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  • Virtualmin & git integration

    - by weby3456
    I've installed virtualmin on my VPS to manage my websites. It's working perfect and as expected nearly a year now. Recently I wanted to add some features to one of my sites, and I need git integration. I've correctly installed git & gitweb on my server, and I can create repositories and watch them under http://sub.domain.com/git/gitweb.cgi Here is the current relevant directory tree: /home/user/domains/sub.domain.com/public_html/git/ drwxr-sr-x user user . drwxr-x--- user user .. -rw-r--r-- user user git-favicon.png -rw-r--r-- user user git-logo.png -rwxr-xr-x user user gitweb.cgi -rw-r--r-- user user gitweb.css drwxrwx--- apache user reponame.git /home/user/domains/sub.domain.com/public_html/git/reponame.git/ drwxrwx--- apache user . drwxr-sr-x user user .. drwxrwx--- apache user branches -rwxrwx--- apache user config -rwxrwx--- user user description -rwxrwx--- apache user HEAD drwxrwx--- apache user hooks drwxrwx--- apache user info drwxrwx--- apache user objects drwxrwx--- apache user refs But I have some questions: When I'm visiting http://sub.domain.com/git/gitweb.cgi, the owner is listed as 'Apache'. why? how can I change that? Usually, to create a new git repository, I'll do something like: $ mkdir proj $ cd proj $ git init Initialized empty Git repository in /home/user/proj/.git/ // here I'm creating the files or copy them from somewhere else $ git add *.php $ git add README $ git commit -m 'initial version' But after creating the repository in virtualmin, I can find a new dir named 'reponame.git' but not the '.git' dir. When I'm trying to run any git command (e.g. git status) I'm receiving "fatal: This operation must be run in a work tree". How can I work with that repository? Currently I need to explicitly grant access for users to be able to view the repositories via gitweb. How can I make certain repositories public?

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  • calculating offer period for subscription

    - by TheVillageIdiot
    I'm maintaining a web application which deals with some kind of subscriptions. Users can to renew their subscriptions from 2 months before expiry (not earlier than that). Sometimes user does not renew before expiry and get grace period which is of 3 months. Now he can renew in these 3 months of grace period. Now the problem part. In the previous transactions of renew requests I have to show what was the offer period for that particular request (subscription start and subscription end period if renew was granted). Things are pretty simple if user renews before expiry, but I'm not able to get things straight if there is grace period specially when the subscriptions is expiring in last months of the year. Also there sometimes calculations go haywire when subscription is ending in jan or feb. All this is happening because offer period is not saved with the application anywhere (I don't know why). so if subscription is ending in 20 October 2008 and renew application is submitted in 16 January 2009 (because of grace period) the offer period should be 21 October 2008 to 20 October 2009.

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  • how to make a software and preserve database integrity and correctness and please help confused

    - by user287745
    i have made an application project in vs 08 c#, sql server from vs 08. the database has like 20 tables and many fields in each have made an interface for adding deleting editting and retrieving data according to predefined needs of the users. now i have to 1) make to project in to a software which i can deliver to professor. that is he can just double click the icon and the software simply starts. no vs 08 needed to start the debugging 2) the database will be on one powerful computer (dual core latest everything win xp) and the user will access it from another computer connected using LAN i am able to change the connection string to the shared database using vs 08/ debugger whenever the server changes but how am i supposed to do that when its a software? 3)there will by many clients am i supposed to give the same software to every one, so they all can connect to the database, how will the integrity and correctness of the database be maintained? i mean the db.mdf file will be in a folder which will be shared with read and write access. so its not necessary that only one user will write at a time. so is there any coding for this or? please help me out here i am stuck do not know what to do i have no practical experience, would appreciate all the help thank you

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  • Users in database server or database tables

    - by Batcat
    Hi all, I came across an interesting issue about client server application design. We have this browser based management application where it has many users using the system. So obvisously within that application we have an user management module within it. I have always thought having an user table in the database to keep all the login details was good enough. However, a senior developer said user management should be done in the database server layer if not then is poorly designed. What he meant was, if a user wants to use the application then a user should be created in the user table AND in the database server as a user account as well. So if I have 50 users using my applications, then I should have 50 database server user logins. I personally think having just one user account in the database server for this database was enough. Just grant this user with the allowed privileges to operate all the necessary operation need by the application. The users that are interacting with the application should have their user accounts created and managed within the database table as they are more related to the application layer. I don't see and agree there is need to create a database server user account for every user created for the application in the user table. A single database server user should be enough to handle all the query sent by the application. Really hope to hear some suggestions / opinions and whether I'm missing something? performance or security issues? Thank you very much.

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  • What is a typical scenario for and end-user reports design?

    - by Sebastian
    Hello! I'm wondering what would be the typical scenario for using an end-user report designer. What I'm thinking of is to have a base report with all the columns that I can have, also with a basic view of the report (formatting, order of columns, etc.) and then let the user to change that format and order, take out or add (from the available columns) data to it, etc. Is that a common way to address what is called end-user designer for reports or I'm off track? I know it depends on the user (if it's someone that can handle SQL or not for example), but is it common to have a scenario where the user can build everthing from the sql query to the formatting? Thanks! Sebastian

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  • can not login to windows

    - by LoRdiE
    Dear, When i login to windows using domain user account, it take a min show welcome and then automatically logoff. I think user profile error, so login with the administrator account and create new local account. When i login using the local user account, it happened the same as domain user account. Only Administrator Level can login to windows. Any know how can i fix this case? Thanks

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  • How do I save user specific data in an asp.net site?

    - by Greg McNulty
    I just set up user profiles using asp.net 3.5 using wvd. For each user I would like to store data that they will be updating every day. For example, every time they go for a run they will update time and distance. I intend to allow them to also look up their history of distance and time from any past date. My question is, what does the database schema usually look like for such a set up? Currently asp.net set up a db for me when I made user profiles. Do I just add an extra table for every user? Should there be one big table with all users data? How do I relate a user I'd to their specific data? Etc.... I have never done this before so any ideas on how this is usually done would be very helpful. Thank you.

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  • UI suggestions on how to display suggested tags for a given text to a user?

    - by Danny
    I am writing a web-app that uses a tagging system to organize the user's submitted reports. Part of it uses ajax to get suggestions for tags to present to the user based on the content of their report. I am looking for suggestions on how to present this information for the user. I'm not quite certain what a friendly way to do this would be. Edit: Well, most of the responses here seem to be focused on the user typing in keywords. The idea I'm trying to define here is more towards presenting the user a set of suggested keywords that they may accept or decline without having to type a tag in manually. (That option is of course still available to them) --------------------------- # say they can checkoff or select tags they like. | o[tag2] x[foo] o[moo] | | x[tag1] o[bar] | ---------------------------

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  • OSX root user keeps re-enabling itself on reboot

    - by geodave
    Running Snow Leopard. Completely inexplicably, I seem to have enabled the OSX root user by accident. I honestly have no idea how it happened, but if memory serves I was looking at the login pane (with my two user accounts) when I must have hit something, and suddenly the two accounts were replaced by one that just said "Other..." Clicking the "Other..." account allows me to type a username and password, but neither of the normal two accounts would work. Since I never set a root password, it wouldn't let me in that way either. So I booted into Single User mode and ran these commands: /sbin/mount -uw / fsck -fy launchctl load /System/Library/LaunchDaemons/com.apple.DirectoryServices.plist dscl . -passwd /Users/root newpassword and that let me login as root. Then, I went to System Preferences, Accounts, Login Options, clicked Join, Open Directory Utility, and lastly in the Edit menu I clicked "Disable Root User" Great, I thought, back to normal. Except rebooting, I still only have the Other... account visible, and the root password I set beforehand doesn't work anymore! I have to reboot into Single User Mode and go through the whole process again just to get back into the system (as root) How on Earth did I accidentally enable this? I didn't even know about the Directory Utility before now. And most importantly, why the heck would it be re-enabling the root user on boot? Thanks in advance to any help!

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  • How to reset the postgres super user password on mac os x

    - by Andrew Barinov
    I installed postgres on my mac running 10.6.8 and I would like to reset the password for the postgres user (I believe this is the super user password) and then restart it. All the directions I found do not work because I think my user name is not recognized by pg as having authority to change the password. (I am on the admin account of my mac) Here is what I tried: Larson-2:~ larson$ psql -U postgres Password for user postgres: psql (9.0.4, server 9.1.2) WARNING: psql version 9.0, server version 9.1. Some psql features might not work. Type "help" for help. postgres=# ALTER USER postgres with password 'mypassword' postgres-# \q and for restart I did: Larson-2:~ larson$ su postgres -c 'pg_ctl -D /opt/local/var/db/postgresql84/defaultdb/ restart > Which didn't work, as the password remained the same as it was before. Can someone provide directions for doing this and for making sure it's recognized by PG? Update I went ahead and edited the pg_hba.conf file located in /Library/PostgreSQL/9.1/data and set the settings as follows: # TYPE DATABASE USER ADDRESS METHOD # "local" is for Unix domain socket connections only local all all trust # IPv4 local connections: host all all 127.0.0.1/32 trust # IPv6 local connections: host all all ::1/128 trust However, like before, the password stayed the same after I changed it. I am not sure what further steps I can take from here.

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  • chrooted sftp user with write permissions to /var/www

    - by matthew
    I am getting confused about this setup that I am trying to deploy. I hope someone of you folks can lend me a hand: much much appreciated. Background info Server is Debian 6.0, ext3, with Apache2/SSL and Nginx at the front as reverse proxy. I need to provide sftp access to the Apache root directory (/var/www), making sure that the sftp user is chrooted to that path with RWX permissions. All this without modifying any default permission in /var/www. drwxr-xr-x 9 root root 4096 Nov 4 22:46 www Inside /var/www -rw-r----- 1 www-data www-data 177 Mar 11 2012 file1 drwxr-x--- 6 www-data www-data 4096 Sep 10 2012 dir1 drwxr-xr-x 7 www-data www-data 4096 Sep 28 2012 dir2 -rw------- 1 root root 19 Apr 6 2012 file2 -rw------- 1 root root 3548528 Sep 28 2012 file3 drwxr-x--- 6 www-data www-data 4096 Aug 22 00:11 dir3 drwxr-x--- 5 www-data www-data 4096 Jul 15 2012 dir4 drwxr-x--- 2 www-data www-data 536576 Nov 24 2012 dir5 drwxr-x--- 2 www-data www-data 4096 Nov 5 00:00 dir6 drwxr-x--- 2 www-data www-data 4096 Nov 4 13:24 dir7 What I have tried created a new group secureftp created a new sftp user, joined to secureftp and www-data groups also with nologin shell. Homedir is / edited sshd_config with Subsystem sftp internal-sftp AllowTcpForwarding no Match Group <secureftp> ChrootDirectory /var/www ForceCommand internal-sftp I can login with the sftp user, list files but no write action is allowed. Sftp user is in the www-data group but permissions in /var/www are read/read+x for the group bit so... It doesn't work. I've also tried with ACL, but as I apply ACL RWX permissions for the sftp user to /var/www (dirs and files recursively), it will change the unix permissions as well which is what I don't want. What can I do here? I was thinking I could enable the user www-data to login as sftp, so that it'll be able to modify files/dirs that www-data owns in /var/www. But for some reason I think this would be a stupid move securitywise.

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  • Managing arbitrary user permissions under PureFTPd

    - by Sebastián Grignoli
    I need to provide an FTP service that needs to be web-managed in the simplest way possible. My customer wants to create folders and users, and give them read only or read/write access arbitrarily. For example: The folder 'Documents' should be read only for several users, writable for internal users, and invisible for the rest. The folder 'Pictures' should be read only for journalists, writable for associates, and invisible for the rest. The folder 'Media' should be read only, writable or invisible for arbitrary users specified on the admin. There could be a large number of users and folders. I can't find a good way to accomplish that. I thought that I could give each user a home folder and put symlinks for the folders he has read access to, and make the user part of the folder's group when he has write access too, but now I think that this wouldn't work, because with PureFTPd (or ProFTPd) I can only specify the virtual user's mapping to a system user, and only one GUID for each virtual user. My approach requires that I could specify several GUIDs for each user (one by each folder he has write access to). I need to start programming this admin and I still don't know wich approach would work, if any. ¿Any ideas?

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  • Win7 - Opening "Programs and Features" as Admin from command line (logged in as regular user)

    - by user1741264
    We have Win7 machines on a domain that we'd like to open the "Programs and Features" control applet via the command line while a regular user is logged in. Heres the catch: I know how to do this using runas from command line BUT after "Programs and Features" opens, I dont truly have the ability to remove a program. I am told that I need to be an Admin to do so. Here are the commands I have tried: runas /user:%computername%\administrator cmd.exe then in the new cmd window running: control appwiz.cpl runas /user:%companydomain%\%domainadminacct% cmd.exe then in the new cmd window running: control appwiz.cpl runas /user:%computername%\administrator cmd.exe then in the new cmd window running: rundll32.exe shell32.dll,Control_RunDLL appwiz.cpl runas /user:%companydomain%\%domainadminacct% cmd.exe then in the new cmd window running: rundll32.exe shell32.dll,Control_RunDLL appwiz.cpl I have also tried all of the above as one long line of code instead of launching a cmd.exe as Admin As you can see, I have tried running the command using both a local admin account (Administrator) AND a domain admin account. I have alos tried launching the runas command as one long command (opening the "programs and features") AND 1st launching a cmd.exe with admin rights and THEN launching the "Prgrams and Features" window. The result is the same: The "Programs and Features" windows opens but when I try to perform an uninstall, I am told I need Admin rights. Thus I am lead to believe that this instance of "Programs and Features" is not truly being run as an admin I am trying to avoid logging the regular user out. I am also aware that every program has its own uninstaller, I do not want to uninstall that way. I want to use the uninstaller in "Programs and Features". Any help is appreciated.

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  • addSubview insertSubview aboveSubview bit confused as to why it does not work

    - by John Smith
    I'm a bit confused all as to why the belowSubview does not work. I'm adding some (subviews) to my navigationController and all of that works well. -(UITableView *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath { ... ... [self.navigationController.view addSubview:ImageView]; [self.navigationController.view addSubview:toolbar]; add some point in my app I wish to add another toolbar or image above my toolbar. so let's say I'm doing something like this -(void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath { ... ... [self.navigationController.view insertSubview:NewImageView aboveSubview:toolbar]; //crucial of course [edit] rvController = [RootViewController alloc] initWithNibName:@"RootViewController" bundle:[NSBundle] mainBundle]; rvController.CurrentLevel += 1; rvController.CurrentTitle = [dictionary objectForKey:@"Title"]; [self.navigationController pushViewController:rvController animated:YES]; rvController.tableDataSource = Children; [rvController release]; However this doesn't work.. Does anyone know what I'm doing wrong here ... Should I have used something else instead of addSubview or is the problem somewhere else?

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  • Confused about home screen widget size in normal screen and larget screen

    - by kknight
    I am designing a home screen widget. The widget layout file is like below. <RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android" android:id="@+id/widget" android:layout_width="240dip" android:layout_height="200dip" android:background="@drawable/base_all" /> I ran this widget on a HTC Hero device, which has a screen of 320 pixels * 480 pixels with mdpi. It ran perfect on HTC Hero. The widget takes 3 cells * 2 cells space, i.e. 240 pixels * 200 pixels. Then I ran this widget on a Nexus One device, which has a screen of 480 pixels * 800 pixels, mdpi. Since Nexus One also is mdpi, so I though 240dip is equivalent to 240 pixels on Nexus One and 200dip is equivalent to 200 pixels on Nexus One, so the widget will not take 3 cells * 2 cells space on Nexus One device. To my surprise, when running on Nexus One device, the widget take exact 3 cells * 2 cells, about 360 pixels * 300 pixels, on Nexus One device. I am confused. The layout xml above specifies 240dip in width and 200dip in height for the widget, but why did it take 360 pixels * 300 pixels on Nexus One Device? What am I missing? Thanks.

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  • Still confused about JavaScript's 'this'.

    - by Nick Lowman
    I've been reading through quite a few articles on the 'this' keyword when using JavaScript objects and I'm still somewhat confused. I'm quite happy writing object orientated Javascript and I get around the 'this' issue by referring the full object path but I don't like the fact I still find 'this' confusing. I found a good answer here which helped me but I'm still not 100% sure. So, onto the example. The following script is linked from test.html with <script src="js/test.js"></script> if (!nick) { var nick = {}; } nick.lowman = function(){ var helloA = 'Hello A'; console.log('1.',this, this.helloA); var init = function(){ var helloB = 'Hello B'; console.log('2.',this, this.helloB); } return { init: init } }(); nick.lowman.init(); What kind of expected to see was 1. Object {} nick.lowman, 'Hello A' 2. Object {} init, 'Hello B' But what I get is this? 1. Window test.html, undefined 2. Object {} init, undefined I think I understand some of what's happening there but I would mind if someone out there explains it to me. Also, I'm not entirely sure why the first 'console.log' is being called at all? If I remove the call to the init function //nick.lowman.init() firebug still outputs 1. Window test.html, undefined. Why is that? Why does nick.lowman() get called by the window object when the html page loads? Many thanks

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  • Confused about Ajax, Basic XMLHTTPRequest

    - by George
    I'm confused about the basics of Ajax. Right now I'm just trying to build a basic Ajax request using plain JavaScript to better understand how things work (as opposed to using Jquery or another library). First off, do you always need to pass a parameter or can you just retrieve data? In its most basic form, could I have an html document (located on the same server) that just has plain text, and another html document retrieve that text and load it on to the page? So I have fox.html with just text that says "The quick brown fox jumped over the lazy dog." and I want to pull in that text into ajax.html on load. I have the following on ajax.html <script type="text/javascript"> function createAJAX() { var ajax = new XMLHttpRequest(); ajax.open('get','fox.html',true); ajax.send(null); ajax = ajax.responseText; return(ajax); } document.write(createAJAX()); </script> This currently writes nothing when I load the page.

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  • Still Confused About Identifying vs. Non-Identifying Relationships

    - by Jason
    So, I've been reading up on identifying vs. non-identifying relationships in my database design, and a number of the answers on SO seem contradicting to me. Here are the two questions I am looking at: What's the Difference Between Identifying and Non-Identifying Relationships Trouble Deciding on Identifying or Non-Identifying Relationship Looking at the top answers from each question, I appear to get two different ideas of what an identifying relationship is. The first question's response says that an identifying relationship "describes a situation in which the existence of a row in the child table depends on a row in the parent table." An example of this that is given is, "An author can write many books (1-to-n relationship), but a book cannot exist without an author." That makes sense to me. However, when I read the response to question two, I get confused as it says, "if a child identifies its parent, it is an identifying relationship." The answer then goes on to give examples such as SSN (is identifying of a Person), but an address is not (because many people can live at an address). To me, this sounds more like a case of the decision between primary key and non-primary key. My own gut feeling (and additional research on other sites) points to the first question and its response being correct. However, I wanted to verify before I continued forward as I don't want to learn something wrong as I am working to understand database design. Thanks in advance.

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  • Qt/C++ - confused about caller/callee, object ownership

    - by Isabel
    I am creating a GUI to manipulate a robot arm. The location of the arm can be described by 6 floats (describing the positions of the various arm joints. The interface consists of a QGraphicsView with a diagram of the arm (which can be clicked to change the arm position - adjusting the 6 floats). The interface also has 6 lineEdit boxes, to also adjust those values separately. When the graphics view is clicked, and when the line edit boxes are changed, I'd like the line edit boxes / graphics view to stay in synchronisation. This brings me to confusion about how to store the 6 floats, and trigger events when they're updated. My current idea is this: The robot arm's location should be represented by a class, RobotArmLocation. Objects of this class then have methods such as obj.ShoulderRotation() and obj.SetShoulderRotation(). The MainWindow has a single instance of RobotArmLocation. Next is the bit I'm more confused about, how to join everything up. I am thinking: The MainWindow has a ArmLocationChanged slot. This is signalled whenever the location object is changed. The diagram class will have a SetRobotArmLocation(RobotArmLocation &loc). When the diagram is changed, it's free to change the location object, and fire a signal to the ArmLocationChanged slot. Likewise, changing any of the text boxes will fire a signal to that ArmLocationChanged slot. The slot then has code to synchronise all the elements. This kind of seems like a mess to me, does anyone have any other suggestions? I've also thought of the following, does it have any merrit? The RobotArmLocation class has a ValueChanged slot, the diagram and textboxes can use that directly, and bypass the MainWindow directly (seems cleaner?) thanks for any wisdom!

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  • I am confused about how to use @SessionAttributes

    - by yusaku
    I am trying to understand architecture of Spring MVC. However, I am completely confused by behavior of @SessionAttributes. Please look at SampleController below , it is handling post method by SuperForm class. In fact, just field of SuperForm class is only binding as I expected. However, After I put @SessionAttributes in Controller, handling method is binding as SubAForm. Can anybody explain me what happened in this binding. ------------------------------------------------------- @Controller @SessionAttributes("form") @RequestMapping(value = "/sample") public class SampleController { @RequestMapping(method = RequestMethod.GET) public String getCreateForm(Model model) { model.addAttribute("form", new SubAForm()); return "sample/input"; } @RequestMapping(method = RequestMethod.POST) public String register(@ModelAttribute("form") SuperForm form, Model model) { return "sample/input"; } } ------------------------------------------------------- public class SuperForm { private Long superId; public Long getSuperId() { return superId; } public void setSuperId(Long superId) { this.superId = superId; } } ------------------------------------------------------- public class SubAForm extends SuperForm { private Long subAId; public Long getSubAId() { return subAId; } public void setSubAId(Long subAId) { this.subAId = subAId; } } ------------------------------------------------------- <form:form modelAttribute="form" method="post"> <fieldset> <legend>SUPER FIELD</legend> <p> SUPER ID:<form:input path="superId" /> </p> </fieldset> <fieldset> <legend>SUB A FIELD</legend> <p> SUB A ID:<form:input path="subAId" /> </p> </fieldset> <p> <input type="submit" value="register" /> </p> </form:form>

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  • Confused about Android API's and compatability

    - by Keith
    I have purchased an HTC Incredible and have dived into the world of android! Only to find myself totally confused about the API levels and backward compatibility. My device runs the 2.1 OS, but I know that most of the devices out there run 1.5 or 1.6; and soon the 2.2 OS will be running on new devices. The SDK has gone through such enormous changes, that even constants have been renamed (from VIEW_ACTION to ACTION_VIEW for example). Methods have been added and removed (onPause replacing the earlier call, etc al). So, If I want to write an application that will work from 1.6+, does that mean I have to install and write my code using the 1.6 API; then test on later versions? Or can I write using the 2.1 SDK and just set the minSDK level and not use "new" features? I have never worked with an SDK that changes SO drastically from release to release! So I am not sure what to do.... I read through an article on the Android Development site(and this posting on stack overflow that references it: http://stackoverflow.com/questions/2076150/should-a-legacy-android-application-be-rebuilt-using-sdk-2-1), but it was still not very clear to me. Any help would be appreciated

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  • So, how is the Oracle HCM Cloud User Experience? In a word, smokin’!

    - by Edith Mireles-Oracle
    By Misha Vaughan, Oracle Applications User Experience Oracle unveiled its game-changing cloud user experience strategy at Oracle OpenWorld 2013 (remember that?) with a new simplified user interface (UI) paradigm.  The Oracle HCM cloud user experience is about light-weight interaction, tailored to the task you are trying to accomplish, on the device you are comfortable working with. A key theme for the Oracle user experience is being able to move from smartphone to tablet to desktop, with all of your data in the cloud. The Oracle HCM Cloud user experience provides designs for better productivity, no matter when and how your employees need to work. Release 8  Oracle recently demonstrated how fast it is moving development forward for our cloud applications, with the availability of release 8.  In release 8, users will see expanded simplicity in the HCM cloud user experience, such as filling out a time card and succession planning. Oracle has also expanded its mobile capabilities with task flows for payslips, managing absences, and advanced analytics. In addition, users will see expanded extensibility with the new structures editor for simplified pages, and the with the user interface text editor, which allows you to update language throughout the UI from one place. If you don’t like calling people who work for you “employees,” you can use this tool to create a term that is suited to your business.  Take a look yourself at what’s available now. What are people saying?Debra Lilley (@debralilley), an Oracle ACE Director who has a long history with Oracle Applications, recently gave her perspective on release 8: “Having had the privilege of seeing a preview of release 8, I am again impressed with the enhancements around simplified UI. Even more so, at a user group event in London this week, an existing Cloud HCM customer speaking publically about his implementation said he was very excited about release 8 as the absence functionality was so superior and simple to use.”  In an interview with Lilley for a blog post by Dennis Howlett  (@dahowlett), we probably couldn’t have asked for a more even-handed look at the Oracle Applications Cloud and the impact of user experience. Take the time to watch all three videos and get the full picture.  In closing, Howlett’s said: “There is always the caveat that getting from the past to Fusion [from the editor: Fusion is now called the Oracle Applications Cloud] is not quite as simple as may be painted, but the outcomes are much better than anticipated in large measure because the user experience is so much better than what went before.” Herman Slange, Technical Manager with Oracle Applications partner Profource, agrees with that comment. “We use on-premise Financials & HCM for internal use. Having a simple user interface that works on a desktop as well as a tablet for (very) non-technical users is a big relief. Coming from E-Business Suite, there is less training (none) required to access HCM content.  From a technical point of view, having the abilities to tailor the simplified UI very easy makes it very efficient for us to adjust to specific customer needs.  When we have a conversation about simplified UI, we just hand over a tablet and ask the customer to just use it. No training and no explanation required.” Finally, in a story by Computer Weekly  about Oracle customer BG Group, a natural gas exploration and production company based in the UK and with a presence in 20 countries, the author states: “The new HR platform has proved to be easier and more intuitive for HR staff to use than the previous SAP-based technology.” What’s Next for Oracle’s Applications Cloud User Experiences? This is the question that Steve Miranda, Oracle Executive Vice President, Applications Development, asks the Applications User Experience team, and we’ve been hard at work for some time now on “what’s next.”  I can’t say too much about it, but I can tell you that we’ve started talking to customers and partners, under non-disclosure agreements, about user experience concepts that we are working on in order to get their feedback. We recently had a chance to talk about possibilities for the Oracle HCM Cloud user experience at an Oracle HCM Southern California Customer Success Summit. This was a fantastic event, hosted by Shane Bliss and Vance Morossi of the Oracle Client Success Team. We got to use the uber-slick facilities of Allergan, our hosts (of Botox fame), headquartered in Irvine, Calif., with a presence in more than 100 countries. Photo by Misha Vaughan, Oracle Applications User Experience Vance Morossi, left, and Shane Bliss, of the Oracle Client Success Team, at an Oracle HCM Southern California Customer Success Summit.  We were treated to a few really excellent talks around human resources (HR). Alice White, VP Human Resources, discussed Allergan's process for global talent acquisition -- how Allergan has designed and deployed a global process, and global tools, along with Oracle and Cognizant, and are now at the end of a global implementation. She shared a couple of insights about the journey for Allergan: “One of the major areas for improvement was on role clarification within the company.” She said the company is “empowering managers and deputizing them as recruiters. Now it is a global process that is nimble and efficient."  Deepak Rammohan, VP Product Management, HCM Cloud, Oracle, also took the stage to talk about pioneering modern HR. He reflected modern HR problems of getting the right data about the workforce, the importance of getting the right talent as a key strategic initiative, and other workforce insights. "How do we design systems to deal with all of this?” he asked. “Make sure the systems are talent-centric. The next piece is collaborative, engaging, and mobile. A lot of this is influenced by what users see today. The last thing is around insight; insight at the point of decision-making." Rammohan showed off some killer HCM Cloud talent demos focused on simplicity and mobility that his team has been cooking up, and closed with a great line about the nature of modern recruiting: "Recruiting is a team sport." Deepak Rammohan, left, and Jake Kuramoto, both of Oracle, debate the merits of a Google Glass concept demo for recruiters on-the-go. Later, in an expo-style format, the Apps UX team showed several concepts for next-generation HCM Cloud user experiences, including demos shown by Jake Kuramoto (@jkuramoto) of The AppsLab, and Aylin Uysal (@aylinuysal), Director, HCM Cloud user experience. We even hauled out our eye-tracker, a research tool used to show where the eye is looking at a particular screen, thanks to teammate Michael LaDuke. Dionne Healy, HCM Client Executive, and Aylin Uysal, Director, HCM Cloud user experiences, Oracle, take a look at new HCM Cloud UX concepts. We closed the day with Jeremy Ashley (@jrwashley), VP, Applications User Experience, who brought it all back together by talking about the big picture for applications cloud user experiences. He covered the trends we are paying attention to now, what users will be expecting of their modern enterprise apps, and what Oracle’s design strategy is around these ideas.   We closed with an excellent reception hosted by ADP Payroll services at Bistango. Want to read more?Want to see where our cloud user experience is going next? Read more on the UsableApps web site about our latest design initiative: “Glance, Scan, Commit.” Or catch up on the back story by looking over our Applications Cloud user experience content on the UsableApps web site.  You can also find out where we’ll be next at the Events page on UsableApps.

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  • Apache-Mina FTPServer Issue — unable to login into apache ftp server while using database user manager

    - by piyush
    I am unable to login into apache ftp server while using database user manager: while entering username and password,I am getting following error in log file: [ INFO] 2013-02-07 20:51:07,779 [] [0:0:0:0:0:0:0:1] RECEIVED: USER piyush [ INFO] 2013-02-07 20:51:07,781 [piyush] [0:0:0:0:0:0:0:1] SENT: 331 User name okay, need password for piyush. [ INFO] 2013-02-07 20:51:07,784 [piyush] [0:0:0:0:0:0:0:1] RECEIVED: PASS ***** [ WARN] 2013-02-07 20:51:07,785 [piyush] [0:0:0:0:0:0:0:1] User failed to log in [ WARN] 2013-02-07 20:51:08,285 [piyush] [0:0:0:0:0:0:0:1] Login failure - piyush [ INFO] 2013-02-07 20:51:08,286 [piyush] [0:0:0:0:0:0:0:1] SENT: 530 Authentication failed. [ INFO] 2013-02-07 20:51:08,286 [piyush] [0:0:0:0:0:0:0:1] RECEIVED: QUIT [ INFO] 2013-02-07 20:51:08,290 [piyush] [0:0:0:0:0:0:0:1] SENT: 221 Goodbye. [ INFO] 2013-02-07 20:51:08,291 [piyush] [0:0:0:0:0:0:0:1] CLOSED here is my xml file ftpd-typical.xml: <?xml version="1.0" encoding="UTF-8"?> <!-- Licensed to the Apache Software Foundation (ASF) under one or more contributor license agreements. See the NOTICE file distributed with this work for additional information regarding copyright ownership. The ASF licenses this file to you under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at http://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. --> <server xmlns="http://mina.apache.org/ftpserver/spring/v1" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:beans="http://www.springframework.org/schema/beans" xsi:schemaLocation=" http://mina.apache.org/ftpserver/spring/v1 http://mina.apache.org/ftpserver/ftpserver-1.0.xsd " id="Prometheus"> <listeners> <nio-listener name="default" port="2121" /> </listeners> <db-user-manager encrypt-passwords="salted"> <data-source> <beans:bean class="org.apache.commons.dbcp.BasicDataSource" > <beans:property name="driverClassName" value="com.mysql.jdbc.Driver" /> <beans:property name="url" value="jdbc:mysql://localhost/apache_test" /> <beans:property name="username" value="amy" /> <beans:property name="password" value="piyush" /> </beans:bean> </data-source> <insert-user>INSERT INTO FTP_USER (userid, userpassword, homedirectory, enableflag, writepermission, idletime, uploadrate, downloadrate) VALUES ('{userid}', '{userpassword}', '{homedirectory}', {enableflag}, {writepermission}, {idletime}, {uploadrate}, {downloadrate}) </insert-user> <update-user>UPDATE FTP_USER SET userpassword='{userpassword}',homedirectory='{homedirectory}',enableflag={enableflag},writepermission={writepermission},idletime={idletime},uploadrate={uploadrate},downloadrate={downloadrate} WHERE userid='{userid}' </update-user> <delete-user>DELETE FROM FTP_USER WHERE userid = '{userid}' </delete-user> <select-user>SELECT userid, userpassword, homedirectory, enableflag, writepermission, idletime, uploadrate, downloadrate, maxloginnumber, maxloginperip FROM FTP_USER WHERE userid = '{userid}' </select-user> <select-all-users>SELECT userid FROM FTP_USER ORDER BY userid </select-all-users> <is-admin>SELECT userid FROM FTP_USER WHERE userid='{userid}' AND userid='admin' </is-admin> <authenticate>SELECT userpassword from FTP_USER WHERE userid='{userid}'</authenticate> </db-user-manager> </server>

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  • mount_afp on linux, user rights

    - by Antonio Sesto
    I need to mount a remote filesystem on a linux box using the afp protocol. The linux box runs an old Debian 4. I downloaded the source code of mount_afp, compiled it and installed it with all the required packages. Then created /etc/fuse with the following command: mknod /dev/fuse c 10 229 (according to the instructions here) I can mount the remote filesystem as root by executing: mount_afp afp://USER:PASSWD@REMOTE_SERVER/FOLDER /mnt/MOUNTPOINT/ but the same command fails when run as normal user (of the local machine). After reading here and there, I created a group fuse, and added my normal user U to the group fuse: [prompt] groups U U fuse Then modified the group of /dev/fuse, that now has the following rights: 0 crwxrwx--- 1 root fuse 10, 229 Feb 8 15:33 /dev/fuse However, if the user U tries to mount the remote filesystem by using the same command as above, U gets the following error: Incorrect permissions on /dev/fuse, mode of device is 20770, uid/gid is 0/1007. But your effective uid/gid is 1004/1004 But the user U with uid 1004 has also gid 1007 (group fuse). I might think the problem is related to real/effective/etc. ID, but I do not know how to proceed and could not find any clear instructions. Could you please help me? There is also another problem. If I mount /mnt/MOUNTPOINT as root and run ls -l /mnt, I get: drwxrwxrwx 15 root root 466 Feb 8 16:34 MONTPOINT If I run ls -l /mnt as normal user U I get: ? ?????????? ? ? ? ? ? MOUNTPOINT in fact when I try to cd /mnt/MOUNTPOINT I get: $-> cd /mnt/MOUNTPOINT -sh: cd: /mnt/MOUNTPOINT: Not a directory Then I unmount /mnt/MOUNTPOINT as root and run again ls -l /mnt as normal user U I get: 0 drwxr-xr-x 2 root root 6 Feb 8 15:32 MOUNTPOINT/ After reading Frank's answer, I killed every shell/process running with privileges of user U. Still U cannot mount the remote filesystem, but the error message has changed. Now it is: "Login error: Authentication failed". The problem is not related to remote login/password since the same command works perfectly when run as root of the local machine. Since I cannot get mount_afp to work with normal users, I decided to follow mgorven's suggestion. So I run the commands: mount_afp -o allow_other afp://USER:PASSWD@REMOTE_SERVER/FOLDER /mnt/MOUNTPOINT/ and mount_afp -o user=U afp://USER:PASSWD@REMOTE_SERVER/FOLDER /mnt/MOUNTPOINT/ The mount succeeds but user U cannot access the mount point. If U executes ls -l in /mnt U@LOCAL_HOST [/mnt] $-> ls -l ls: cannot access MOUNT_POINT: Permission denied total 0 ? ?????????? ? ? ? ? ? MOUNT_POINT Is it so hard to have this utility working?

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  • database replication for new user signup

    - by Jeff Storey
    I have a database that stores the users of my application. When a new user signs up, a record is inserted into the database for that user. I have a replicated version (slave) of this database (using mysql for now). What I'm concerned about is this scenario: step 1: user signs up and user record is inserted into the database step 2: user then tries to login, and the login process queries the database for the user. however, this query hits the slave database, but the user record has not yet been replicated in the slave and it returns an error that the user does not exist. This is a pretty trivial example, but I can see how it can apply to a lot of cases. Is there a strategy for configuring replicated databases to help prevent this situation?

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