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  • RDP suggesting a user name when connecting to a server

    - by Neolisk
    Prior to Event X, RDPing to Server 2003 always caused the user name appear blank and Login to be enabled, so you could pick to which domain you would log in. For us it's either local or our domain. Since a recent Event X a domain + user name is being suggested for every server and it's not the most recently used user name. If you remove it manually from RDP dialog, it's still being pre-populated for you, and then at the next available opportunity it returns into General/User name option of RDP dialog. So user name field comes pre-populated and you cannot change to log in locally (only if you manually erase domain specifier - everything before \) - Log in to option is disabled by default. We did not do any changes to our domain or client machines, so I am suspecting some Windows update caused it (and this being Event X). Interesting fact - it does not consistently happen on all machines, and some can login to some servers fine, while other servers keep suggesting a default user name. What could be that Event X and is there a way to fix it? EDIT: I tried this - How to clear remote desktop connections history and specifically this part of it: reg delete "HKEY_CURRENT_USER\Software\Microsoft\Terminal Server Client\UsernameHint" /f The problem still persists.

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  • Oracle Event: Database Enterprise User Security

    - by user12603048
    One of the high-value benefits of an integrated Identity and Access Management platform is the ability to leverage a unified corporate directory as the primary authentication source for database access. On July 11, 2012 at 08:00 am PDT, Oracle will host a webcast showing how Enterprise User Security (EUS) can be used to externalize and centrally manage database users in a directory server. The webcast will briefly introduce EUS, followed by a detailed discussion about the various directory options that are supported, including integration with Microsoft Active Directory. We'll conclude how to avoid common pitfalls deploying EUS with directory services. Discussion topics will include Understanding EUS basics Understanding EUS and directory integration options Avoiding common EUS deployment mistakes Make sure to register and mark this date on your calendar! - Click here to register.

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  • Calling Knowledge Workers: Make a Difference in the User Experience

    - by Charles Knapp
    Do you consider yourself a knowledge worker? Do you have ideas of how to make CRM software work smarter so you don't have to work harder? The Oracle Middleware User Experience team will be conducting customer feedback focus groups at Oracle OpenWorld, October 1-3. All it takes is a couple of hours or less for us to learn from you. Customer participation helps Oracle develop outstanding products and solutions. Knowledge workers of all types are invited to participate: Finance, Sales, Human Resources, Marketing, Recruiters, Budget Managers, Project & Product Managers and more. To participate in these sessions you do not have to be registered for Oracle OpenWorld. If you or someone you know is interested in participating, please email muxtesting_us at oracle.com with your name, company, job title, work and mobile phone numbers with country code, and email address.

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  • Ubuntu user expectations from 12.04 and future releases

    - by Rick Green - Turbo
    How much further ahead is 12.10 vs 12.04 in respect to kernel updates and applications? Example: Gimp's newest release is 2.8 which runs equally as well in both 12.04 and 12.10 and probably will in 13.04. What restricts 12.04 from having "the same" look, feel, applications and kernel as 12.10 or the upcoming 13.04? I know that it's more than a name change.....it's whats under the hood that counts. Incrementally upgrading, I feel is safer than radical changes from release to release. Trying to keep a stable desktop and current user experience, how far can I take updating applications before I absolutely have to make a distro upgrade from 12.04LTS

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  • disable password prompt on user switch?

    - by matthewn
    I've got 11.10 on a desktop machine with two users. Both users have "Password" set to "Not asked on login" in Users Settings. At startup, either user can log in without a password. But once both users are logged in, it takes a password to switch between users. In previous Ubuntus, you could override this by setting /desktop/gnome/lockdown/disable_lock_screen to True in gconf-editor. That is ignored in the Gnome 3 / Unity era. Does anyone know a way to disable the password prompt when switching between users in Oneiric?

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  • Restrict user to folder (not root) on VSFTPD in Ubuntu

    - by omega1
    I am a new Linux (Ubuntu) user and have a VPS where I am setting up a backup FTP service. I have followed this guide, which I have managed to do correctly and it works. I have two users (user1 and user2) with the same directory /home/users/test. user1 can read/write and user2 can only read. This works OK. When the users log in, they go straight into the correct directory /home/users/test, but they can navigate back down to the home directory, which I do not want to happen. I cannot seem to find out how to not allow this, and have them not be able to navigate back to the /home/ or /home/test/ directories.

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  • TGT validation fails, but only for one user

    - by wzzrd
    I'm seeing the weirdest thing here. I have a couple of RHEL3, 4 and 5 machines that validate user credentials through Kerberos with an Active Directoy domain controller as their KDC. This works for all of my users, save one. There is one account that is unable to log into RHEL3 Linux machines and generates the following errors there: May 31 13:53:19 mybox sshd(pam_unix)[7186]: authentication failure; logname= uid=0 euid=0 tty=ssh ruser= rhost=10.0.0.1 user=user May 31 13:53:20 mybox sshd[7186]: pam_krb5: TGT verification failed for `user' May 31 13:53:20 mybox sshd[7186]: pam_krb5: authentication fails for `user' Other accounts, like my own, are fine: May 31 17:25:30 mybox sshd(pam_unix)[12913]: authentication failure; logname= uid=0 euid=0 tty=ssh ruser= rhost=10.0.0.1 user=myuser May 31 17:25:31 mybox sshd[12913]: pam_krb5: TGT for myuser successfully verified May 31 17:25:31 mybox sshd[12913]: pam_krb5: authentication succeeds for `myuser' May 31 17:25:31 mybox sshd(pam_unix)[12915]: session opened for user myuser by (uid=0) As you can see, TGT validation fails. This only happens for this specific account, not for any other. The failing useraccount's password has been reset, I inspected both user objects in Active Directory, but I see nothing out of the ordinary. If I have the failing useraccount log into a RHEL4 or 5 box, there is not problem, so it must be RHEL3 specific, but the fact that only one account suffers from this, alludes me. Maybe someone has seen this before?

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  • Completing a Basic ASP.NET 3.5 User Input Validation Project

    You learned the basics and configuration steps of the most common types of validation web controls used in ASP.NET 3.5 in the first two parts of this tutorial series. In this last part you will learn how to integrate all of these validation web controls in a working ASP.NET project. You will also learn not only how to validate user input in the client side but also how to validate the page on the server side.... Microsoft SQL Server? Value Calculator Reduce Costs & Increase Value with Microsoft SQL Server? 2008. Download Today!

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  • API to UPDATE Oracle FND User

    - by PRajkumar
    API - fnd_user_pkg.updateuser Example -- Consider a FND User having following Details --     Lets Try to Update its Email Id from [email protected] to [email protected]   -- ------------------------------------------------ -- API to UPDATE Oracle FND User -- ------------------------------------------------ DECLARE     lc_user_name                           VARCHAR2(100)   := 'PRAJ_TEST';     lc_user_password                   VARCHAR2(100)   := 'Oracle123';     ld_user_start_date                   DATE                      := TO_DATE('23-JUN-2012');     ld_user_end_date                    VARCHAR2(100)  := NULL;     ld_password_date                   VARCHAR2(100)  := TO_DATE('23-JUN-2012');     ld_password_lifespan_days  NUMBER               := 90;     ln_person_id                            NUMBER                := 32979;     lc_email_address                     VARCHAR2(100)  := '[email protected]'; BEGIN    fnd_user_pkg.updateuser    (  x_user_name                           => lc_user_name,       x_owner                                   => NULL,       x_unencrypted_password    => lc_user_password,       x_start_date                             => ld_user_start_date,       x_end_date                              => ld_user_end_date,       x_password_date                   => ld_password_date,       x_password_lifespan_days  => ld_password_lifespan_days,       x_employee_id                       => ln_person_id,       x_email_address                    => lc_email_address    );  COMMIT; EXCEPTION     WHEN OTHERS THEN                       ROLLBACK;                       DBMS_OUTPUT.PUT_LINE(SQLERRM); END; / SHOW ERR;       

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  • User input and automated input separation

    - by tpaksu
    I have a MySQL database and an automation script which modifies the data inside once a day. And these columns may have changed by an user manually. What is the best approach to make the system only update the automated data, not the manually edited ones? I mean yes, flagging the cell which is manually edited is one way to do it, but I want to know if there's another way to accomplish this? Just curiosity. BTW, the question is about cell values, not rows.

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  • User Password I didn't even know of [duplicate]

    - by KiriharaAkaya
    This question already has an answer here: How do I reset a lost administrative password? 13 answers I'm a new user of Ubuntu and I installed it on my computer using Virtual Box. I've been trying to install Java JDK for the past two hours without any success because when I try to do so, Terminal asks for my password, which I don't even remember creating. Can someone please help and tell me how am I suppose to change the password in order to install JDK? It will be very much appreciated, thank you.

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  • Queyring container with Linq + group by ?

    - by Prix
    public class ItemList { public int GuID { get; set; } public int ItemID { get; set; } public string Name { get; set; } public entityType Status { get; set; } public class Waypoint { public int Zone { get; set; } public int SubID { get; set; } public int Heading { get; set; } public float PosX { get; set; } public float PosY { get; set; } public float PosZ { get; set; } } public List<Waypoint> Routes = new List<Waypoint>(); } I have a list of items using the above class and now I need to group it by ItemID and join the first entry of Routes of each iqual ItemID. So for example, let's say on my list I have: GUID ItemID ListOfRoutes 1 23 first entry only 2 23 first entry only 3 23 first entry only 4 23 first entry only 5 23 first entry only 6 23 first entry only 7 23 first entry only Means I have to group entries 1 to 7 as 1 Item with all the Routes entries. So I would have one ItemID 23 with 7 Routes on it where those routes are the first element of that given GUID Routes List. My question is if it is possible using LINQ to make a statment to do something like that this: var query = from ItemList entry in myList where status.Contains(entry.Status) group entry by entry.ItemID into result select new { items = new { ID = entry.ItemID, Name = entry.Name }, routes = from ItemList m in entry group m.Routes.FirstOrDefault() by n.NpcID into m2 }; So basicly I would have list of unique IDS information with a inner list of all the first entry of each GUID route that had the same ItemID. <<< UPDATE: This would be an example of public List<ItemList> myList = new List<ItemList>(); data: GUID ItemID ListOfRoutes 1 20 Routes[1] 2 20 Routes[2] 3 20 Routes[3] 4 20 Routes[4] 5 20 Routes[5] 6 55 Routes[6] 7 55 Routes[7] 8 55 Routes[8] 9 1 Routes[9] 10 1 Routes[10] As you can see GUID is unique, ItemID can reapeat it self. Each GUID has a Routes list and all routes list have a minimum of 1 entry and above. Routes example. Routes[1] have: Entry Zone SubID Heading PosX PosY PosZ 1 1200 0 100 1029.32 837.21 29.10 2 1200 0 120 1129.32 537.21 29.10 3 1200 0 180 1229.32 137.21 29.10 4 1200 0 360 1329.32 437.21 29.10 5 1200 0 100 1429.32 637.21 29.10 Routes[2] have: Entry Zone SubID Heading PosX PosY PosZ 1 100 0 10 129.32 437.21 29.10 So what I want to do is a list of all entries I have on myList grouped by ItemID maintainning the fields ItemID and Name ... and a new field or item that will have all the first elements of Routes of those GUIDs. For example ItemID 20 would produce the follow result: ItemID, Name, ListOfRoutes This ItemID ListOfRoutes would contain Routes[1] first entry: Entry Zone SubID Heading PosX PosY PosZ 1 1200 0 100 1029.32 837.21 29.10 Routes[2] first entry: Entry Zone SubID Heading PosX PosY PosZ 1 100 0 10 129.32 437.21 29.10 Routes[3], Routes[4], Routes[5] first entries. Example of how myList is feeded: ItemList newItem = new ItemList(); newItem.GUID = GUID; newItem.ItemID = ItemID; newItem.Name = Name; newItem.Status = Status; // Item location ItemList.Waypoint itemLocation = new ItemList.Waypoint(); itemLocation.SubID = SubID; itemLocation.Zone = Zone; itemLocation.Heading = convertHeading(Heading); itemLocation.PosX = PosX; itemLocation.PosY = PosY; itemLocation.PosZ = PosZ; itemLocation.Rest = Rest; newItem.Routes.Add(itemLocation); myList.Add(newItem);

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  • Efficient representation of Hierarchies in Hibernate.

    - by Alison G
    I'm having some trouble representing an object hierarchy in Hibernate. I've searched around, and haven't managed to find any examples doing this or similar - you have my apologies if this is a common question. I have two types which I'd like to persist using Hibernate: Groups and Items. * Groups are identified uniquely by a combination of their name and their parent. * The groups are arranged in a number of trees, such that every Group has zero or one parent Group. * Each Item can be a member of zero or more Groups. Ideally, I'd like a bi-directional relationship allowing me to get: * all Groups that an Item is a member of * all Items that are a member of a particular Group or its descendants. I also need to be able to traverse the Group tree from the top in order to display it on the UI. The basic object structure would ideally look like this: class Group { ... /** @return all items in this group and its descendants */ Set<Item> getAllItems() { ... } /** @return all direct children of this group */ Set<Group> getChildren() { ... } ... } class Item { ... /** @return all groups that this Item is a direct member of */ Set<Group> getGroups() { ... } ... } Originally, I had just made a simple bi-directional many-to-many relationship between Items and Groups, such that fetching all items in a group hierarchy required recursion down the tree, and fetching groups for an Item was a simple getter, i.e.: class Group { ... private Set<Item> items; private Set<Group> children; ... /** @return all items in this group and its descendants */ Set<Item> getAllItems() { Set<Item> allItems = new HashSet<Item>(); allItems.addAll(this.items); for(Group child : this.getChildren()) { allItems.addAll(child.getAllItems()); } return allItems; } /** @return all direct children of this group */ Set<Group> getChildren() { return this.children; } ... } class Item { ... private Set<Group> groups; /** @return all groups that this Item is a direct member of */ Set<Group> getGroups() { return this.groups; } ... } However, this resulted in multiple database requests to fetch the Items in a Group with many descendants, or for retrieving the entire Group tree to display in the UI. This seems very inefficient, especially with deeper, larger group trees. Is there a better or standard way of representing this relationship in Hibernate? Am I doing anything obviously wrong or stupid? My only other thought so far was this: Replace the group's id, parent and name fields with a unique "path" String which specifies the whole ancestry of a group, e.g.: /rootGroup /rootGroup/aChild /rootGroup/aChild/aGrandChild The join table between Groups and Items would then contain group_path and item_id. This immediately solves the two issues I was suffering previously: 1. The entire group hierarchy can be fetched from the database in a single query and reconstructed in-memory. 2. To retrieve all Items in a group or its descendants, we can select from group_item where group_path='N' or group_path like 'N/%' However, this seems to defeat the point of using Hibernate. All thoughts welcome!

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  • Facebook Invalid OAuth access token signature trying to post an attachment to group wall from PHP

    - by Volodymyr B
    I am an administrator (manager role) of a Facebook Group. I created an app, and stored its id and secret. I want my app to be able to post something on the Facebook group's feed. But when I attempt to post, I get the error 190 Invalid OAuth access token signature, even though I able to successfully obtain the access_token with publish_stream and offline_access scopes. It has the form of NNNNNNNNNNNNNNN|XXXXXXXXXXXXXXXXXXXXXXXXXXX, where N is a number (15) and X is a letter or a number (27). What should I do more to get this accomplished? Here is the code I am using: public static function postToFB($message, $image, $link) { //Get App Token $token = self::getFacebookToken(); // Create FB Object Instance $facebook = new Facebook(array( 'appId' => self::fb_appid, 'secret' => self::fb_secret, 'cookie' => true )); //$token = $facebook->getAccessToken(); //Try to Publish on wall or catch the Facebook exception try { $attachment = array('access_token' => $token, 'message' => $message, 'picture' => $image, 'link' => $link, //'name' => '', //'caption' => '', 'description' => 'More...', //'actions' => array(array('name' => 'Action Text', 'link' => 'http://apps.facebook.com/xxxxxx/')) ); $result = $facebook->api('/'.self::fb_groupid.'/feed/', 'post', $attachment); } catch (FacebookApiException $e) { //If the post is not published, print error details echo '<pre>'; print_r($e); echo '</pre>'; } } Code which returns the token //Function to Get Access Token public static function getFacebookToken($appid = self::fb_appid, $appsecret = self::fb_secret) { $args = array( 'grant_type' => 'client_credentials', 'client_id' => $appid, 'client_secret' => $appsecret, 'redirect_uri' => 'https://www.facebook.com/connect/login_success.html', 'scope' => 'publish_stream,offline_access' ); $ch = curl_init(); $url = 'https://graph.facebook.com/oauth/access_token'; curl_setopt($ch, CURLOPT_URL, $url); curl_setopt($ch, CURLOPT_HEADER, false); curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); curl_setopt($ch, CURLOPT_POST, true); curl_setopt($ch, CURLOPT_POSTFIELDS, $args); try { $data = curl_exec($ch); } catch (Exception $exc) { error_log($exc->getMessage()); } return json_encode($data); } If I uncomment $token = $facebook->getAccessToken(); in the posting code, it gives me yet another error (#200) The user hasn't authorized the application to perform this action. The token I get using developers.facebook.com/tools/explorer/ is of another form, much longer and with it I am able to post to the group page feed. How do I do it without copy/paste from Graph API Explorer and how do I post as a group instead of posting as a user? Thanks.

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  • Listing issue, GROUP mysql

    - by SethCodes
    Here is a mock-up example of Mysql table: | ID | Country | City | ________________________________ | 1 | Sweden | Stockholm | | 2 | Sweden | Stockholm | | 3 | Sweden | Lund | | 4 | Sweden | Lund | | 5 | Germany | Berlin | | 6 | Germany | Berlin | | 7 | Germany | Hamburg | | 8 | Germany | Hamburg | Notice how both rows Country and city have repeated values inside them. Using GROUP BY country, city in my PDO query, the values will not repeat while in loop. Here is PDO for this: $query = "SELECT id, city, country FROM table GROUP BY country, city"; $stmt = $db->query($query); while($row = $stmt->fetch(PDO::FETCH_ASSOC)) : The above code will result in an output like this (some editing in-between). GROUP BY works but the country repeats: Sweden - Stockholm Sweden - Lund Germany - Berlin Germany - Hamburg Using bootstrap collapse and above code, I separate the country from the city with a simple drop down collopase. Here is code: <li> <a data-toggle="collapse" data-target="#<?= $row['id']; ?>" href="search.php?country=<?= $row['country']; ?>"> <?= $row['country']; ?> </a> <div id ="<?= $row['id']; ?>" class="collapse in"> //collapse div here <a href="search.php?city=<?= $row['city']; ?>"> <?= $row['city']; ?><br></a> </div> //end </li> It then looks something like this (once collapse is initiated): Sweden > Stockholm Sweden > Lund Germany >Berlin Germany >Hamburg Here is where I face the problem. The above lists the values Sweden and Germany 2 times. I want Sweden and Germany to only list one time, and the cities listed below, so the desired look is to be this: Sweden // Lists one time > Stockholm > Lund Germany // Lists one time >Berlin >Hamburg I have tried using DISTINCT, GROUP_CONTACT and other methods, yet none get my desired output (above). Suggestions? Below is my current full code in action: <? $query = "SELECT id, city, country FROM table GROUP BY country, city"; $stmt = $db->query($query); while($row = $stmt->fetch(PDO::FETCH_ASSOC)) : ?> <li> <a data-toggle="collapse" data-target="#<?= $row['id']; ?>" href="search.php?country=<?= $row['country']; ?>"> <?= $row['country']; ?> </a> <div id ="<?= $row['id']; ?>" class="collapse in"> //collapse div here <a href="search.php?city=<?= $row['city']; ?>"> <?= $row['city']; ?><br></a> </div> //end </li> <? endwhile ?>

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  • Implementing EAV pattern with Hibernate for User -> Settings relationship

    - by Trevor
    I'm trying to setup a simple EAV pattern in my web app using Java/Spring MVC and Hibernate. I can't seem to figure out the magic behind the hibernate XML setup for this scenario. My database table "SETUP" has three columns: user_id (FK) setup_item setup_value The database composite key is made up of user_id | setup_item Here's the Setup.java class: public class Setup implements CommonFormElements, Serializable { private Map data = new HashMap(); private String saveAction; private Integer speciesNamingList; private User user; Logger log = LoggerFactory.getLogger(Setup.class); public String getSaveAction() { return saveAction; } public void setSaveAction(String action) { this.saveAction = action; } public User getUser() { return user; } public void setUser(User user) { this.user = user; } public Integer getSpeciesNamingList() { return speciesNamingList; } public void setSpeciesNamingList(Integer speciesNamingList) { this.speciesNamingList = speciesNamingList; } public Map getData() { return data; } public void setData(Map data) { this.data = data; } } My problem with the Hibernate setup, is that I can't seem to figure out how to map out the fact that a foreign key and the key of a map will construct the composite key of the table... this is due to a lack of experience using Hibernate. Here's my initial attempt at getting this to work: <composite-id> <key-many-to-one foreign-key="id" name="user" column="user_id" class="Business.User"> <meta attribute="use-in-equals">true</meta> </key-many-to-one> </composite-id> <map lazy="false" name="data" table="setup"> <key column="user_id" property-ref="user"/> <composite-map-key class="Command.Setup"> <key-property name="data" column="setup_item" type="string"/> </composite-map-key> <element column="setup_value" not-null="true" type="string"/> </map> Any insight into how to properly map this common scenario would be most appreciated!

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  • User Mailer Failing

    - by Trevor Nederlof
    I have setup a process in my User model to send a bunch of @users to a mailing script, user_mailer.rb I am using the http://postageapp.com app to send out emails. The users are getting to the User_mailer but I am getting an error from there. Can anyone please point me in the right direction. User Model: class User < ActiveRecord::Base acts_as_authentic def self.mail_out weekday = Date.today.strftime('%A').downcase @users = find(:all, :conditions => {"#{weekday}sub".to_sym => 't'}) UserMailer.deliver_mail_out(@users) end end User_mailer.rb class UserMailer < ActionMailer::Base def mail_out(users) @recipients = { } users.each do |user| @recipients[user.email] = { :zipcode => user.zipcode } end from "[email protected]" subject "Check out the trailer of the day!" body :user => user end end mail_out.html.erb {{zipcode}}, Please check out the trailer of the day at http://www.dailytrailer.net Thank you! -- The DailyTrailer.net Team User db schema create_table "users", :force => true do |t| t.string "email" t.date "birthday" t.string "gender" t.string "zipcode" t.datetime "created_at" t.datetime "updated_at" t.string "crypted_password" t.string "password_salt" t.string "persistence_token" t.string "mondaysub", :default => "f", :null => false t.string "tuesdaysub", :default => "f", :null => false t.string "wednesdaysub", :default => "f", :null => false t.string "thursdaysub", :default => "f", :null => false t.string "fridaysub", :default => "f", :null => false t.string "saturdaysub", :default => "f", :null => false t.string "sundaysub", :default => "f", :null => false end Error: /var/lib/gems/1.8/gems/rails-2.3.5/lib/commands/runner.rb:48: undefined method `name' for #<User:0xb6e8ae48> (NoMethodError) from /home/tnederlof/Dropbox/Ruby/daily_trailer/app/models/user_mailer.rb:5:in `mail_out' from /home/tnederlof/Dropbox/Ruby/daily_trailer/app/models/user_mailer.rb:4:in `each' from /home/tnederlof/Dropbox/Ruby/daily_trailer/app/models/user_mailer.rb:4:in `mail_out' from /home/tnederlof/.gem/ruby/1.8/gems/actionmailer-2.3.5/lib/action_mailer/base.rb:459:in `__send__' from /home/tnederlof/.gem/ruby/1.8/gems/actionmailer-2.3.5/lib/action_mailer/base.rb:459:in `create!' from /home/tnederlof/.gem/ruby/1.8/gems/actionmailer-2.3.5/lib/action_mailer/base.rb:452:in `initialize' from /home/tnederlof/.gem/ruby/1.8/gems/actionmailer-2.3.5/lib/action_mailer/base.rb:395:in `new' from /home/tnederlof/.gem/ruby/1.8/gems/actionmailer-2.3.5/lib/action_mailer/base.rb:395:in `method_missing' from /home/tnederlof/Dropbox/Ruby/daily_trailer/app/models/user.rb:13:in `mail_out' from (eval):1 from /usr/lib/ruby/1.8/rubygems/custom_require.rb:31:in `eval' from /var/lib/gems/1.8/gems/rails-2.3.5/lib/commands/runner.rb:48 from /usr/lib/ruby/1.8/rubygems/custom_require.rb:31:in `gem_original_require' from /usr/lib/ruby/1.8/rubygems/custom_require.rb:31:in `require' from script/runner:3

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  • Best way to ask confirmation from user before leaving the page

    - by JohnathanKong
    Hey Everyone, I am currently building a registration page where if the user leaves, I want to pop up a CSS box asking him if he is sure or not. I can accomplish this feat using confirm boxes, but the client says that they are too ugly. I've tried using unload and beforeunload, but both cannot stop the page from being redirected. Using those to events, I return false, so maybe there's a way to cancel other than returning false? Another solution that I've had was redirecting them to another page that has my popup, but the problem with that is that if they do want to leave the page, and it wasn't a mistake, they lose the page they were originally trying to go to. If I was a user, that would irritate me. The last solution was real popup window. The only thing I don't like about that is that the main winow will have their destination page while the pop will have my page. In my opinion it looks disjoint. On top of that, I'd be worried about popup blockers. Just to add to everyones comments. I understand that it is irritating to prevent users from exiting the page, and in my opinion it should not be done. Right now I am using a confirm box at this point. What happens is that it's not actually "preventing" the user from leaving, what the client actually wants to do is make a suggestion if the user is having doubts about registering. If the user is halfway through the registraiton process and leaves for some reason, the client wants to offer the user a free coupon to a seminar (this client is selling seminars) to hopefully persuade the user to register. The client is under the impression that since the user is already on the form, he is thinking of registering, and therefore maybe a seminar of what he is registering for would be the final push to get the user to register. Ideally I don't have to prevent the user from leaving, what would be just as good, and in my opinion better is if I can pause the unload process. Maybe a sleep command? I don't really have to keep the user on the page because either way they will be leaving to go to a different page. Also, as people have stated, this is a terriable title, so if someone knows a better one, I'd really appreciate it if they could change the title to something no so spammer inviting.

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  • Sharepoint AD imported users are becomming sporadically corrupted, causing us to have to create a new account

    - by TrevJen
    Sharepoint 2007 MOSS with AD imported users. All servers are 2008. ***UPDATE More details in testing. This Sharepoint is in an AD Child domain (clients.mycompany.local), which is sub to the root of the AD tree (mycompany.local). The user is in the parent tree (as are half of the other functional users. I have elevated the user rights to Domain. In looking at the logs, it seems that the Sharepoint server is trying to authenticate them by querying the DC for the clients domain (which is the way it normally works and still works for all existing identically configured users). I think if I could force it to authenticate up to the top domain DC then it would be ok?? I have around 50 users, over the past 2 months, I have had a handful of the users suddenly unable to login to Sharepoint. When they login, they either get a blank screen or they are repropmted. These users are using accounts that have been used for many months, sometimes the problem originates with a password change. In all cases, the users account works on every other Active Directory authenticated resource (domain, exchange, LDAP). In the most recent case, last night I was forced deleted a user ("John smith") because of corruption. The orifinal account name was jsmith. I deleted him from active directory, then deleted him from the profile list in Sharepoint Shared Services. I could not find a way to delete him from the Sharepoint user list, but I reran the import after recreating his account (renamed it too just to be sure to "smithj"). At first, this did not wor, the user could still access all other resources but Sharepoint. then, some 30 minutes later it inexplicably started working. This morning, the user changed passwords, which immediatly broke the login on Sharepoint again. Logs by request from matt b Office SharePoint Server Date: 4/13/2010 2:00:00 PM Event ID: 7888 Task Category: Office Server General Level: Error Keywords: Classic User: N/A Computer: nb-portal-01.clients.netboundary.local Description: A runtime exception was detected. Details follow. Message: Access is denied. (Exception from HRESULT: 0x80070005 (E_ACCESSDENIED)) – TrevJen 19 hours ago Techinal Details: System.UnauthorizedAccessException: Access is denied. (Exception from HRESULT: 0x80070005 (E_ACCESSDENIED)) at Microsoft.SharePoint.SPGlobal.HandleUnauthorizedAccessException(UnauthorizedAccessException ex) at Microsoft.SharePoint.Library.SPRequest.UpdateField(String bstrUrl, String bstrListName, String bstrXML) at Microsoft.SharePoint.SPField.UpdateCore(Boolean bToggleSealed) – TrevJen 19 hours ago at Microsoft.SharePoint.SPField.Update() at Microsoft.Office.Server.UserProfiles.SiteSynchronizer.UserSynchronizer.PushSchemaToList(Boolean& bAddedColumn) at Microsoft.Office.Server.UserProfiles.SiteSynchronizer.UserSynchronizer.SynchFull() at Microsoft.Office.Server.UserProfiles.SiteSynchronizer.Synch() at Microsoft.Office.Server.Diagnostics.FirstChanceHandler.ExceptionFilter(Boolean fRethrowException, TryBlock tryBlock, FilterBlock filter, CatchBlock catchBlock, FinallyBlock finallyBlock) – TrevJen 19 hours ago Log Name: Application Source: Office SharePoint Server Date: 4/13/2010 2:00:00 PM Event ID: 5553 Task Category: User Profiles Level: Error Keywords: Classic User: N/A Computer: nb-portal-01.clients.netboundary.local Description: failure trying to synch site 6fea15e2-0899-4c19-9016-44d77834c018 for ContentDB b2002b0b-3d4c-411a-8c4f-3d047ca9322c WebApp 3aff7051-455d-4a70-a377-5b1c36df618e. Exception message was Access is denied. (Exception from HRESULT: 0x80070005 (E_ACCESSDENIED)). – TrevJen 18 hours ago

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  • File Not Found error launching Guest OS in a non-administrator user account

    - by ToreTrygg
    Hi, I am running Fusion 2.0.6 (196839) on an iMac running 10.6.2 with 3 user accounts (1 Administrator). I have Fusion set up to share the Guest OS, and it's been working splendidly for nearly a year. Within the Guest OS (Windows XP PRO), there are also 3 user accounts (1 Administrator). Last night I went to back up my VM to an external drive, and to minimize the file size and transfer time, I deleted all Snapshots but the most-recent one. I then backed up the VM externally (28.23 GB). Today, one of my users tried to launch the Guest OS from within her user account, and received the following error message: "File Not Found: Windows XP Pro-000006.vmdk "This file is required to power on this virtual machine. If this file was moved, please provide its new location." My two choices are Cancel and Browse. When I browse, I can locate the Windows XP Pro-000006.vmdk file, which appears to be contained within the VM file (Windows XP Pro.vmwarevm). However, it still won't launch from a non-Admin user account. If I view the package contents of the VM file from the user account, the above file is present and appears to be created upon each launch of the Guest OS. If I go back to my Administrator account on the Mac and then launch Fusion, the Guest OS works perfectly for all 3 user accounts within XP Pro. I have tried to delete the Guest OS from Fusion's Library within the problem user account, then re-connect it to that Library, but the result is the same. The Guest OS data integrity is 100% -- but is accessible only from the OS X Administrator account. This problem only surfaced after deleting several older Snapshots. Again, the data is there, the Guest OS powers up normally in the Mac's Administrator account, but persistently returns the above error when attempting to power on from a non-Admin account on the Mac. I'm not sure how this is affecting the error, but when I look at Hard Disk Settings, the "unable-to-locate" file is the filename of the virtual HD. I don't want to make any changes to my (working) VM without any advice from the knowledgeable people on this forum. Any help will be greatly appreciated, thanks!

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  • Add user in CentOS 5

    - by Ron
    I created a new user in my CentOS web server with useradd. Added a password with passwd. But I can't log in with the user via SSH. I keep getting 'access denied'. I checked to make sure that the password was assigned and that the account is active. /var/log/secure shows the following error: Aug 13 03:41:40 server1 su: pam_unix(su:auth): authentication failure; logname= uid=500 euid=0 tty=pts/0 ruser=rwade rhost= user=root Please help, Thanks Thanks for the responses so far: I should add that it is a VPS on a remote computer, fresh out of the box. I can log in as the root user quite fine. I can also su to the new user, but I cannot log in as the new user. Here is my sshd_config file: # $OpenBSD: sshd_config,v 1.73 2005/12/06 22:38:28 reyk Exp $ # This is the sshd server system-wide configuration file. See # sshd_config(5) for more information. # This sshd was compiled with PATH=/usr/local/bin:/bin:/usr/bin # The strategy used for options in the default sshd_config shipped with # OpenSSH is to specify options with their default value where # possible, but leave them commented. Uncommented options change a # default value. #Port 22 #Protocol 2,1 Protocol 2 #AddressFamily any #ListenAddress 0.0.0.0 #ListenAddress :: # HostKey for protocol version 1 #HostKey /etc/ssh/ssh_host_key # HostKeys for protocol version 2 #HostKey /etc/ssh/ssh_host_rsa_key #HostKey /etc/ssh/ssh_host_dsa_key # Lifetime and size of ephemeral version 1 server key #KeyRegenerationInterval 1h #ServerKeyBits 768 # Logging # obsoletes QuietMode and FascistLogging #SyslogFacility AUTH SyslogFacility AUTHPRIV #LogLevel INFO # Authentication: #LoginGraceTime 2m #PermitRootLogin yes #StrictModes yes #MaxAuthTries 6 #RSAAuthentication yes #PubkeyAuthentication yes #AuthorizedKeysFile .ssh/authorized_keys # For this to work you will also need host keys in /etc/ssh/ssh_known_hosts #RhostsRSAAuthentication no # similar for protocol version 2 #HostbasedAuthentication no # Change to yes if you don't trust ~/.ssh/known_hosts for # RhostsRSAAuthentication and HostbasedAuthentication #IgnoreUserKnownHosts no # Don't read the user's ~/.rhosts and ~/.shosts files #IgnoreRhosts yes # To disable tunneled clear text passwords, change to no here! #PasswordAuthentication yes #PermitEmptyPasswords no PasswordAuthentication yes # Change to no to disable s/key passwords #ChallengeResponseAuthentication yes ChallengeResponseAuthentication no # Kerberos options #KerberosAuthentication no #KerberosOrLocalPasswd yes #KerberosTicketCleanup yes #KerberosGetAFSToken no # GSSAPI options #GSSAPIAuthentication no GSSAPIAuthentication yes #GSSAPICleanupCredentials yes GSSAPICleanupCredentials yes # Set this to 'yes' to enable PAM authentication, account processing, # and session processing. If this is enabled, PAM authentication will # be allowed through the ChallengeResponseAuthentication mechanism. # Depending on your PAM configuration, this may bypass the setting of # PasswordAuthentication, PermitEmptyPasswords, and # "PermitRootLogin without-password". If you just want the PAM account and # session checks to run without PAM authentication, then enable this but set # ChallengeResponseAuthentication=no #UsePAM no UsePAM yes # Accept locale-related environment variables AcceptEnv LANG LC_CTYPE LC_NUMERIC LC_TIME LC_COLLATE LC_MONETARY LC_MESSAGES AcceptEnv LC_PAPER LC_NAME LC_ADDRESS LC_TELEPHONE LC_MEASUREMENT AcceptEnv LC_IDENTIFICATION LC_ALL #AllowTcpForwarding yes #GatewayPorts no #X11Forwarding no X11Forwarding yes #X11DisplayOffset 10 #X11UseLocalhost yes #PrintMotd yes #PrintLastLog yes #TCPKeepAlive yes #UseLogin no #UsePrivilegeSeparation yes #PermitUserEnvironment no #Compression delayed #ClientAliveInterval 0 #ClientAliveCountMax 3 #ShowPatchLevel no #UseDNS yes #PidFile /var/run/sshd.pid #MaxStartups 10 #PermitTunnel no #ChrootDirectory none # no default banner path #Banner /some/path # override default of no subsystems Subsystem sftp /usr/libexec/openssh/sftp-server

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  • Add user in CentOS 5

    - by Ron
    I created a new user in my CentOS web server with useradd. Added a password with passwd. But I can't log in with the user via SSH. I keep getting 'access denied'. I checked to make sure that the password was assigned and that the account is active. /var/log/secure shows the following error: Aug 13 03:41:40 server1 su: pam_unix(su:auth): authentication failure; logname= uid=500 euid=0 tty=pts/0 ruser=rwade rhost= user=root Please help, Thanks Thanks for the responses so far: I should add that it is a VPS on a remote computer, fresh out of the box. I can log in as the root user quite fine. I can also su to the new user, but I cannot log in as the new user. Here is my sshd_config file: # $OpenBSD: sshd_config,v 1.73 2005/12/06 22:38:28 reyk Exp $ # This is the sshd server system-wide configuration file. See # sshd_config(5) for more information. # This sshd was compiled with PATH=/usr/local/bin:/bin:/usr/bin # The strategy used for options in the default sshd_config shipped with # OpenSSH is to specify options with their default value where # possible, but leave them commented. Uncommented options change a # default value. #Port 22 #Protocol 2,1 Protocol 2 #AddressFamily any #ListenAddress 0.0.0.0 #ListenAddress :: # HostKey for protocol version 1 #HostKey /etc/ssh/ssh_host_key # HostKeys for protocol version 2 #HostKey /etc/ssh/ssh_host_rsa_key #HostKey /etc/ssh/ssh_host_dsa_key # Lifetime and size of ephemeral version 1 server key #KeyRegenerationInterval 1h #ServerKeyBits 768 # Logging # obsoletes QuietMode and FascistLogging #SyslogFacility AUTH SyslogFacility AUTHPRIV #LogLevel INFO # Authentication: #LoginGraceTime 2m #PermitRootLogin yes #StrictModes yes #MaxAuthTries 6 #RSAAuthentication yes #PubkeyAuthentication yes #AuthorizedKeysFile .ssh/authorized_keys # For this to work you will also need host keys in /etc/ssh/ssh_known_hosts #RhostsRSAAuthentication no # similar for protocol version 2 #HostbasedAuthentication no # Change to yes if you don't trust ~/.ssh/known_hosts for # RhostsRSAAuthentication and HostbasedAuthentication #IgnoreUserKnownHosts no # Don't read the user's ~/.rhosts and ~/.shosts files #IgnoreRhosts yes # To disable tunneled clear text passwords, change to no here! #PasswordAuthentication yes #PermitEmptyPasswords no PasswordAuthentication yes # Change to no to disable s/key passwords #ChallengeResponseAuthentication yes ChallengeResponseAuthentication no # Kerberos options #KerberosAuthentication no #KerberosOrLocalPasswd yes #KerberosTicketCleanup yes #KerberosGetAFSToken no # GSSAPI options #GSSAPIAuthentication no GSSAPIAuthentication yes #GSSAPICleanupCredentials yes GSSAPICleanupCredentials yes # Set this to 'yes' to enable PAM authentication, account processing, # and session processing. If this is enabled, PAM authentication will # be allowed through the ChallengeResponseAuthentication mechanism. # Depending on your PAM configuration, this may bypass the setting of # PasswordAuthentication, PermitEmptyPasswords, and # "PermitRootLogin without-password". If you just want the PAM account and # session checks to run without PAM authentication, then enable this but set # ChallengeResponseAuthentication=no #UsePAM no UsePAM yes # Accept locale-related environment variables AcceptEnv LANG LC_CTYPE LC_NUMERIC LC_TIME LC_COLLATE LC_MONETARY LC_MESSAGES AcceptEnv LC_PAPER LC_NAME LC_ADDRESS LC_TELEPHONE LC_MEASUREMENT AcceptEnv LC_IDENTIFICATION LC_ALL #AllowTcpForwarding yes #GatewayPorts no #X11Forwarding no X11Forwarding yes #X11DisplayOffset 10 #X11UseLocalhost yes #PrintMotd yes #PrintLastLog yes #TCPKeepAlive yes #UseLogin no #UsePrivilegeSeparation yes #PermitUserEnvironment no #Compression delayed #ClientAliveInterval 0 #ClientAliveCountMax 3 #ShowPatchLevel no #UseDNS yes #PidFile /var/run/sshd.pid #MaxStartups 10 #PermitTunnel no #ChrootDirectory none # no default banner path #Banner /some/path # override default of no subsystems Subsystem sftp /usr/libexec/openssh/sftp-server

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  • Very simply, how can check if a user exists against my MySQL database?

    - by Sergio Tapia
    Here's what I have but nothing is output to the screen. :\ <html> <head> </head> <body> <? mysql_connect(localhost, "sergio", "123"); @mysql_select_db("multas") or die( "Unable to select database"); $query="SELECT * FROM usuario"; $result=mysql_query($query); $num=mysql_numrows($result); $i=0; $username=GET["u"]; $password=GET["p"]; while ($i < $num) { $dbusername=mysql_result($result,$i,"username"); $dbpassword=mysql_result($result,$i,"password"); if(($username == $dbusername) && ($password == $dbpassword)){ echo "si"; } $i++; } ?> </body> </html> I'm iterating through all users and seeing if there is a match for user && password. Any guidance?

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  • SQL - Count grouped entries and then get the max values grouped by date

    - by Marcus
    hello, I am out of any logic how to write the right sql statment. I've got a sqlite table holding every played track in a row with played date/time Now I will count the plays of all artists, grouped by day and then find the artist with the max playcount per day. I used this Query SELECT COUNT(ARTISTID) AS artistcount, ARTIST AS artistname,strftime('%Y-%m-%d', playtime) AS day_played FROM playcount GROUP BY artistname to get this result "93"|"The Skygreen Leopards"|"2010-06-16" "2" |"Arcade Fire" |"2010-06-15" "2" |"Dead Kennedys" |"2010-06-15" "2" |"Wolf People" |"2010-06-15" "3" |"16 Horsepower" |"2010-06-15" "3" |"Alela Diane" |"2010-06-15" "46"|"Motorama" |"2010-06-15" "1" |"Ariel Pink's Haunted" |"2010-06-14" I tried then to query this virtual table but I always get false results in artistname. SELECT MAX(artistcount), artistname , day_played FROM ( SELECT COUNT(ARTISTID) AS artistcount, ARTIST AS artistname,strftime('%Y-%m-%d', playtime) AS day_played FROM playcount GROUP BY artistname ) GROUP BY strftime('%Y-%m-%d',day_played) result in this "93"|"lilium" |"2010-06-16" "46"|"Wolf People"|"2010-06-15" "30"|"of Montreal"|"2010-06-14" but the artist name is false. I think through the grouping by day, it just use the last artist, or so. I tested stuff like INNER JOIN or GROUP BY ... HAVING in trial and error, I read examples of similar issues but always get lost in columnnames and stuff (I am a bit burned out) I hope someone can give me a hint. thanks m

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  • Delight and Excite

    - by Applications User Experience
    Mick McGee, CEO & President, EchoUser Editor’s Note: EchoUser is a User Experience design firm in San Francisco and a member of the Oracle Usability Advisory Board. Mick and his staff regularly consult on Oracle Applications UX projects. Being part of a user experience design firm, we have the luxury of working with a lot of great people across many great companies. We get to help people solve their problems.  At least we used to. The basic design challenge is still the same; however, the goal is not necessarily to solve “problems” anymore; it is, “I want our products to delight and excite!” The question for us as UX professionals is how to design to those goals, and then how to assess them from a usability perspective. I’m not sure where I first heard “delight and excite” (A book? blog post? Facebook  status? Steve Jobs quote?), but now I hear these listed as user experience goals all the time. In particular, somewhat paradoxically, I routinely hear them in enterprise software conversations. And when asking these same enterprise companies what will make the project successful, we very often hear, “Make it like Apple.” In past days, it was “make it like Yahoo (or Amazon or Google“) but now Apple is the common benchmark. Steve Jobs and Apple were not secrets, but with Jobs’ passing and Apple becoming the world’s most valuable company in the last year, the impact of great design and experience is suddenly very widespread. In particular, users’ expectations have gone way up. Being an enterprise company is no shield to the general expectations that users now have, for all products. Designing a “Minimum Viable Product” The user experience challenge has historically been, to echo the words of Eric Ries (author of Lean Startup) , to create a “minimum viable product”: the proverbial, “make it good enough”. But, in our profession, the “minimum viable” part of that phrase has oftentimes, unfortunately, referred to the design and user experience. Technology typically dominated the focus of the biggest, most successful companies. Few have had the laser focus of Apple to also create and sell design and user experience alongside great technology. But now that Apple is the most valuable company in the world, copying their success is a common undertaking. Great design is now a premium offering that everyone wants, from the one-person startup to the largest companies, consumer and enterprise. This emerging business paradigm will have significant impact across the user experience design process and profession. One area that particularly interests me is, how are we going to evaluate these new emerging “delight and excite” experiences, which are further customized to each particular domain? How to Measure “Delight and Excite” Traditional usability measures of task completion rate, assists, time, and errors are still extremely useful in many situations; however, they are too blunt to offer much insight into emerging experiences “Satisfaction” is usually assessed in user testing, in roughly equivalent importance to the above objective metrics. Various surveys and scales have provided ways to measure satisfying UX, with whatever questions they include. However, to meet the demands of new business goals and keep users at the center of design and development processes, we have to explore new methods to better capture custom-experience goals and emotion-driven user responses. We have had success assessing custom experiences, including “delight and excite”, by employing a variety of user testing methods that tend to combine formative and summative techniques (formative being focused more on identifying usability issues and ways to improve design, and summative focused more on metrics). Our most successful tool has been one we’ve been using for a long time, Magnitude Estimation Technique (MET). But it’s not necessarily about MET as a measure, rather how it is created. Caption: For one client, EchoUser did two rounds of testing.  Each test was a mix of performing representative tasks and gathering qualitative impressions. Each user participated in an in-person moderated 1-on-1 session for 1 hour, using a testing set-up where they held the phone. The primary goal was to identify usability issues and recommend design improvements. MET is based on a definition of the desired experience, which users will then use to rate items of interest (usually tasks in a usability test). In other words, a custom experience definition needs to be created. This can then be used to measure satisfaction in accomplishing tasks; “delight and excite”; or anything else from strategic goals, user demands, or elsewhere. For reference, our standard MET definition in usability testing is: “User experience is your perception of how easy to use, well designed and productive an interface is to complete tasks.” Articulating the User Experience We’ve helped construct experience definitions for several clients to better match their business goals. One example is a modification of the above that was needed for a company that makes medical-related products: “User experience is your perception of how easy to use, well-designed, productive and safe an interface is for conducting tasks. ‘Safe’ is how free an environment (including devices, software, facilities, people, etc.) is from danger, risk, and injury.” Another example is from a company that is pushing hard to incorporate “delight” into their enterprise business line: “User experience is your perception of a product’s ease of use and learning, satisfaction and delight in design, and ability to accomplish objectives.” I find the last one particularly compelling in that there is little that identifies the experience as being for a highly technical enterprise application. That definition could easily be applied to any number of consumer products. We have gone further than the above, including “sexy” and “cool” where decision-makers insisted they were part of the desired experience. We also applied it to completely different experiences where the “interface” was, for example, riding public transit, the “tasks” were train rides, and we followed the participants through the train-riding journey and rated various aspects accordingly: “A good public transportation experience is a cost-effective way of reliably, conveniently, and safely getting me to my intended destination on time.” To construct these definitions, we’ve employed both bottom-up and top-down approaches, depending on circumstances. For bottom-up, user inputs help dictate the terms that best fit the desired experience (usually by way of cluster and factor analysis). Top-down depends on strategic, visionary goals expressed by upper management that we then attempt to integrate into product development (e.g., “delight and excite”). We like a combination of both approaches to push the innovation envelope, but still be mindful of current user concerns. Hopefully the idea of crafting your own custom experience, and a way to measure it, can provide you with some ideas how you can adapt your user experience needs to whatever company you are in. Whether product-development or service-oriented, nearly every company is ultimately providing a user experience. The Bottom Line Creating great experiences may have been popularized by Steve Jobs and Apple, but I’ll be honest, it’s a good feeling to be moving from “good enough” to “delight and excite,” despite the challenge that entails. In fact, it’s because of that challenge that we will expand what we do as UX professionals to help deliver and assess those experiences. I’m excited to see how we, Oracle, and the rest of the industry will live up to that challenge.

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