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  • nothing happen after a command mount -t (worked before)

    - by user3449429
    i'm having a weird problem. i used to lauch manually the mount command to link a folder on my PLEX server with a folder on my NAS since yesterday it was ok, but i had to halt my plex server and when i tried to mount again the folder, nothing happen. it ask me the su password and that's all. here the command i use in my fstab: //192.168.1.2/Series_TV /home/cidou/Series_TV cifs _netdev,credentials=/home/cidou/.smbcredentials 0 0 //192.168.1.2/films /home/cidou/Films cifs _netdev,credentials=/home/cidou/.smbcredentials 0 0 i tried this command too: sudo mount -t smbfs //192.168.1.2/Films /home/cidou/Films -o user=myname,password=mypass,sec=ntlm --verbose i run an ubuntu 12.04 LTS uname -a Linux plex 3.8.0-29-generic #42~precise1-Ubuntu SMP Wed Aug 14 15:31:16 UTC 2013 i686 i686 i386 GNU/Linux Welcome to Ubuntu 12.04.4 LTS (GNU/Linux 3.8.0-29-generic i686) * Documentation: https://help.ubuntu.com/ System information disabled due to load higher than 2.0 Thanks for reading

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  • Directory access control with Apache: do I need to use a specific .htaccess?

    - by Mirror51
    I have an Apache webserver, and in the Apache configuration, I have Alias /backups "/backups" <Directory "/backups"> AllowOverride None Options Indexes Order allow,deny Allow from all </Directory> I can access files via http://127.0.0.1/backups. The problem is everyone can access that. I have a web interface, e.g. http://localhost/adminm that is protected with htaccess and password. Now I don't want separate .htaccess and .htpasswd for /backups, and I don't want a second password prompt when a user clicks on /backups in the web interface. Is there any way to use same .htaccess and .htpasswd for the backups directory?

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  • Nomodeset Installation

    - by Camacho3112
    I were following the address from Coldfish on How to set nomodeset, but I don't know how to "save" the changes made to the line GRUB_CMDLINE_LINUX_DEFAULT="quiet splash nomodeset" I hit CTRL+O to save and get File Name to write: /etc/default/grub AND typed sudo update-grub AND hit ENTER. After that, I open another Terminal an type: sudo update-grub (ask me for password) and them I got this: joseluis@ubuntu:~$ sudo update-grub [sudo] password for joseluis: Generating grub.cfg ... cat: /boot/grub/video.lst: No such file or directory Found linux image: /boot/vmlinuz-2.6.38-12-generic Found initrd image: /boot/initrd.img-2.6.38-12-generic Found linux image: /boot/vmlinuz-2.6.38-8-generic Found initrd image: /boot/initrd.img-2.6.38-8-generic Found Windows 7 (loader) on /dev/sda1 Found Ubuntu 10.04 LTS (10.04) on /dev/sda6 done joseluis@ubuntu:~$ SO: Were I'm? Were is my direction now? Thanks for the help.

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  • Encrypted usb stick not booting anymore how can I resolve the Disaster

    - by statquant
    Guys I am experiencing massive problem here. I created a bootable usb stick that I "full disk" encrypted with Ubuntu 13.04. I was using it when it froze. I had to reboot it manually and now I cannot boot on it anymore (It is not listed in the boot menu anymore) If I put when I run Ubuntu on my laptop ubuntu can see it and I am asked for a password But the correct password does not seem to work (I assume that this passphrase is the one I was using previously at startup) Can you please help me figure this thing out, this is massive problem to me...

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  • Ubuntu 12.04 on Vm Player showing Wired Network Instead of Wireless Network

    - by Fak365
    I am new to Ubuntu, recently I installed ubuntu 12.04 in Vm Player (Virtual Machine) on my Dell laptop having windows 7 ultimate 32 bit for just to check the security of my wireless network and want to crack the WiFi (WPA-PSK) password but in ubuntu it does not show the wireless network it shows the 2 arrow sign as i have not connect the ethernet cable to my laptop and connected through WiFi on my main OS (windows 7) but it shows the wired network and internet is working but it does not show wifi connection. On windows 7 WiFi is connected and showing the WiFi connection and working correctly.But my main motive is to crack the WiFi password as it can't detect WiFi network so what to do? Please Help.!!thanks My Laptop Specification : Laptop : Dell Latitude D620 OS : Windows 7 Ultimate 32 bit Processor : Core Duo 2 T7200 @ 2Ghz Ram : 2 GB WiFi card : Intel Pro/Wireless 3945 ABG Virtual Machine : Vm Player V 5.0.1 If Need to Install Drivers Please Give Me Full Information how to install and which driver I should install. Thanks In Advance.

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  • How to determine if someone is accessing our database remotely?

    - by Vednor
    I own a content publishing website developed using CakePHP(tm) v 2.1.2 and 5.1.63 MySQL. It was developed by a freelance developer who kept remote access to the database which I wasn’t aware of. One day he accessed to the site and overwrote all the data. After the attack, my hosting provider disabled the remote access to our database and changed the password. But somehow he accessed the site database again and overwrote some information. We’ve managed to stop the attack second time by taking the site down immediately. But now we’re suspecting that he’ll attack again. What we could identified that he’s running a query and changing every information from the database in matter of a sec. Is there any possible way to detect the way he’s accessing our database without remote access or knowing our Cpanel password? Or to identify whether he has left something inside the site that granting him access to our database?

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  • Unable to connect FileZilla to ubuntu ec2

    - by user1775063
    I have a micro ubuntu instance on ec2. I have done a passwd to set it to simple password. I have installed vsftpd on the ec2 instance. And imported the ec2 pem file via FileZilla-Settings-SFTP, and configured vsftpd.conf with following listen=YES anonymous_enable=NO local_enable=YES write_enable=YES local_umask=022 dirmessage_enable=YES use_localtime=YES xferlog_enable=YES connect_from_port_20=YES secure_chroot_dir=/var/run/vsftpd/empty pam_service_name=vsftpd rsa_cert_file=/etc/ssl/private/vsftpd.pem local_root=/home/ubuntu pasv_enable=YES pasv_max_port=12100 pasv_min_port=12000 port_enable=YES I am using username ubuntu, password that_i_set, port 21. I get the following error Error: Critical error Error: Could not connect to server

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  • Get root access for copying files to /usr/share/...?

    - by Vinaychalluru
    To be short, I want to copy a folder to a location /usr/share/screenlets/..... I don't know how to do it. I am using Ubuntu 10.04. I tried by logging in as root from terminal giving "su with my password". I even changed my user account type to ADMINISTRATOR by giving the root password when asked, yet, no use. Think all of you know that, even the option PASTE in the context menu's list in the folder "/usr/share/..." is INACTIVE. How can I copy those files?

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  • How can I run everything as root

    - by Hermione
    I have dual booted to lubuntu (with Windows XP) and everytime and then I'm getting asked for my password. How do I run everything as root and not ask a password again? Ideally I wanted to run nginx but it has permission denied issues: apathetic@ubuntu:~$ service nginx start Starting nginx: nginx: [alert] could not open error log file: open() "/var/log/nginx/error.log" failed (13: Permission denied) 2012/08/03 20:06:25 [warn] 4762#0: the "user" directive makes sense only if the master process runs with super-user privileges, ignored in /etc/nginx/nginx.conf:1 nginx: the configuration file /etc/nginx/nginx.conf syntax is ok 2012/08/03 20:06:25 [emerg] 4762#0: open() "/var/run/nginx.pid" failed (13: Permission denied) nginx: configuration file /etc/nginx/nginx.conf test failed

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  • Stuck in screen saver during upgrade

    - by vinni_f
    I'm upgrading from 10.04 to 12.04. The screen saver is now on and I can't exit it. If I type my password or move the mouse the screen flickers. Regardless of what I do the screen saver stays and the login prompt never shows. If I type my password and then enter it, it does nothing. Last time I checked the packages were all downloaded and I was in the install process. I really want this upgrade to work. It would be really bad to have to restart from a fresh install. Cheers.

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  • Webcam Q&A page with guest "speakers" ideas?

    - by myladeybugg
    I want to host one way webcam Q&A's/AMA's and embed it on my sites webpage. I must be able to have specified hosts for these Q&A's connect quickly and easily to the webpage, and read user posted messages (questions) for interaction. I know there are some systems out there, like tinychat, but I'm unaware if there is anything exactly used for what I am looking for. Perhaps requiring a specific account/password before allowing streaming on the page, where I can email the host a password or directions to create an account to begin live streaming. Most of these people doing the question and answers on webcam will have little technical knowledge, so extra cookie points goes to ease of use! Thanks for ideas in advance! PS: I apologize if this is the incorrect section for posting this question, this is my first.

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  • 12.04 Login gone wrong

    - by Mark H
    I seem to have a fault in the login procedure somewhere. When I boooted up I found that if I selected Guest I could use the computer. If I selected my administrator account I was taken straight to the terminal. So I opened a guest session, went to user settings, and unlocked my admin account (it accepted the password!), and amended it to show "password None" and automatic login. I now find that when I boot up I am taken straight to the terminal - if I exit terminal I get the login screen - if I select the administrator login I go back to terminal - if i select guest I can use the PC, and if I select mu normal user account I can use the PC. So I cannot login as an administrator - so admin functions such as update are no longer accessible. Sorry to be so long winde but I am stuck. Can anyone suggest anything - I am a beginner with this

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  • Private downloads [on hold]

    - by user1314836
    I am setting up a personal website and I would like to be able to share certain files with friends (e.g. photo albums or documents). Of course, each of the files should only be accessible to certain people, for example with a password or a private address. I would like to have a simple system, easy to maintain. I'm quite sure that Wordpress would let me know do something but keeping a Wordpress installation just for sharing a few files per month seems too much work. I am thinking of using the Apache capabilities for folder or file password protecting. Is this the best idea? Or would it be better just to block the directory navigation so that only people with the full path for each of the files can download them? Thanks!

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  • Can't Login to my PC

    - by user68775
    Hope u solve my problem. I am new to Ubuntu. I tried to install Ubuntu 12.04 in my pc which is way old because friends told me that it could be installed in any machine. I dually installed ubuntu along with Windows XP. After installation i get a message "Run ubuntu in low graphics" I gave yes and then i was asked for userid and Password. But i keep getting the same UserID page even though i gave the corrent UserID and Password. Is there any solution you can provide. This is my first experience with Ubuntu.

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  • Script to connect to hidden wireless network with static IP?

    - by nLinked
    Would like a script, when run, it should connect to a hidden wireless network with these details: SSID is "Wireless" Network is not broadcasting its SSID above (is hidden) WPA2-PSK, AES, password is "password" Static IP: 192.168.1.1 Def. gateway: 192.168.1.254 DNS: 192.168.1.254 No idea how to do this. But I do know the wireless interface is called wlan0 and I'm on Ubuntu 10.10. I don't want to use the built in Network Manager as it never auto-connects on startup. WICD doesn't work either. Any ideas most welcome.

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  • KSoap2 Android not valid SOAP

    - by Rogier21
    Hello all, I am trying to post to my own test soap server (C#) with Android in combination with KSOAP2. Now I have the specifications from the SOAP server, it expects: POST /SharingpointCheckBarcode.asmx HTTP/1.1 Host: awc.test.trin-it.nl Content-Type: text/xml; charset=utf-8 Content-Length: length SOAPAction: "http://tempuri.org/checkBarcode" <?xml version="1.0" encoding="utf-8"?> <soap:Envelope xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"> <soap:Header> <AuthHeader xmlns="http://tempuri.org/"> <username>string</username> <password>string</password> </AuthHeader> </soap:Header> <soap:Body> <checkBarcode xmlns="http://tempuri.org/"> <barcode>string</barcode> </checkBarcode> </soap:Body> </soap:Envelope> But what Android KSOAP2 sends out: <?xml version="1.0" encoding="utf-8"?> <v:Envelope xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns:d="http://www.w3.org/2001/XMLSchema" xmlns:c="http://schemas.xmlsoap.org/soap/encoding/" xmlns:v="http://schemas.xmlsoap.org/soap/envelope/"> <v:Header /> <v:Body> <checkBarcode xmlns="http://tempuri.org" id="o0" c:root="1"> <username i:type="d:string">test</username> <password i:type="d:string">test</password> <barcode i:type="d:string">2620813000301</barcode> </checkBarcode> </v:Body> </v:Envelope> With this code: try { SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); request.addProperty("username", "test"); request.addProperty("password", "test"); request.addProperty("barcode", "2620813000301"); SoapSerializationEnvelope envelope = new SoapSerializationEnvelope(SoapEnvelope.VER11); envelope.dotNet = true; envelope.encodingStyle = "test"; envelope.setOutputSoapObject(request); AndroidHttpTransport androidHttpTransport = new AndroidHttpTransport (URL); androidHttpTransport.debug = true; androidHttpTransport.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"utf-8\"?>"); androidHttpTransport.call(SOAP_ACTION, envelope); Log.d("MyAPP", "----------------- " + androidHttpTransport.requestDump +"\r\n\r\n" + androidHttpTransport.responseDump); ((TextView)findViewById(R.id.lblStatus)).setText(androidHttpTransport.requestDump +"\r\n\r\n" + androidHttpTransport.responseDump); } catch(Exception E) { ((TextView)findViewById(R.id.lblStatus)).setText("ERROR:" + E.getClass().getName() + ": " + E.getMessage()); } The response I get back from the server is that there are no results found, so not an error, but when I test it with another App or PHP, it with the same data, it says it's OK. I think it's because of the

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  • How to use NSMutableDictionary to store and retrieve data

    - by TechFusion
    I have created Window Based application and tab bar controller as root controller. My objective is to store Text Field data values in one tab bar VC and will be accessible and editable by other VC and also retrievable when application start. I am looking to use NSMutableDictionary class in AppDelegate so that I can access stored Data Values with keys. //TestAppDelegate.h extern NSString *kNamekey ; extern NSString *kUserIDkey ; extern NSString *kPasswordkey ; @interface TestAppDelegate :NSObject{ UIWindow *window; IBOutlet UITabBarController *rootController; NSMutableDictionary *outlineData ; } @property(nonatomic,retain)IBOutlet UIWindow *window; @property(nonatomic,retain)IBOutlet UITabBarController *rootController; @property(nonatomic,retain) NSMutableDictionary *outlineData ; @end //TestAppDelegate.m import "TestAppDelegate.h" NSString *kNamekey =@"Namekey"; NSString *kUserIDkey =@"UserIDkey"; NSString *kPasswordkey =@"Passwordkey"; @implemetation TestAppDelegate @synthesize outlineData ; -(void)applicationDidFinishLaunching:(UIApplication)application { NSMutableDictionary *tempMutableCopy = [[[NSUserDefaults standardUserDefaults] objectForKey:kRestoreLocationKey] mutableCopy]; self.outlineData = tempMutableCopy; [tempMutableCopy release]; if(outlineData == nil){ NSString *NameDefault = NULL; NSString *UserIDdefault= NULL; NSString *Passworddefault= NULL; NSMutableDictionary *appDefaults = [NSMutableDictionary dictionaryWithObjectsAndKeys: NameDefault, kNamekey , UserIDdefault, kUserIDkey , Passworddefault, kPasswordkey , nil]; self.outlineData = appDefaults; [appDefaults release]; } [window addSubview:rootController.view]; [window makeKeyAndVisible]; NSMutableDictionary *savedLocationDict = [NSMutableDictionary dictionaryWithObject:outlineData forKey:kRestoreLocationKey]; [[NSUserDefaults standardUserDefaults] registerDefaults:savedLocationDict]; [[NSUserDefaults standardUserDefaults] synchronize]; } -(void)applicationWillTerminate:(UIApplication *)application { [[NSUserDefaults standardUserDefaults] setObject:outlineData forKey:kRestoreLocationKey]; } @end Here ViewController is ViewController of Navigation Controller which is attached with one tab bar.. I have attached xib file with ViewController //ViewController.h @interface IBOutlet UITextField *Name; IBOutlet UITextField *UserId; IBOutlet UITextField *Password; } @property(retain,nonatomic) IBOutlet UITextField *Name @property(retain,nonatomic) IBOutlet UITextField *UserId; @property(retain,nonatomic) IBOutlet UITextField *Password; -(IBAction)Save:(id)sender; @end Here in ViewController.m, I am storing object values with keys. /ViewController.m -(IBAction)Save:(id)sender{ TestAppDelegate appDelegate = (TestAppDelegate)[[UIApplication sharedApplication] delegate]; [appDelegate.outlineData setObject:Name.text forKey:kNamekey ]; [appDelegate.outlineData setObject:UserId.text forKey:kUserIDkey ]; [appDelegate.outlineData setObject:Password.text forKey:kPasswordkey]; [[NSUserDefaults standardUserDefaults] synchronize]; } I am accessing stored object using following method. -(void)loadData { TabBarAppDelegate *appDelegate = (TabBarAppDelegate *)[[UIApplication sharedApplication] delegate]; Name = [appDelegate.outlineData objectForKey:kNamekey ]; UserId = [appDelegate.outlineData objectForKey:kUserIDkey ]; Password = [appDelegate.outlineData objectForKey:kPasswordkey]; [Name release]; [UserId release]; [Password release]; } I am getting EXEC_BAD_ACCESS in application. Where I am making mistake ? Thanks,

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  • How to prevent jQuery FancyBox from closing immediately after submit?

    - by Dimitri
    Hi! I'm loading an inline registration form in a FancyBox from jQuery. However after submitting the form, the box immediately closes while there is some feedback that I want to show the user in the FancyBox itself. This feedback is generated on the server side and is printed in the FancyBox. How can I make the box only closing when their is no feedback anymore? I was thinking about using ajax to just refresh the FancyBox itself and not the whole page after refreshing. But I just can't figure out how this ajax $.ajax({type, cache, url, data, success}); works... Also it seems like there's no reaction from the 'submit bind' in the javascript. I hope someone can help me with this problem. I paste my code below. If any questions, plz ask.. Thx in advance! This is the javascript: <script type="text/javascript"> $(document).ready(function() { $("#various1").fancybox({ 'transitionIn' : 'none', 'transitionOut' : 'none', 'scrolling' : 'no', 'titleShow' : false, 'onClosed' : function() { $("#registration_error").hide(); } }); }); $("#registration_form").bind("submit", function() { if ($("#registration_error").val() != "Registration succeeded!") { $("#registration_error").show(); $.fancybox.resize(); return false; } $.fancybox.showActivity(); $.ajax({ type : "POST", cache : false, url : "/data/login.php", data : $(this).serializeArray(), success : function(data) { $.fancybox(data); } }); return false; }); This is the inline form that I show in the FancyBox: <div style="display: none;"> <div id="registration" style="width:227px;height:250px;overflow:auto;padding:7px;"> <?php echo "<p id=\"registration_error\">".$feed."</p>"; ?> <form id="registration_form" action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post"> <p> <label for="username">Username: </label> <input type="text" id="login_name" name="username" size="30" /> </p> <p> <label for="password">Password: </label> <input type="password" id="pass" name="pw" size="30" /> </p> <p> <label for="repeat_password">Repeat password: </label> <input type="password" id="rep_pass" name="rep_pw" size="30" /> </p> <p> <input type="submit" value="Register" name="register" id="reg" /> </p> <p> <em></em> </p> </form> </div> </div>

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  • Android using ksoap calling PHP SOAP webservice fails: 'procedure 'CheckLogin' not found

    - by AmazingDreams
    I'm trying to call a PHP SOAP webservice. I know my webservice functions correctly because I use it in a WPF project succesfully. I'm also building an app in android, using the same webservice. The WSDL file can be found here: http://www.wegotcha.nl/servicehandler/service.wsdl This is my code in the android app: String SOAP_ACTION = "http://www.wegotcha.nl/servicehandler/CheckLogin"; String NAMESPACE = "http://www.wegotcha.nl/servicehandler"; String METHOD_NAME = "CheckLogin"; String URL = "http://www.wegotcha.nl/servicehandler/servicehandler.php"; String resultData = ""; SoapSerializationEnvelope soapEnv = new SoapSerializationEnvelope(SoapEnvelope.VER11); SoapObject request = new SoapObject(NAMESPACE, METHOD_NAME); SoapObject UserCredentials = new SoapObject("Types", "UserCredentials6"); UserCredentials.addProperty("mail", params[0]); UserCredentials.addProperty("password", md5(params[1])); request.addSoapObject(UserCredentials); soapEnv.setOutputSoapObject(request); HttpTransportSE http = new HttpTransportSE(URL); http.setXmlVersionTag("<?xml version=\"1.0\" encoding=\"utf-8\"?>"); http.debug = true; try { http.call(SOAP_ACTION, soapEnv); } catch (IOException e) { e.printStackTrace(); } catch (XmlPullParserException e) { e.printStackTrace(); } SoapObject results = null; results = (SoapObject)soapEnv.bodyOut; if(results != null) resultData = results.getProperty(0).toString(); return resultData; Using fiddler I got the following: Android request: <?xml version="1.0" encoding="utf-8"?> <v:Envelope xmlns:i="http://www.w3.org/2001/XMLSchema-instance" xmlns:d="http://www.w3.org/2001/XMLSchema" xmlns:c="http://schemas.xmlsoap.org/soap/encoding/" xmlns:v="http://schemas.xmlsoap.org/soap/envelope/"><v:Header /> <v:Body> <n0:CheckLogin id="o0" c:root="1" xmlns:n0="http://www.wegotcha.nl/servicehandler"> <n1:UserCredentials6 i:type="n1:UserCredentials6" xmlns:n1="Types"> <mail i:type="d:string">myemail</mail> <password i:type="d:string">myhashedpass</password> </n1:UserCredentials6> </n0:CheckLogin> </v:Body> </v:Envelope> Getting the following response: Procedure 'CheckLogin' not present My request produced by my WPF app looks completely different: <s:Envelope xmlns:s="http://schemas.xmlsoap.org/soap/envelope/"> <s:Body xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"> <UserCredentials6 xmlns="Types"> <mail>mymail</mail> <password>mypassword</password> </UserCredentials6> </s:Body> </s:Envelope> After googling my ass off I was not able to solve this problem by myself. It may be possible there are some weird things in my Java code because I changed a lot since. I hope you guys will be able to help me, thanks. EDIT: My webservice is of document/literal encoding style, after doing some research I found I should be able to use SoepEnvelope Instead of SoapSerializationEnvelope Though when I replace this I get an error before the try cache block, causing my app to crash. Error: 11-04 16:23:26.786: E/AndroidRuntime(26447): Caused by: java.lang.ClassCastException: org.ksoap2.serialization.SoapObject cannot be cast to org.kxml2.kdom.Node Which is caused by these lines: request.addSoapObject(UserCredentials); soapEnv.setOutputSoapObject(request); This may be a solution though, how do I go about this? I found nothing about using a SoapEnvelope instead of a SoapSerializationEnvelope except for this awesome tutorial: http://ksoap.objectweb.org/project/mailingLists/ksoap/msg00849.html

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  • Entity with Guid ID is not inserted by NHibernate

    - by DanK
    I am experimenting with NHibernate (version 2.1.0.4000) with Fluent NHibernate Automapping. My test set of entities persists fine with default integer IDs I am now trying to use Guid IDs with the entities. Unfortunately changing the Id property to a Guid seems to stop NHibernate inserting objects. Here is the entity class: public class User { public virtual int Id { get; private set; } public virtual string FirstName { get; set; } public virtual string LastName { get; set; } public virtual string Email { get; set; } public virtual string Password { get; set; } public virtual List<UserGroup> Groups { get; set; } } And here is the Fluent NHibernate configuration I am using: SessionFactory = Fluently.Configure() //.Database(SQLiteConfiguration.Standard.InMemory) .Database(MsSqlConfiguration.MsSql2008.ConnectionString(@"Data Source=.\SQLEXPRESS;Initial Catalog=NHibernateTest;Uid=NHibernateTest;Password=password").ShowSql()) .Mappings(m => m.AutoMappings.Add( AutoMap.AssemblyOf<TestEntities.User>() .UseOverridesFromAssemblyOf<UserGroupMappingOverride>())) .ExposeConfiguration(x => { x.SetProperty("current_session_context_class","web"); }) .ExposeConfiguration(Cfg => _configuration = Cfg) .BuildSessionFactory(); Here is the log output when using an integer ID: 16:23:14.287 [4] DEBUG NHibernate.Event.Default.DefaultSaveOrUpdateEventListener - saving transient instance 16:23:14.291 [4] DEBUG NHibernate.Event.Default.AbstractSaveEventListener - saving [TestEntities.User#<null>] 16:23:14.299 [4] DEBUG NHibernate.Event.Default.AbstractSaveEventListener - executing insertions 16:23:14.309 [4] DEBUG NHibernate.Event.Default.AbstractSaveEventListener - executing identity-insert immediately 16:23:14.313 [4] DEBUG NHibernate.Persister.Entity.AbstractEntityPersister - Inserting entity: TestEntities.User (native id) 16:23:14.321 [4] DEBUG NHibernate.AdoNet.AbstractBatcher - Opened new IDbCommand, open IDbCommands: 1 16:23:14.321 [4] DEBUG NHibernate.AdoNet.AbstractBatcher - Building an IDbCommand object for the SqlString: INSERT INTO [User] (FirstName, LastName, Email, Password) VALUES (?, ?, ?, ?); select SCOPE_IDENTITY() 16:23:14.322 [4] DEBUG NHibernate.Persister.Entity.AbstractEntityPersister - Dehydrating entity: [TestEntities.User#<null>] 16:23:14.323 [4] DEBUG NHibernate.Type.StringType - binding null to parameter: 0 16:23:14.323 [4] DEBUG NHibernate.Type.StringType - binding null to parameter: 1 16:23:14.323 [4] DEBUG NHibernate.Type.StringType - binding 'ertr' to parameter: 2 16:23:14.324 [4] DEBUG NHibernate.Type.StringType - binding 'tretret' to parameter: 3 16:23:14.329 [4] DEBUG NHibernate.SQL - INSERT INTO [User] (FirstName, LastName, Email, Password) VALUES (@p0, @p1, @p2, @p3); select SCOPE_IDENTITY();@p0 = NULL, @p1 = NULL, @p2 = 'ertr', @p3 = 'tretret' and here is the output when using a Guid: 16:50:14.008 [4] DEBUG NHibernate.Event.Default.DefaultSaveOrUpdateEventListener - saving transient instance 16:50:14.012 [4] DEBUG NHibernate.Event.Default.AbstractSaveEventListener - generated identifier: d74e1bd3-1c01-46c8-996c-9d370115780d, using strategy: NHibernate.Id.GuidCombGenerator 16:50:14.013 [4] DEBUG NHibernate.Event.Default.AbstractSaveEventListener - saving [TestEntities.User#d74e1bd3-1c01-46c8-996c-9d370115780d] This is where it silently fails, with no exception thrown or further log entries. It looks like it is generating the Guid ID correctly for the new object, but is just not getting any further than that. Is there something I need to do differently in order to use Guid IDs? Thanks, Dan.

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  • How to include the login form on the Home index page in MVC

    - by Bernard Larouche
    Hi guys I really need your help for this. I am relatively new to programming and I need help to something that could be easy for a experienced programmer. I would like to get the login form that we get for free in an MVC application on the left sidebar of my Home index page instead of the usual Account/Login page. I am facing some problems. First I need a product object to be displayed on my Home Index page as well. What I did is that I added a product object to the LogOnModel that they provide in the AccountModels class and I created a UserControl (partial view) copying the content of the LogOn.aspx view. Now my Home index.aspx as well as my partial view inherits the LogOnModel class. I can see the Login form on my Home Index page as well as my product object BUT the login Form is never empty. The last username and password always appear there. I know I must have forgotten something or have done something wrong or the way did it is completely wrong !! Please could you give me some advice Thks <%@ Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<CoderForTradersSite.Models.LogOnModel>" %> <h4>Login Form</h4> <p> Please enter your username and password. <%= Html.ActionLink("Register", "Register") %> if you don't have an account. </p> <% using (Html.BeginForm()) { %> <%= Html.ValidationSummary(true, "Login was unsuccessful. Please correct the errors and try again.") %> <div> <fieldset> <legend>Account Information</legend> <div class="editor-label"> <%= Html.LabelFor(m => m.UserName) %> </div> <div class="editor-field"> <%= Html.TextBoxFor(m => m.UserName) %> <%= Html.ValidationMessageFor(m => m.UserName) %> </div> <div class="editor-label"> <%= Html.LabelFor(m => m.Password) %> </div> <div class="editor-field"> <%= Html.PasswordFor(m => m.Password) %> <%= Html.ValidationMessageFor(m => m.Password) %> </div> <div class="editor-label"> <%= Html.CheckBoxFor(m => m.RememberMe) %> <%= Html.LabelFor(m => m.RememberMe) %> </div> <p> <input type="submit" value="Log On" /> </p> </fieldset> </div> <% } %>

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  • Rails controller method going to plural form

    - by Jty.tan
    I'm new to rails! Ok, I am trying to set up a user signup form. It is mapped as a singular resource in the routes map.resource :user And trying to create the user through the console works fine. the controller code for user's signup is as follows: def signup @user = User.new#(params[:user]) end def create @user = User.new(params[:user]) #debugger if request.post? if @user.save session[:user] = User.authenticate(@user.login, @user.password) flash[:message] = "Signup successful" redirect_to registries_path else flash[:warning] = "Signup unsuccessful" #redirect_to 'user/signup' end end end The signup view is as follows (and this is where i think something is going wrong) <% form_tag user_path do %> <p>User creation.</p> <p><%= error_messages_for 'user' %></p> <p> <label>Username:</label> <%= text_field_tag 'login', nil, :size => 20, :maxlength => 20 %> </p> <p> <label>Password:</label> <%= password_field_tag 'password', nil, :size => 20, :maxlength => 20 %> </p> <p> <label>Password confirmation:</label> <%= password_field_tag 'password_confirmation', nil, :size => 20, :maxlength => 20 %> </p> <p> <label>Email:</label> <%= text_field_tag 'email' %> </p> <p><%= submit_tag 'Signup' %></p> <% end %> Now, that page renders just fine. I've called the form on the "user_path" which is singular (i think?). But when I hit the submit button, it gives me an error saying that uninitialized constant UsersController the occurence of the error makes sense, since User is meant to be singular, so if it is trying to call the Users controller, it should be chucking an error. When I checked the server log, it shows this message: Processing ApplicationController#create (for 127.0.0.1 at 2010-05-08 16:26:14) [POST] Parameters: {"commit"=>"Signup", "password_confirmation"=>"[FILTERED]", "action"=>"create", "authenticity_token"=>"yOcHY+rMjaqmu9HS8EwnDqJKbc0Zxictc0y4dtD26ac=", "controller"=>"users", "login"=>"bob", "password"=>"[FILTERED]", "email"=>"[email protected]"} NameError (uninitialized constant UsersController): In the params, I can see that it is calling the "users" controller. But I'm not sure how to fix that, or what is causing it to call the "users" controller as opposed to the "user" controller. Any ideas? Thanks in advance!

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  • Tomcat does save logged users during restart

    - by mabuzer
    How to force Tomcat to save logged users, so that the they kept logged in even after Tomcat has restarted? Right now the user has to login again everytime. Added the following lines into web-app context.xml: <Manager className="org.apache.catalina.session.PersistentManager"> <Store className="org.apache.catalina.session.FileStore"/> </Manager> but still I see login page after Tomcat restart, I use Tomcat 6.0.26 Update I managed to solve it like this: 1) Make my own extended version of FormAuthentication class: package com.alz.tomcat; import java.io.IOException; import java.security.Principal; import org.apache.catalina.Session; import org.apache.catalina.deploy.LoginConfig; import org.apache.catalina.connector.Request; import org.apache.catalina.connector.Response; import org.apache.catalina.authenticator.Constants; import org.apache.catalina.authenticator.FormAuthenticator; /** * * @author mabuzer */ public class Authenticator extends FormAuthenticator { @Override public boolean authenticate(Request request, Response response, LoginConfig config) throws IOException { String username = (String) request.getSession().getAttribute("USERNAME"); String password = (String) request.getSession().getAttribute("PASSWORD"); Principal principal = request.getUserPrincipal(); Session session = request.getSessionInternal(true); if (request.getUserPrincipal() == null && !isNull(username) && !isNull(password)) { principal = context.getRealm().authenticate(username, password); if (principal != null) { session.setNote(Constants.FORM_PRINCIPAL_NOTE, principal); if (!matchRequest(request)) { register(request, response, principal, Constants.FORM_METHOD, username, password); return (true); } } return super.authenticate(request, response, config); } else { return super.authenticate(request, response, config); } } private boolean isNull(String str) { if (str == null || "".equals(str)) { return true; } else { return false; } } } 2) Have your own ContextConfig class: package com.alz.tomcat; import java.util.HashMap; import org.apache.catalina.Valve; /** * * @author [email protected] */ public class ContextConfig extends org.apache.catalina.startup.ContextConfig { public ContextConfig() { super(); // we need to append our authenticator setCustomAuthenticators(customAuthenticators); customAuthenticators = new HashMap(); customAuthenticators.put("Authenticator" , new Authenticator()); } } 3) Have a class extends LifeCycleListener to set replace default ContextConfig the one you made: package com.alz.tomcat; import org.apache.catalina.Lifecycle; import org.apache.catalina.LifecycleEvent; import org.apache.catalina.core.StandardHost; /** * * @author [email protected] */ public class LifeCycleListener implements org.apache.catalina.LifecycleListener { public void lifecycleEvent(LifecycleEvent lifeCycleEvent) { if (Lifecycle.BEFORE_START_EVENT.equals(lifeCycleEvent.getType())) { StandardHost aStandardHost = (StandardHost) lifeCycleEvent.getLifecycle(); aStandardHost.setConfigClass("com.alz.tomcat.ContextConfig"); } } } 4) Final step which is to add your LifeCycleListener to server.xml in Host tag like this: <Host appBase="webapps" autoDeploy="true" name="localhost" unpackWARs="true" xmlNamespaceAware="false" xmlValidation="false"> <Listener className="com.alz.tomcat.LifeCycleListener"/> </Host>

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