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  • MVC Pages that require the user to be logged in

    - by keithjgrant
    I'm working on a little MVC framework and I'm wondering what the "best way" is to structure things so secure pages/controllers always ensure the user is logged in (and thus automatically redirects to a login page--or elsewhere--if not). Obviously, there are a lot of ways to do it, but I'm wondering what solution(s) are the most common or are considered the best practice. Some ideas I had: Explicitly call user->isLoggedIn() at the beginning of your controller action method? (Seems far too easy to forget and leave an important page unsecure on accident) Make your controller extend a secureController that always checks for login in the constructor? Do this check in the model when secure information is requested? (Seems like redundant calls would be made) Something else entirely? Note: I'm working in PHP, though the question is not language-dependent.

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  • PHP, login to pop3-server

    - by Ockonal
    Hi guys, I have another one problem with pop3. Here is connection to pop3-server: $pop3Server = '62.113.86.215'; // mail.roller.ru $pop3User = 'mail-robot%roller.ru'; $pop_conn = fsockopen($pop3Server, 110, $errno, $errstr, 30); echo fgets($pop_conn, 1024); It returns OK. The next step is login: fputs($pop_conn, 'USER '.$pop3User.'\r\n'); //stream_set_timeout($pop_conn, 3); print fgets($pop_conn, 1024); And I get time-out. Why? p.s. Here is full code: http://pastie.org/934170

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  • Offer access to a private page without login

    - by dccarmo
    So I've been struggling with a nice and easy way to allow users to access a private page without asking them to fill out a login/password form. What I'm thinking about using right now is for each private page I generate a uniqueid (using php uniqid function) and then send the URI to the user. He would access his private page as "www.mywebsite.com/private_page/13ffa2c4a". I think it's relatively safe and user friendly, without asking too much of information. I thought maybe when the user access this page it would ask for it's e-mail just to be sure, but the best would be nothing at all. Is this really safe? I mean not internet banking safe, but enough for a simple access? Do you think there's a better solution? Thanks. :)

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  • SEO on a login based website

    - by Paul
    Ok so say I have a website that the majority of content is behind a closed door, login screen for paying customers only. How can I let google know what is there so people can find the site through a search engine. What programming techniques could I use to give more content to the google bot. I will have some overview and demonstration screens to tempt the users but will these five to ten pages of content be enough for the search engines. The key work is quite a highly sort one and there is quite alot of competition. And before anyone asks the site is not a subscription porn site (Sadly)

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  • Facebook login within an iframe (but outside FB)

    - by Cystack
    At some point in my application, I use an iframe. In this iframe, some Facebook stuff could be useful and for that, I use the graph API a lot. Problem is : when someone isn't logged in to facebook or hasn't allowed my application yet, they are prompted to click the "connect with facebook" button. But as soon as they click it, the iframe gets destroyed and the top page is replaced with the Facebook login page. Eventually the top page is redireccted to the previous iframe URL, but the former parent page is lost forever. Is this the intended behaviour of FB connect ? Is there a way to avoid it or to hack around it ? (maybe using a popup instead of an iframe, but that sounds ugly (uglier than an iframe)). I am currently using the PHP SDK Thanks a lot

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  • login with users, groups and permissions

    - by Dan Bemowski
    OK, I have a set of tables that I want to use for my user logins. I am guessing that I need a separate model for each table in the database. My tables are as follows: Users - user information such as first and last name, groups_id, status, etc... groups - defines the user groups with id, name, description permissions - defines a list of permissions that a group can have permission_assignments - groups_id and permissions_id. many to many relationship table I am not sure how to go about populating an array that would contain the list of permissions that a user would have based on the group they are in after a successful login. Basically, a user belongs to a group, and the group gets assigned permissions. I want to then be able to validate functions/methods based on weather the logged in user has certain permissions. Any help is appreciated

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  • Are two database trips reasonable for a login system?

    - by Randolph Potter
    I am designing a login system for a project, and have an issue about it requiring two trips to the database when a user logs in. User types in username and password Database is polled and password hash is retrieved for comparative purposes (first trip) Code tests hash against entered password (and salt), and if verified, resets the session ID New session ID and username are sent back to the database to write a row to the login table, and generate a login ID for that session. EDIT: I am using a random salt. Does this design make sense? Am I missing something? Is my concern about two trips unfounded? Comments and suggestions are welcome.

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  • Music player freezes when I switch to one of the terminal window until I login to that terminal

    - by Guanidene
    Whenever I switch to a terminal window from my X window, (by pressing Ctrl+Alt+1) my music player (banshee) running in X freezes until I login to the terminal window. Once I login into it (or either switch back to my X with or without logging into it) the music player resumes from the point it freezed. However, I observed that when am transferring files from my laptop to another computer over ssh and if I switch to one of the terminal windows, the transfer does not pause even if I don`t login to the termianl window. I just wanted to know what could possibly be the reason for such a discrimination.

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  • How to layer if statements when order of logic is irrelevant?

    - by jimmyjimmy
    Basically I have a series of logic in my website that can lead to 5 total outcomes. Basically two different if tests and then a catch all else statement. For example: if cond1: if mod1: #do things elif mod2: #do things elif cond2: if mod1: #do things elif mod2 #do things else: #do things I was thinking about rewriting it like this: if cond1 and mod1: #do things elif cond1 and mod2: #do things elif cond2 and mod1: #do things elif cond2 and mod2: #do things else: #do things Is there any real difference in these two coding options/a better choice for this kind of logic testing?

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  • How do I install OSQA with limited access to the server?

    - by Noir
    I would very much like to install OSQA on my host (provided/owned by my friend), but I can not make heads or tails of how I would go about doing it; a lot of the dependancies and such can't be installed with the current level of access (or most probably just knowledge) that I have. I can provide any further information required, and attempt to contact (friend) to see if he can do any of it for me, but I'm unsure as to how often I will be able to contact him. exec and shell_exec are not disabled on the server. If there are any other alternatives that I could use, please let me know! I tried Qwench but that was pretty buggy. I can usually deal with these kind of things, but this is my first foray into Python stuff!

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  • I cannot login to my ubuntu admin!!!! HELP!

    - by Spinz01
    I used lightdm to hide the users at the login prompt of ubuntu and it hid my account. The only one visible is "other" and it doesn't even have a known password. So, I have accessed the terminal and created a guest account and another account without passwords and they are not visible. I am new to using the terminal so I don't know how to write the commands for either: making all users/accounts visible to the login and/or making the guest accounts visible, and then once in the GUI change the login settings to see my admin account. I have no idea of what else I could do, or even how to do it. I would appreciate any help, I'm in a desperate situation because I really need to access my desktop! THANK YOU!!! Any ideas are appreciated!!!

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  • AWS Amazon EC2 - password-less SSH login for non-root users using PEM keypairs

    - by Mark White
    We've got a couple of clusters running on AWS (HAProxy/Solr, PGPool/PostgreSQL) and we've setup scripts to allow new slave instances to be auto-included into the clusters by updating their IPs to config files held on S3, then SSHing to the master instance to kick them to download the revised config and restart the service. It's all working nicely, but in testing we're using our master pem for SSH which means it needs to be stored on an instance. Not good. I want a non-root user that can use an AWS keypair who will have sudo access to run the download-config-and-restart scripts, but nothing else. rbash seems to be the way to go, but I understand this can be insecure unless setup correctly. So what security holes are there in this approach: New AWS keypair created for user.pem (not really called 'user') New user on instances: user Public key for user is in ~user/.ssh/authorized_keys (taken by creating new instance with user.pem, and copying it from /root/.ssh/authorized_keys) Private key for user is in ~user/.ssh/user.pem 'user' has login shell of /home/user/bin/rbash ~user/bin/ contains symbolic links to /bin/rbash and /usr/bin/sudo /etc/sudoers has entry "user ALL=(root) NOPASSWD: ~user/.bashrc sets PATH to /home/user/bin/ only ~user/.inputrc has 'set disable-completion on' to prevent double tabbing from 'sudo /' to find paths. ~user/ -R is owned by root with read-only access to user, except for ~user/.ssh which has write access for user (for writing known_hosts), and ~user/bin/* which are +x Inter-instance communication uses 'ssh -o StrictHostKeyChecking=no -i ~user/.ssh/user.pem user@ sudo ' Any thoughts would be welcome. Mark...

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  • ASP.NET / Active Directory - Supporting auto login for domain users

    - by Krisc
    I am developing a simple ASP.NET website that will run on the intranet on a WS2008(IIS7) box and respond to users running XP/IE8. Everything is domain connected and I am trying to automatically login the users much like SharePoint does. On my dev machine (XP), when running the site through VS, everything works. I can pickup on the user perfectly. I am using the following settings: <authentication mode="Windows"/> <identity impersonate="true"/> <anonymousIdentification enabled="false"/> <authorization> <allow users="*"/> <deny users="?"/> </authorization> However, when I publish to the WS2008 box, it doesn't work. Clearly I am missing a setting in IIS7 to support this. I have the following set for Authentication on the site: Anon Auth - Enabled ASP.NET Impersonation - Enabled Basic Auth - Disabled Forms Auth - Disabled Windows Auth - Disabled What am I missing? Thanks

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  • Paypal (sandbox) buy now link redirects to paypal (sandbox) login page instead of order summary

    - by Nicolas
    Hi, Before going live I try to test the paypal process against the paypal sandbox mode, but after the summary of what the user is going to pay on my website(buy now button), the link does not redirect to a paypal summary of the prder but ask the user to connect to paypal. Even after logging into the buyer sandbox account there's no summary of the order. It just disappears. Here's is the code I use on the checkout page: <form action="https://www.sandbox.paypal.com/cgi-bin/webscr" method="post"> <input type="hidden" name="cmd" value="_s-xclick"> <input type="hidden" name="hosted_button_id" value="XXXXXXXXXXXXXXXX"> <input type="hidden" name="notify_url" value="http://www.website.com/paypal/"> <div class="suggestion"> <input type="image" src="https://www.sandbox.paypal.com/en_GB/i/btn/btn_paynowCC_LG.gif" border="0" name="submit" alt="PayPal - The safer, easier way to pay online!"> <img alt="" border="0" src="https://www.sandbox.paypal.com/en_GB/i/scr/pixel.gif" width="1" height="1"> </div> </form> Any idea why it redirects to the payapl login page instead of the order one? Btw I'm using the Website Basic Payment (not PRO then). Cheers, Nicolas.

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  • Another Security Exception on GoDaddy after Login attempt

    - by Brian Boatright
    Host: GoDaddy Shared Hosting Trust Level: Medium The following happens after I submit a valid user/pass. The database has read/write permissions and when I remove the login requirement on an admin page that updates the database work as expected. Has anyone else had this issue or know what the problem is? Anyone? Server Error in '/' Application. Security Exception Description: The application attempted to perform an operation not allowed by the security policy. To grant this application the required permission please contact your system administrator or change the application's trust level in the configuration file. Exception Details: System.Security.SecurityException: Request for the permission of type 'System.Security.Permissions.FileIOPermission, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089' failed. Source Error: An unhandled exception was generated during the execution of the current web request. Information regarding the origin and location of the exception can be identified using the exception stack trace below. Stack Trace: [SecurityException: Request for the permission of type 'System.Security.Permissions.FileIOPermission, mscorlib, Version=2.0.0.0, Culture=neutral, PublicKeyToken=b77a5c561934e089' failed.] System.Security.CodeAccessSecurityEngine.Check(Object demand, StackCrawlMark& stackMark, Boolean isPermSet) +0 System.Security.CodeAccessPermission.Demand() +59 System.IO.FileStream.Init(String path, FileMode mode, FileAccess access, Int32 rights, Boolean useRights, FileShare share, Int32 bufferSize, FileOptions options, SECURITY_ATTRIBUTES secAttrs, String msgPath, Boolean bFromProxy) +684 System.IO.FileStream..ctor(String path, FileMode mode, FileAccess access, FileShare share) +114 System.Configuration.Internal.InternalConfigHost.StaticOpenStreamForRead(String streamName) +80 System.Configuration.Internal.InternalConfigHost.System.Configuration.Internal.IInternalConfigHost.OpenStreamForRead(String streamName, Boolean assertPermissions) +115 System.Configuration.Internal.InternalConfigHost.System.Configuration.Internal.IInternalConfigHost.OpenStreamForRead(String streamName) +7 System.Configuration.Internal.DelegatingConfigHost.OpenStreamForRead(String streamName) +10 System.Configuration.UpdateConfigHost.OpenStreamForRead(String streamName) +42 System.Configuration.BaseConfigurationRecord.InitConfigFromFile() +437 Version Information: Microsoft .NET Framework Version:2.0.50727.1433; ASP.NET Version:2.0.50727.1433

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  • Unable to login using OpenID for google apps using vanity URL

    - by GeekTantra
    Unable to login using OpenID for google apps using vanity URL I keep getting the following error whenever I use ajatus.co.in/openid as the openid url: The Allow Access screen appears but followed by this error Unable to log in with your OpenID provider: The OpenID Provider issued an assertion for an Identifier whose discovery information did not match. Assertion endpoint info: ClaimedIdentifier: http://ajatus.co.in/openid?id=1134xxxxxxxxxxxxxxx39 ProviderLocalIdentifier: http://ajatus.co.in/openid?id=1134xxxxxxxxxxxxxxx39 ProviderEndpoint: https://www.google.com/a/ajatus.co.in/o8/ud?be=o8 OpenID version: 2.0 Service Type URIs: Discovered endpoint info: [{ ClaimedIdentifier: http://specs.openid.net/auth/2.0/identifier_select ProviderLocalIdentifier: http://specs.openid.net/auth/2.0/identifier_select ProviderEndpoint: https://www.google.com/a/ajatus.co.in/o8/ud?be=o8 OpenID version: 2.0 Service Type URIs: http://specs.openid.net/auth/2.0/server },] Contents of ajatus.co.in/openid <?xml version="1.0" encoding="UTF-8"?> <xrds:XRDS xmlns:xrds="xri://$xrds" xmlns="xri://$xrd*($v*2.0)"> <XRD> <Service priority="0"> <Type>http://specs.openid.net/auth/2.0/signon</Type> <URI>https://www.google.com/a/ajatus.co.in/o8/ud?be=o8</URI> </Service> <Service priority="0"> <Type>http://specs.openid.net/auth/2.0/server</Type> <URI>https://www.google.com/a/ajatus.co.in/o8/ud?be=o8</URI> </Service> </XRD> </xrds:XRDS> contents of ajatus.co.in/.well-known/host-meta is Link: <https://www.google.com/accounts/o8/site-xrds?hd=ajatus.co.in>; rel="describedby http://reltype.google.com/openid/xrd-op"; type="application/xrds+xml"

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  • Crystal Reports Failed Database Login

    - by Marlon
    Hello, After spending a good 3 to 4 hours on google trying to find any solution to my problem I haven't had much luck. Basically, we use crystal reports for our .NET applications with a sql server back end, we have many clients each with their own server and so our reports need to have their connections dynamically set. Up until a week ago this worked fine. However a few days ago a client reported they were getting a database login prompt for a report (only one report, the rest worked fine). We were quite stumped but we managed to reproduce it on a netbook which didn't have visual studio or sql server installed. In the end the dev decided to reproduce the report in the hope it was just an oddity in that particular report. Unfortunately a new client today also experienced the same problem, but this time for every crystal report they had - and also they worked on the netbook, so we are really quite lost here. Below is a screenshot of what our clients get presented with - and here is the code that I use to set the connection information in the report cI.ServerName = (string)builder["Data Source"]; cI.DatabaseName = (string)builder["Initial Catalog"]; cI.UserID = (string)builder["User ID"]; cI.Password = (string)builder["Password"]; foreach (IConnectionInfo info in cryRpt.DataSourceConnections) { info.SetConnection(cI.ServerName, cI.DatabaseName, cI.UserID, cI.Password); } foreach (ReportDocument sub in cryRpt.Subreports) { foreach (IConnectionInfo info in sub.DataSourceConnections) { info.SetConnection(cI.ServerName, cI.DatabaseName, cI.UserID, cI.Password); } } As always, any help much appreciated.

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  • ajax-based login fails: IE6 and IE7 ajax calls delete session data (session_id however is kept)

    - by mateipavel
    This question comes after two days of testing and debugging, right after the shock I had seeing that none of the websites i build using ajax-based login work in IE<8 The most simplified scenario si this: 1. mypage.php : session_start(); $_SESSION['mytest'] = 'x'; <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.3.2/jquery.min.js"> </script> <script type="text/javascript"> function loadit() { $.post('http://www.mysite.com/myajax.php', {action: 'test'}, function(result){alert(result);}, 'html'); } </script> <a href="javascript:void(0);" onclick="loadit(); return false;">test link</a> 2. myajax.php session_start(); print_r($_SESSION); print session_id(); When I click the "test link", the ajax call is made and the result is alert()-ed: IE6: weird bullet-character (&bull;) IE7: Array( ) <session_id> IE8/FF (Expected behaviour): Array( [mytest] => 'x' ) <session_id> I would really appreciate some pointers regarding: 1. why this happens 2. how to fix it Thank you.

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  • Multiuser login into winforms application

    - by schoetbi
    Hi there, i have a winforms app in C# that needs access control for certain forms. That means, the application is running under the same (default) user at system startup, but certain forms need to be secured, so that only certain windows users could have access to the additional functions after identifying themself with username and password. For that step windows authentication should be used. Now the tricky part. Although the application was started under a "normal" user I would like the superusers to "login" into the special form without restarting the entiere application. My question now is. Is this possible (i.e. create one thread with the credentials of an administrator). Or do I need to setup another appdomain for that? Please give me a hint wather the user of a running application could be changed somehow. Thank you. EDIT I replaced administrators by "certain users" since the privileged user could be just another ordinary user that is granted access to the special functionality by the configuration of the installation.

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  • php curl login problem

    - by user331329
    <?php // create a new CURL resource $file_path = '/mail'; define("COOKIE_FILE", "c:\cookie.txt"); $ch = curl_init(); // set URL and other appropriate options curl_setopt($ch, CURLOPT_URL, "https://mail.gov.in/iwc/signin"); curl_setopt($ch, CURLOPT_HEADER, 0); curl_setopt ($ch, CURLOPT_SSL_VERIFYPEER, FALSE); curl_setopt ($ch, CURLOPT_USERAGENT, "Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6"); curl_setopt ($ch, CURLOPT_TIMEOUT, 60); curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1); curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); curl_setopt ($ch, CURLOPT_COOKIESESSION, TRUE); session_write_close(); $strCookie = 'PHPSESSID=d095af0e30afc021dd3652734009' . $_COOKIE['PHPSESSID'] . '; path=/mail'; curl_setopt( $ch, CURLOPT_COOKIE, $strCookie ); curl_setopt ($ch, CURLOPT_COOKIEJAR, COOKIE_FILE); curl_setopt($ch, CURLOPT_COOKIEFILE, COOKIE_FILE); curl_setopt ($ch, CURLOPT_POST, 1); curl_setopt ($ch, CURLOPT_POSTFIELDS,'fromLogin=true&domainName=nic.in&username=&password=&button=Sign%20In'); $url = curl_getinfo($ch); // grab URL and pass it to the browser $data = curl_exec($ch); echo $data."<pre>"; echo "<pre>"; print_r($url); // close CURL resource, and free up system resources curl_close($ch); ?> whats wrong with my code why i am not able to login directly in their mail

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  • broken SQL 2008 SP1 Web Edition (can not login with SSMS)

    - by gerryLowry
    Scenario: My installation of SQL Server 2008 Web Edition SP1 was working properly. Since I've recently joined Microsoft's Website Spark*, I removed SQL2008 and installed SQL 2008 again using my Website Spark edition and license from the MSDN download site. Next, I updated SQL 2008 to SP1 (this is required because I'm running Windows 2008 Server R2 Web edition). When I launch SSMS (SQL Server Management Studio), "User name" is "myhost\Administrator" and is greyed out so it can not be changed. When I installed my Website Spark version, I did not include "myhost\Administrator" when I was configuring SQL 2008 service accounts. Instead I created an administrator account "myhost\mySQLaccount". ERROR MESSAGE: Connect to Server (X) Cannot connect to (local) Additional information: Login failed for user 'myhost'Admistrator' (Microsoft SQL Server, Error: 18456) I tried to use the SQL Server Configuration Manager to correct this problem but could not find any useful way to fix this issue. How to I fix this problem? Connect to Server ... Server type: Database Engine Server name: (local) Authentication: Windows Authentication Please advise. Thank you. Gerry * http://www.microsoft.com/web/websitespark/default.aspx

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  • how to login to criaglist through c#

    - by kosikiza
    i m using the following code to login to criaglist, but hav't successsed yet. string formParams = string.Format("inputEmailHandle={0}&inputPassword={1}", "[email protected]", "pakistan01"); //string postData = "[email protected]&inputPassword=pakistan01"; string uri = "https://accounts.craigslist.org/"; HttpWebRequest request = (HttpWebRequest)WebRequest.Create(uri); request.KeepAlive = true; request.ProtocolVersion = HttpVersion.Version10; request.Method = "POST"; byte[] postBytes = Encoding.ASCII.GetBytes(formParams); request.ContentType = "application/x-www-form-urlencoded"; request.ContentLength = postBytes.Length; Stream requestStream = request.GetRequestStream(); requestStream.Write(postBytes, 0, postBytes.Length); requestStream.Close(); HttpWebResponse response = (HttpWebResponse)request.GetResponse(); cookyHeader = response.Headers["Set-cookie"]; string pageSource; string getUrl = "https://post.craigslist.org/del"; WebRequest getRequest = WebRequest.Create(getUrl); getRequest.Headers.Add("Cookie", cookyHeader); WebResponse getResponse = getRequest.GetResponse(); using (StreamReader sr = new StreamReader(getResponse.GetResponseStream())) { pageSource = sr.ReadToEnd(); }

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  • Asp.Net Login Control very slow initial connection to Non-Trusted AD Domain

    - by Eric Brown - Cal
    ASP.NET Login control is very slow making the initial connection to AD when authenticating to a different domain than the domain the web server is a member of. Problem occurs for the IIS server and when using with the Visual Studio's built in web server. It takes about 30 seconds the first time when attempting to use the control to connect against another domain. There is no trust relationship bewteen the web server's domain and the other domains (attempted connecting to several different domains). Subsequent connections execute quickly until the connection times out. Using Systernals Process Monitor to troubleshoot, there are two OpenQuery operations right before the delay to "C:\WINDOWS\asembly\GAC_MSIL\System.DirectoryServices\2.0.0.0_b03f5f7f11d50a3a\Netapi32.dll with a result NAME NOT FOUND" and right after the 30 second delay the TCP Send and TCP Recieves indicate communication begins with the AD server. Things we have tried: Impersonating an administrator on the web server in the web.config; Granting permissions to the CryptoKeys to the NetworkService and ASPNET; Specifying by IP instead of DNS name; Multiple variations of specifying the name and ldap server with domains and OU's; Local host entries; Looked for ports being blocked (SYN_SENT) with netstat -an. Nslookup resolves all the domains and systems involved correectly. TraceRt shows the Correct routes Any Idea or hints are greately appreicated.

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  • tastypie posting and full example

    - by Justin M
    Is there a full tastypie django example site and setup available for download? I have been wrestling with wrapping my head around it all day. I have the following code. Basically, I have a POST form that is handled with ajax. When I click "submit" on my form and the ajax request runs, the call returns "POST http://192.168.1.110:8000/api/private/client_basic_info/ 404 (NOT FOUND)" I have the URL configured alright, I think. I can access http://192.168.1.110:8000/api/private/client_basic_info/?format=json just fine. Am I missing some settings or making some fundamental errors in my methods? My intent is that each user can fill out/modify one and only one "client basic information" form/model. a page: {% extends "layout-column-100.html" %} {% load uni_form_tags sekizai_tags %} {% block title %}Basic Information{% endblock %} {% block main_content %} {% addtoblock "js" %} <script language="JavaScript"> $(document).ready( function() { $('#client_basic_info_form').submit(function (e) { form = $(this) form.find('span.error-message, span.success-message').remove() form.find('.invalid').removeClass('invalid') form.find('input[type="submit"]').attr('disabled', 'disabled') e.preventDefault(); var values = {} $.each($(this).serializeArray(), function(i, field) { values[field.name] = field.value; }) $.ajax({ type: 'POST', contentType: 'application/json', data: JSON.stringify(values), dataType: 'json', processData: false, url: '/api/private/client_basic_info/', success: function(data, status, jqXHR) { form.find('input[type="submit"]') .after('<span class="success-message">Saved successfully!</span>') .removeAttr('disabled') }, error: function(jqXHR, textStatus, errorThrown) { console.log(jqXHR) console.log(textStatus) console.log(errorThrown) var errors = JSON.parse(jqXHR.responseText) for (field in errors) { var field_error = errors[field][0] $('#id_' + field).addClass('invalid') .after('<span class="error-message">'+ field_error +'</span>') } form.find('input[type="submit"]').removeAttr('disabled') } }) // end $.ajax() }) // end $('#client_basic_info_form').submit() }) // end $(document).ready() </script> {% endaddtoblock %} {% uni_form form form.helper %} {% endblock %} resources from residence.models import ClientBasicInfo from residence.forms.profiler import ClientBasicInfoForm from tastypie import fields from tastypie.resources import ModelResource from tastypie.authentication import BasicAuthentication from tastypie.authorization import DjangoAuthorization, Authorization from tastypie.validation import FormValidation from tastypie.resources import ModelResource, ALL, ALL_WITH_RELATIONS from django.core.urlresolvers import reverse from django.contrib.auth.models import User class UserResource(ModelResource): class Meta: queryset = User.objects.all() resource_name = 'user' fields = ['username'] filtering = { 'username': ALL, } include_resource_uri = False authentication = BasicAuthentication() authorization = DjangoAuthorization() def dehydrate(self, bundle): forms_incomplete = [] if ClientBasicInfo.objects.filter(user=bundle.request.user).count() < 1: forms_incomplete.append({'name': 'Basic Information', 'url': reverse('client_basic_info')}) bundle.data['forms_incomplete'] = forms_incomplete return bundle class ClientBasicInfoResource(ModelResource): user = fields.ForeignKey(UserResource, 'user') class Meta: authentication = BasicAuthentication() authorization = DjangoAuthorization() include_resource_uri = False queryset = ClientBasicInfo.objects.all() resource_name = 'client_basic_info' validation = FormValidation(form_class=ClientBasicInfoForm) list_allowed_methods = ['get', 'post', ] detail_allowed_methods = ['get', 'post', 'put', 'delete'] Edit: My resources file is now: from residence.models import ClientBasicInfo from residence.forms.profiler import ClientBasicInfoForm from tastypie import fields from tastypie.resources import ModelResource from tastypie.authentication import BasicAuthentication from tastypie.authorization import DjangoAuthorization, Authorization from tastypie.validation import FormValidation from tastypie.resources import ModelResource, ALL, ALL_WITH_RELATIONS from django.core.urlresolvers import reverse from django.contrib.auth.models import User class UserResource(ModelResource): class Meta: queryset = User.objects.all() resource_name = 'user' fields = ['username'] filtering = { 'username': ALL, } include_resource_uri = False authentication = BasicAuthentication() authorization = DjangoAuthorization() #def apply_authorization_limits(self, request, object_list): # return object_list.filter(username=request.user) def dehydrate(self, bundle): forms_incomplete = [] if ClientBasicInfo.objects.filter(user=bundle.request.user).count() < 1: forms_incomplete.append({'name': 'Basic Information', 'url': reverse('client_basic_info')}) bundle.data['forms_incomplete'] = forms_incomplete return bundle class ClientBasicInfoResource(ModelResource): # user = fields.ForeignKey(UserResource, 'user') class Meta: authentication = BasicAuthentication() authorization = DjangoAuthorization() include_resource_uri = False queryset = ClientBasicInfo.objects.all() resource_name = 'client_basic_info' validation = FormValidation(form_class=ClientBasicInfoForm) #list_allowed_methods = ['get', 'post', ] #detail_allowed_methods = ['get', 'post', 'put', 'delete'] def apply_authorization_limits(self, request, object_list): return object_list.filter(user=request.user) I made the user field of the ClientBasicInfo nullable and the POST seems to work. I want to try updating the entry now. Would that just be appending the pk to the ajax url? For example /api/private/client_basic_info/21/? When I submit that form I get a 501 NOT IMPLEMENTED message. What exactly haven't I implemented? I am subclassing ModelResource, which should have all the ORM-related functions implemented according to the docs.

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  • how can I know current viewcontroller name in iphone

    - by Shikhar
    I have BaseView which implement UIViewController. Every view in project must implement this BaseView. In BaseView, I have method: -(void) checkLoginStatus { defaults = [[NSUserDefaults alloc] init]; if(![[defaults objectForKey:@"USERID"] length] > 0 ) { Login *login=[[Login alloc] initWithNibName:@"Login" bundle:nil]; [self.navigationController pushViewController:login animated:TRUE]; [login release]; } [defaults release]; } The problem is my Login view also implement BaseView, checks for login, and again open LoginView i.e. stuck in to recursive calling. Can I check in checkLoginStatus method if request is from LoginView then take no action else check login. Ex: (void) checkLoginStatus { if(SubView is NOT Login){ defaults = [[NSUserDefaults alloc] init]; if(![[defaults objectForKey:@"USERID"] length] 0 ) { Login *login=[[Login alloc] initWithNibName:@"Login" bundle:nil]; [self.navigationController pushViewController:login animated:TRUE]; [login release]; } [defaults release]; } } Please help..

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