form is dynamically built depending on selection through jquery. So the form is called quite a lot and different tokens created every-time. So.. if the same user calls the form 3 times the session would be rewritten? Would it help at all to check if the token exists and not create one unless its empty? or perhaps someone knows of a good way to work with form tokens? If 3 users are on then the token is created 3 times with different values, right? If I check for the token and 3 users are on then the session is created 3 times with the same values?

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  • setting up Ubuntu 10.10 as paravirtualized guest in Xen on RHEL5 host - what kernel?

    - by kostmo
    I've discovered the tool ubuntu-vm-builder, which I've installed and then invoked on an Ubuntu workstation as: sudo vmbuilder xen ubuntu --suite maverick --flavour virtual --arch amd64 --mem=512 --rootsize 8192 This workstation is not the intended target host of the virtual machine, however; I would like to host the guest on a Red Hat Enterprise Linux 5 machine that is running Xen 3.0.3. The output of this command appears to be a folder named ubuntu-xen containing three files: tmpXXXXXX, a very large file which I assume is the root partition image tmpYYYYYY, a somewhat large file which I assume is the swap partition image xen.conf, a text file I have copied the xen.conf file to the RHEL server's /etc/xen directory under the new name newvm, adjusting the paths of tempXXXXXX and tempYYYYYYin the file after also copying them from my local workstation to the RHEL server. When I launch the Virtual Machine Manager virt-manager, I can see the newvm virtual machine listed underneath the Dom0 machine. When I try to start newvm, I get the error: Error starting domain: virDomainCreate() failed POST operation failed: (xend.err 'Error creating domain: Kernel image does not exist: None') Indeed, there exists an entry kernel = 'None' in the xen.conf file. How do I find out what the path of the kernel should be? Is this path supposed to be to a kernel stored on the local filesystem of the RHEL5 host, or is it supposed to be a path inside the guest image? I see that the vmbuilder command provides for a --xen-kernel option, along with a --xen-ramdisk option, but I'm not sure what to use for either. I think I should be able to get this to work, since Ubuntu is said to be supported as a Xen guest, even though the Xen 4.0.1 docs state support for only a limited set of distributions, Ubuntu excluded. Update 1 When running vmbuilder on my local workstation, I did observe an output line saying: Calling hook: install_kernel and later, output lines saying: update-initramfs: Generating /boot/initrd.img-2.6.35-23-virtual [...] run-parts: executing /etc/kernel/postinst.d/initramfs-tools 2.6.35-23-virtual /boot/vmlinuz-2.6.35-23-virtual So in the xen.conf file, I tried setting the lines: kernel = '/boot/vmlinuz-2.6.35-23-virtual' ramdisk = '/boot/initrd.img-2.6.35-23-virtual' When trying to start the VM, I got an error similar to last time: Error starting domain: virDomainCreate() failed POST operation failed: (xend.err 'Error creating domain: Kernel image does not exist: /boot/vmlinuz-2.6.35-23-virtual') This makes me think that the RHEL5 machine is looking for local files, rather than a file within the binary guest disk image. After running sudo updatedb on my workstation, neither of those files were found. If the vmbuilder tool had tried to install them, it must have failed. Update 2 I was able to extract the kernel and initrd images from the guest disk binary by mounting it: mkdir mnt_tmp sudo mount ubuntu-xen/tmpXXXXXX mnt_tmp/ -o loop cp mnt_tmp/boot/vmlinuz-2.6.35-23-virtual virtual_kernel_ubuntu cp mnt_tmp/boot/initrd.img-2.6.35-23-virtual virtual_initrd_ubuntu These two files I copied to the RHEL5 server, and edited the xen.conf file to point to them as kernel and ramdisk. With this done, I could "run" the newvm virtual machine from within virt-manager, but was met with the message Console Not Configured For Guest when I double clicked the entry to open the Virtual Machine Console. As suggested by a forum, I then added the line vfb = [ 'type=vnc' ] to the configuration file, recreated the virtual machine (a ~10 min process), and this time got the message: Connecting to console for guest This remained indefinitely; after selecting View - Serial Console, I found a kernel panic: [5442621.272173] Kernel panic - not syncing: Attempted to kill the idle task! [5442621.272179] Pid: 0, comm: swapper Tainted: G D 2.6.35-23-virtual #41-Ubuntu [5442621.272184] Call Trace: [5442621.272191] [<ffffffff815a1b81>] panic+0x90/0x111 [5442621.272199] [<ffffffff810652ee>] do_exit+0x3be/0x3f0 [5442621.272204] [<ffffffff815a5e20>] oops_end+0xb0/0xf0 [5442621.272211] [<ffffffff8100ddeb>] die+0x5b/0x90 [5442621.272216] [<ffffffff815a56c4>] do_trap+0xc4/0x170 [5442621.272221] [<ffffffff8100ba35>] do_invalid_op+0x95/0xb0 [5442621.272227] [<ffffffff8130851c>] ? intel_idle+0xac/0x180 [5442621.272232] [<ffffffff810072bf>] ? xen_restore_fl_direct_end+0x0/0x1 [5442621.272239] [<ffffffff815a48fe>] ? _raw_spin_unlock_irqrestore+0x1e/0x30 [5442621.272247] [<ffffffff8108dfb7>] ? tick_broadcast_oneshot_control+0xc7/0x120 [5442621.272253] [<ffffffff8100ad5b>] invalid_op+0x1b/0x20 [5442621.272259] [<ffffffff8130851c>] ? intel_idle+0xac/0x180 [5442621.272264] [<ffffffff813084e0>] ? intel_idle+0x70/0x180 [5442621.272269] [<ffffffff810072bf>] ? xen_restore_fl_direct_end+0x0/0x1 [5442621.272275] [<ffffffff8148a147>] cpuidle_idle_call+0xa7/0x140 [5442621.272281] [<ffffffff81008d93>] cpu_idle+0xb3/0x110 [5442621.272286] [<ffffffff815873aa>] rest_init+0x8a/0x90 [5442621.272291] [<ffffffff81b04c9d>] start_kernel+0x387/0x390 [5442621.272297] [<ffffffff81b04341>] x86_64_start_reservations+0x12c/0x130 [5442621.272303] [<ffffffff81b08002>] xen_start_kernel+0x55d/0x561 Update 3 I tried an i386 architecture instead of amd64, but got the same kernel panic. Also, it seems the Virtual Machine Manager pays attention to the format of the filename of the kernel; for the same kernel binary, I tried simply naming it vmlinuz-virtual, which threw out an error box about an invalid kernel. When I named it vmlinuz-2.6.35-23-virtual, it did not throw the error, but it did still result in the kernel panic shortly thereafter.

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