Search Results

Search found 90397 results on 3616 pages for 'user preferences'.

Page 42/3616 | < Previous Page | 38 39 40 41 42 43 44 45 46 47 48 49  | Next Page >

  • adding user to windows make the administrator icon to disappear !

    - by user283322
    Hello I used the command "net user" to add a new admin user to windows like that: net user myuser11 myuser11 /add net localgroup Administrators myuser11 /add the problem that after restart windows I only see the icon of the "myuser" and the default windows "Administrator" login icon disappeared !! the "Administrator" files still exists and I can login as "Administrator" after ctrl+alt+del but I need of course to display the "Administrator" icon in welcome screen how I fix that ? I use windows xp sp3 thanks

    Read the article

  • How to manage end user documentation for a project under continuous integration?

    - by mcdon
    I have a project under continuous integration and would like to add end user documentation to the project. The end user documentation is a user manual, not API documentation. In our environment we use windows, c#, msbuild, cruisecontrol.net and subversion. We are currently using DocToHelp to create our help file, which is based on an msword document. I'm looking for some guidance on how to manage the end user documentation. What documentation tools should I use? Should any of the documentation tools be part of the build script? Should the output files from the documentation tool be stored in subversion? What type of help files would be best to use?

    Read the article

  • How can I check if a user has written his username and password correctly?

    - by Sergio Tapia
    I'm using a Linq-to-SQL class called Scans.dbml. In that class I've dragged a table called Users (username, password, role) onto the graphic area and now I can access User object via a UserRepository class: using System; using System.Collections.Generic; using System.Linq; using System.Text; namespace Scanner.Classes { public class UserRepository { private ScansDataContext db = new ScansDataContext(); public User getUser(string username) { return db.Users.SingleOrDefault(x => x.username == username); } public bool exists(string username) { } } } Now in my Login form, I want to use this Linq-to-SQL goodness to do all the data related activities. UserRepository users = new UserRepository(); private void btnLogin_Click(object sender, EventArgs e) { loginToSystem(); } private void loginToSystem() { if (users.getUser(txtUsername.Text)) { } //If txtUsername exists && User.password == Salt(txtPassword) //then Show.MainForm() with User.accountType in constructor to set permissions. } I need help with verifying that a user exists && that that users.password is equal to SALT(txtpassword.text). Any guidance please?

    Read the article

  • How can I isolate the form controls in a ASP Web User Control from the rest of the page's form contr

    - by Justin808
    I have a Web User Control I created for authentication. The web user control is inside the box below. Clicking any button (1 or 2) below works correct as it goes to the correct c# button click event in the code behind file. If I press enter on fields a or b it goes to the correct callback (button1's) if I press enter on field c it still goes to button1's callback, not button2's How can I give my web user control a nice self contained for and view state etc, so it wont mess with the remainder of the page's form? +--------------+ | User: __a___ | | Pass: __b___ | | [button1]| +--------------+ Prompt:______c______ [button2]

    Read the article

  • UX Design Question: Should a multi step wizard save the form contents when the user clicks 'go back'

    - by Ashwin Prabhu
    I am developing a web application that collects data over multiple steps through a wizard. Steps are generally not interdependent, in that data input at each step has little or no effect on the consequent steps. However each step may have a set of validations which determine whether the user can progress to the next step by clicking 'continue' What should be the behavior when the user clicks previous? a Quickly move to the previous page, thus losing all the unsaved data in the form. Prompting the user with a warning is an option, but it can become irritating quite soon. b Move to the previous page saving all the data in the current step - without triggering validations, so that when the user comes back she sees the form in the same state that she left it in. c any other behaviour All opinions are welcome :)

    Read the article

  • How to visually reject user input in a table?

    - by FX
    In the programming of a table-based application module (i.e. the user mostly enters tabular data in an already laid-out table), how would you reject user input for a given cell? The scenario is: the user edits the cell, enters something (text, picture, ...) and you want them to notice when they finish editing (hitting enter, for example) that their entry is not valid for your given "format" (in the wider meaning: it can be that they entered a string instead of a number, that their entry is too long, too short, they include a picture while it's not acceptable, ...). I can see two different things happening: You can rather easily fit their entry into your format, and you do so, but you want them to notice it so they can change if your guess is not good enough (example: they entered "15.47" in a field that needs to be an integer, so your program makes it "15") You cannot guess what to do with their entry, and want to inform them that it's not valid. My question specifically is: what visual display can you offer to inform the user that his input is invalid? Is it preferable to refuse to leave the editing mode, or not? The two things I can imagine are: using colors (red background if invalid, yellow background for my case 1 above) when you reject an input, do something like Apple does for password entry of user accounts: you make the cell "shaking" (i.e. oscillating left and right) for one second, and keep the focus/editing in their so they don't loose what they've typed. Let's hear your suggestions. PS: This question is, at least in my thought process, somehow a continuation and a specialization of my previous question on getting users to read error messages. PPS: Made this community wiki, was that the right thing to do on this kind of question or not?

    Read the article

  • Do you think its user unfriendly to show error message in tooltips ?

    - by msfanboy
    Hello, when my user enters data validated as wrong a red circle with a white exclamation mark is shown in the right part of the textbox with the wrong data. The error message is only shown when the user hovers the textbox with wrong data. Do you think that is a bad User experience ? I could show the red error message text to the right side of the textboxes if there would still be space...

    Read the article

  • User control event or method override where custom properties are valid?

    - by Curtis White
    I have an ASP.NET user control that is used in another use control. The parent user control uses data-binding to bind to a custom property of the child user control. What method can I override or page event where I am ensured that the property state is set? I think in a page it is PageLoaded versus the Page_Load override? I am looking for this in the user control because my property is always null even though it is set. Thanks.

    Read the article

  • multi-user rvm gem install failure when called from CloudFormation::Init

    - by Peter Mounce
    I've taken an Amazon Linux AMI (based on CentOS) and installed RVM (1.10.3) to it in multi-user fashion (see {1} below). I used that to install ruby 1.9.3-p125, rubygems 1.8.17, and bundler 1.1 as the baseline requirements for most things I'm going to be using the instances for. I've captured that instance to an AMI, and am now launching it via CloudFormation, with some CloudFormation::Init commands. One of them is to use s3cmd to pull down a private gem from S3, and the next one, the one that fails, is to install that gem. It fails with an error message 2012-03-15 16:53:20,201 [ERROR] Command 20_install_gems (/usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem install ./*.gem) failed 2012-03-15 16:53:20,202 [DEBUG] Command 20_install_gems output: /usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem:12:in `require': no such file to load -- rubygems (LoadError) from /usr/local/rvm/rubies/ruby-1.9.3-p125/bin/gem:12 Now, that happens during the cfn-init execution - I assume, but haven't checked yet, that cfn-init is being run with an environment different from that of ec2-user (there are no other users on the instance). If I run gem install mygem.gem in an interactive session then that works fine. So, my question really, is what should I do to make this work for cfn-init? Have I correctly set up rvm as multi-user? I've confirmed that cfn-init is being run as the root user, with his restricted environment. How should I source the /etc/profile.d/rvm.sh into root's sessions? {1} My semi-automated rvm installation steps (run in interactive session as ec2-user): sudo bash -s stable < <(curl -s https://raw.github.com/wayneeseguin/rvm/master/binscripts/rvm-installer ) sudo gpasswd -a ec2-user rvm # iconv-devel is baked into centos' glibc sudo yum install -y autoconf automake bison bzip2 gcc-c++ git libffi-devel libtool libxml2-devel libxslt-devel libyaml-devel make openssl-devel patch readline readline-devel zlib zlib-devel source /etc/profile.d/rvm.sh rvm list known # in a new session: rvm install ruby-1.9.3-p125 rvm use 1.9.3 --default gem update --system # gems required by public_web-awareness gem install aws-sdk bundler cocaine sinatra echo -e "gem: --no-ri --no-rdoc\n" > /home/ec2-user/.gemrc # delete unnecessary documentation files rm -rf `gem env gemdir`/doc sudo -s sudo echo -e "gem: --no-ri --no-rdoc\n" > /etc/skel/.gemrc sudo echo -e "gem: --no-ri --no-rdoc\n" > /etc/gemrc # ctrl + d out of the sudo session Some environment information: [ec2-user@ip ~]$ echo $PATH /usr/local/rvm/gems/ruby-1.9.3-p125/bin:/usr/local/rvm/gems/ruby-1.9.3-p125@global/bin:/usr/local/rvm/rubies/ruby-1.9.3-p125/bin:/usr/local/rvm/bin:/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/sbin:/opt/aws/bin:/home/ec2-user/bin [ec2-user@ip ~]$ echo $GEM_HOME /usr/local/rvm/gems/ruby-1.9.3-p125 [ec2-user@ip ~]$ echo $GEM_PATH /usr/local/rvm/gems/ruby-1.9.3-p125:/usr/local/rvm/gems/ruby-1.9.3-p125@global [ec2-user@ip ~]$ echo $BUNDLE_PATH [ec2-user@ip ~]$ gem list *** LOCAL GEMS *** aws-sdk (1.3.6) bundler (1.1.0) cocaine (0.2.1) httparty (0.8.1) json (1.6.5) multi_json (1.1.0) multi_xml (0.4.1) nokogiri (1.5.1, 1.5.0) rack (1.4.1) rack-protection (1.2.0) rake (0.9.2) sinatra (1.3.2) tilt (1.3.3) uuidtools (2.1.2) yamler (0.1.0)

    Read the article

  • Django User M2M relationship

    - by Antonio
    When trying to syncdb with the following models: class Contact(models.Model): user_from = models.ForeignKey(User,related_name='from_user') user_to = models.ForeignKey(User, related_name='to_user') class Meta: unique_together = (('user_from', 'user_to'),) User.add_to_class('following', models.ManyToManyField('self', through=Contact, related_name='followers', symmetrical=False)) I get the following error: Error: One or more models did not validate: auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: The model User has two manually-defined m2m relations through the model Contact, which is not permitted. Please consider using an extra field on your intermediary model instead. auth.user: Accessor for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'. auth.user: Reverse query name for m2m field 'following' clashes with related m2m field 'User.followers'. Add a related_name argument to the definition for 'following'.

    Read the article

  • How to change a Linux user password from python

    - by Vaulor
    I'm having problems with changing a Linux user's password from python. I've tried so many things, but I couldn't manage to solve the issue, here is the sample of things I've already tried: sudo_password is the password for sudo, sudo_command is the command I want the system to run, user is get from a List and is the user who I want to change the password for, and newpass is the pass I want to assign to 'user' user = list.get(ANCHOR) sudo_command = 'passwd' f = open("passwordusu.tmp", "w") f.write("%s\n%s" % (newpass, newpass)) f.close() A=os.system('echo -e %s|sudo -S %s < %s %s' % (sudo_password, sudo_command,'passwordusu.tmp', user)) print A windowpass.destroy() 'A' is the return value for the execution of os.system, in this case 256. I tried also A=os.system('echo %s|sudo -S %s < %s %s' % (sudo_password, sudo_command,'passwordusu.tmp', user)) but it returns the same error code. I tried several other ways with 'passwd' command, but whithout succes. With 'chpasswd' command I 've tried this: user = list.get(ANCHOR) sudo_command = 'chpasswd' f = open("passwordusu.tmp", "w") f.write("%s:%s" % (user, newpass)) f.close() A=os.system('echo %s|sudo -S %s < %s %s' % (sudo_password, sudo_command,'passwordusu.tmp', user)) print A windowpass.destroy() also with: A=os.system('echo %s|sudo -S %s:%s|%s' % (sudo_password, user, newpass, sudo_command)) @;which returns 32512 A=os.system("echo %s | sudo -S %s < \"%s\"" % (sudo_password, sudo_command, "passwordusu.tmp")) @;which returns 256 I tried 'mkpasswd' and 'usermod' too like this: user = list.get(ANCHOR) sudo_command = 'mkpasswd -m sha-512' os.system("echo %s | sudo -S %s %s > passwd.tmp" % (sudo_password,sudo_command, newpass)) sudo_command="usermod -p" f = open('passwd.tmp', 'r') for line in f.readlines(): newpassencryp=line f.close() A=os.system("echo %s | sudo -S %s %s %s" % (sudo_password, sudo_command, newpassencryp, user)) @;which returns 32512 but, if you go to https://www.mkpasswd.net , hash the 'newpass' and substitute for 'newpassencryp', it returns 0 which theoretically means it has gone right, but so far it doesn't changes the password. I've searched on internet and stackoverflow for this issue or similar and tried what solutions exposed, but again,without success. I would really apreciate any help, and of course, if you need more info i'll be glad to supply it! Thanks in advance.

    Read the article

  • apache sendmail: trying to change user "from" address from apache to domain account

    - by Wes
    I apologize if I am asking a question already answered, but my problem isn't really that I haven't found an answer. I have, in fact, found a half-dozen different "solutions" to my problem, tried them all, in various combinations, and have been consistently unsuccessful. The goal All I want to do is change the envelope "from" address for all email sent from [email protected] to [email protected], always. What I've already done I am running Apache, PHP, and sendmail on CentOS 5.5, [email protected]. We have an SMTP server at 192.168.0.4. The domain's email accounts are all at @domain.org. I have successfully set up "smart host" using this line in the sendmail.mc file: define(`SMART_HOST', `192.168.0.4')dnl Then I set up masquerading, and was hopeful this would solve it. I have this in the .mc file: FEATURE(`masquerade_entire_domain')dnl FEATURE(`masquerade_envelope')dnl FEATURE(`allmasquerade')dnl MASQUERADE_AS(`domain.org')dnl MASQUERADE_DOMAIN(`domain.org.')dnl MASQUERADE_DOMAIN(`localhost.localdomain.')dnl This rewrites "to" addresses, but not "from" addresses. Testing from the command line: sendmail -v [email protected] Always is shown from the local user (in this case root, or my local user account). I had read that "sendmail" command sometimes bypasses masquerading. Nevertheless, using the "mail" command has the same result. After that, I have explored several "solutions", including: mailertable virtusertable FEATURE(`accept_unresolvable_domains')dnl LOCAL_DOMAIN(`localhost.localdomain')dnl FEATURE(`genericstable')dnl /etc/mail/access file /etc/mail/local-host-names file /etc/mail/trusted-users file All to no affect. The last thing I've tried So, I decided to go in a different direction, and try to set the envelope "from" address via PHP, using either the configuration in /etc/php.ini, or adding the -f parameter to the mail() function or to sendmail command. If I run this command: sendmail -v -f [email protected] [email protected] I get this error in /var/log/maillog: Mar 30 08:56:16 localhost sendmail[24022]: p2UCuE8w024022: [email protected], size=5, class=0, nrcpts=1, msgid=<[email protected]>, relay=user@localhost Mar 30 08:56:19 localhost sendmail[24022]: p2UCuE8w024022: [email protected], [email protected] (500/502), delay=00:00:05, xdelay=00:00:03, mailer=relay, pri=30005, relay=[192.168.0.4] [192.168.0.4], dsn=5.1.1, stat=User unknown Mar 30 08:56:19 localhost sendmail[24022]: p2UCuE8w024022: p2UCuE8x024022: DSN: User unknown Mar 30 08:56:23 localhost sendmail[24022]: p2UCuE8x024022: [email protected], delay=00:00:04, xdelay=00:00:04, mailer=relay, pri=31029, relay=[192.168.0.4] [192.168.0.4], dsn=2.0.0, stat=Sent (Ok: queued as B5E2E40E0A2) Which is basically a "User unknown" 550 error. Help Please help. What do I need to change? Should I just start over in the sendmail.mc file? It has a ton of config options stuffed in it, over days of trying things. Why is changing the envelope "from" address via the command line generating a "User unknown" error?

    Read the article

  • How to get Alfresco login ticket without user password, but with impersonating user with user principal name (UPN)

    - by dok
    I'm writing a DLL that has function for getting Alfresco login ticket without using user password, using only a user principal name (UPN). I’m calling alfresco REST API service /wcservice. I use NTLM in Alfresco. I’m impersonating users using WindowsIdentity constructor as explained here http://msdn.microsoft.com/en-us/library/ms998351.aspx#paght000023_impersonatingbyusingwindowsidentity. I checked and user is properly impersonated (I checked WindowsIdentity.GetCurrent().Name property). After impersonating a user, I try to make HttpWebRequest and set its credentials with CredentialsCache.DefaultNetworkCredentials. I get the error: The remote server returned an error: (401) Unauthorized. at System.Net.HttpWebRequest.GetResponse() When I use new NetworkCredential("username", "P@ssw0rd") to set request credentials, I get Alfresco login ticket (HttpStatusCode.OK, 200). Is there any way that I can get Alfresco login ticket without user password? Here is the code that I'm using: private string GetTicket(string UPN) { WindowsIdentity identity = new WindowsIdentity(UPN); WindowsImpersonationContext context = null; try { context = identity.Impersonate(); MakeWebRequest(); } catch (Exception e) { return e.Message + Environment.NewLine + e.StackTrace; } finally { if (context != null) { context.Undo(); } } } private string MakeWebRequest() { string URI = "http://alfrescoserver/alfresco/wcservice/mg/util/login"; HttpWebRequest request = WebRequest.Create(URI) as HttpWebRequest; request.CookieContainer = new CookieContainer(1); //request.Credentials = new NetworkCredential("username", "p@ssw0rd"); // It works with this request.Credentials = CredentialCache.DefaultNetworkCredentials; // It doesn’t work with this //request.Credentials = CredentialCache.DefaultCredentials; // It doesn’t work with this either try { using (HttpWebResponse response = request.GetResponse() as HttpWebResponse) { StreamReader sr = new StreamReader(response.GetResponseStream()); return sr.ReadToEnd(); } } catch (Exception e) { return (e.Message + Environment.NewLine + e.StackTrace); } } Here are records from Alfresco stdout.log (if it helps in any way): 17:18:04,550 DEBUG [app.servlet.NTLMAuthenticationFilter] Processing request: /alfresco/wcservice/mg/util/login SID:7453F7BD4FD2E6A61AD40A31A37733A5 17:18:04,550 DEBUG [web.scripts.DeclarativeRegistry] Web Script index lookup for uri /mg/util/login took 0.526239ms 17:18:04,550 DEBUG [app.servlet.NTLMAuthenticationFilter] New NTLM auth request from 10.**.**.** (10.**.**.**:1229) 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Processing request: /alfresco/wcservice/mg/util/login SID:7453F7BD4FD2E6A61AD40A31A37733A5 17:18:04,566 DEBUG [web.scripts.DeclarativeRegistry] Web Script index lookup for uri /mg/util/login took 0.400909ms 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Received type1 [Type1:0xe20882b7,Domain:<NotSet>,Wks:<NotSet>] 17:18:04,566 DEBUG [app.servlet.NTLMAuthenticationFilter] Client domain null 17:18:04,675 DEBUG [app.servlet.NTLMAuthenticationFilter] Sending NTLM type2 to client - [Type2:0x80000283,Target:AlfrescoServerA,Ch:197e2631cc3f9e0a]

    Read the article

  • Security Tips for super user in Ubuntu 11.10?

    - by Gaurav_Java
    Ubuntu is a relatively safe but this should not prevent you to be vigilant. I'm working on securing Ubuntu 11.10 and was looking for every tip to secure my Ubuntu 11.10 and upcoming version 12.04. I go through this question . But my Question is Begin as Super User and as Normal user also. Which is Best tools which I can use for preventing my data? What Services and features I Enable Manually after installing ubuntu ? What security measure I should take? What i don't do ? What are other precaution some takes as super user? what are Normal security stay patched in ubuntu? and if i am missing something then please add them

    Read the article

  • My session at the Vancouver Silverlight User Group

    - by pluginbaby
    Next week I will be in Vancouver and talk at the local User Group: the Vancouver Silverlight User Group. Title: HTML5 and Silverlight 5: facts, assumptions and near future Abstract: In this session, I will try to clarify what we hear (and not hear) around these technologies, maybe add a few guess on their role in Windows 8... as well as presenting a technical comparison between HTML5 and Silverlight 5: HTML vs XAML, tools, languages, databinding, performance, etc. Date: Wednesday, July 6, 2011 Thanks Telerik to sponsor the room for this event. More details and registration: http://www.meetup.com/Vancouver-Silverlight-User-Group/events/22849231/

    Read the article

  • T-SQL User-Defined Functions: the good, the bad, and the ugly (part 1)

    - by Hugo Kornelis
    So you thought that encapsulating code in user-defined functions for easy reuse is a good idea? Think again! SQL Server supports three types of user-defined functions. Only one of them qualifies as good. The other two – well, the title says it all, doesn’t it? The bad: scalar functions A scalar user-defined function (UDF) is very much like a stored procedure, except that it always returns a single value of a predefined data type – and because of that property, it isn’t invoked with an EXECUTE statement,...(read more)

    Read the article

  • Default file permissions for php user www-data

    - by John Isaacks
    I have a php installed on my ubuntu machine. The web root is /var/www I set the permissions for this folder like so: sudo chown -R ftpuser:www-data /var/www ftpuser is the user I set up so I can ftp to /var/www from another machine on the network. www-data is the user php uses. I double checked using whoami from php. Whenever I ftp upload a new file to the machine the group has no permissions to the file. So when I try to access it in my browser via machine-name/new-file.php I am told permission denied and I have to go and chmod the new file. I am wondering if there is a way I can default the www-data user/group to have access permissions to new files so I don't have to keep chmod every new file?

    Read the article

  • T-SQL User-Defined Functions: the good, the bad, and the ugly (part 1)

    - by Hugo Kornelis
    So you thought that encapsulating code in user-defined functions for easy reuse is a good idea? Think again! SQL Server supports three types of user-defined functions. Only one of them qualifies as good. The other two – well, the title says it all, doesn’t it? The bad: scalar functions A scalar user-defined function (UDF) is very much like a stored procedure, except that it always returns a single value of a predefined data type – and because of that property, it isn’t invoked with an EXECUTE statement,...(read more)

    Read the article

  • Problem with gnome-shell-extensions-user-theme

    - by sodiumnitrate
    I'm trying to install themes with gnome tweak, and I need to install gnome-shell-extensions-user-theme because otherwise I cannot see the shell extensions tab. However, I cannot install shell extensions. I have tried to install by adding the PPA with the following: sudo add-apt-repository ppa:webupd8team/gnome3 Then, sudo apt-get update Finally, when I try to install: sudo apt-get install gnome-shell-extensions-user-theme It gives an error: The following packages have unmet dependencies: gnome-shell-extensions-user-theme : Depends: gnome-shell-extensions-common but it is not going to be installed E: Unable to correct problems, you have held broken packages. I am convinced that there is a problem with the package. So I went on and tried to install the extensions from the website: https://extensions.gnome.org/ But even though I use Firefox (15.0), I cannot see the "switch" that is being mentioned to install the extension. Maybe the version of Firefox is too new. Is there any workaround that you know of? (By the way, I use Ubuntu 12.04, freshly downloaded and installed.)

    Read the article

  • Operation not permitted for chown : For a domain user directory

    - by Lunar Mushrooms
    I am trying to change ownership of a domain user home directory, which is mounted over nfs. Current user/group for that folder is nobody/nogroup. The following chown command is issued from "root" user shell. But I am getting permission error. How to resolve this ? sudo chown -Rv VANILLA\\userone:VANILLA\\domain^users /lhome/VANILLA/userone chown: changing ownership of `/lhome/VANILLA/userone': Operation not permitted failed to change ownership of `/lhome/VANILLA/userone' from nobody:nogroup to VANILLA\userone:VANILLA\domain^users My OS is Ubuntu LTS 12.04 32 bit.

    Read the article

  • User prompts (MessageBox) with MVVM

    - by mukapu
    The problem statement: I am tired of thinking how to show a simple message box or user prompt and act based on the response in Model-View-View-Model (MVVM). Common approaches: - It's ok, let's just do this one thing from ViewModel and mock this out for unit testing - Design my own dialog, then what to do from there - Can I write something in view code behind, ah yes, that seems to be the only way out, as anyway MVVM is still not matured...  - and what not?   I am pretty much one among the few frustrated out in this world looking for some convincing answers. I think we can do it a little neater without having the feeling of violating any of our self defined rules! Solution: The Control - Implement a simple control with no designer visibility. - Allow a property to be bound to tell when to show the MessageBox - Provide command binding for possible user actions, Yes, No, Cancel... How do I Use? - Just place the necessary XAML tags in the view - Implement the command for all user actions in the View Model - Run unit tests on the commands

    Read the article

  • Entity Framework: Connecting to a mdf user database file via localDB during script execution

    - by Marko Apfel
    Problem If you run the “Generate database from model” wizard and execute the generated script the destination database could be the wrong one (for instance master of the SQL Server). Solution To use an own mdf attachable user database some connection information must specified during script execution. Execute your script opens the dialog “Connect to Server”. Press “Options” and go to the second tab “Connection Properties”. Select “Browse server” in the “Connect to database” dropdown box: Confirm the information dialog with “Yes”. In the following dialog you could choose your user database. Now the schema is created in the user database.

    Read the article

  • Product Support Webcast for Existing Customers: Oracle Webcenter Portal 11g User & Administration Tips

    - by John Klinke
    Register for our upcoming Advisor Webcast 'Oracle WebCenter Portal 11g: User & Administration Tips' scheduled for November 12, 2013 at 11:00 am, Eastern Standard Time (8:00 am Pacific Standard Time, 4:00 pm GMT Time, 5:00 pm Europe Time). This 1-hour session is recommended for technical and functional users who use Oracle WebCenter Portal to build company portals using run-time tools.Topics will include:• Whats new in 11.1.1.8 of WebCenter Portal• Terminology Changes• Using the Portal once its built• Setting up Self Registration (Admins)• End User Experience• Development Environment• Patching InformationFor more information and to register for this Advisor Webcast, please see Oracle WebCenter Portal 11g: User & Administration Tips (Doc ID 1585902.1).

    Read the article

  • Locate a user's bashrc file

    - by Starkers
    Really confused. Upon running cat etc/passwd I have found this: postgres:x:117:126:PostgreSQL administrator,,,:/var/lib/postgresql:/bin/bash meaning I have a postgres user, right? I want to change the bashrc environment file of this user to make commands available to it. /var/lib/postgresql doesn't contain a bashrc file, and /bin/bash doesn't contain it either, so I don't really know what's going on. All I know is a created postgres using the useradd command, so why do I have some weird user with no home directory? So confused :(

    Read the article

  • I have lost sudo privileges in Ubuntu 13.04 [duplicate]

    - by Fredca
    This question already has an answer here: How do I add myself back as a sudo user? 3 answers I have lost sudo privileges in Ubuntu 13.04 these are the responses I get user@user-desktop:~$ sudo sudo: effective uid is not 0, is sudo installed setuid root? user@user-desktop:~$ groups user user : user adm cdrom sudo dip plugdev lpadmin sambashare user@user-desktop:~$ su user Password: su: Authentication failure user@user-desktop:~$ who am i user pts/0 2013-10-24 08:54 (:0.0) user@user-desktop:~$ why can't I invoke sudo if one of my groups is sudo? also I have noticed that /etc/sudoers needs sudo privileges. sudoers.so doesn't exist in /usr/lib but does in /usr/lib/sudo is this correct in 13.04? Please note that the user is already a member of both sudo and adm groups.

    Read the article

< Previous Page | 38 39 40 41 42 43 44 45 46 47 48 49  | Next Page >