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See the latest Applications Cloud user experiences at Oracle OpenWorld 2014

- by mvaughan

By Misha Vaughan, Oracle Applications User Experience OAUX Day: Oracle Applications Cloud User Experience Strategy & Roadmap?. This event is for partners, Oracle sales, and customers who are passionate about Oracle’s commitment to the ongoing user experience investment in Oracle’s Applications Cloud. If you want to see where we are going firsthand, contact the Applications UX team to attend this special event, scheduled the week before Oracle OpenWorld.All attendees must be approved to attend and have signed Oracle’s non-disclosure agreement. Register HERE.Date and time: 8 a.m. - 5 p.m. Wednesday, Sept. 24, 2014 Location: Oracle Conference Center, Redwood City, Calif. Oracle Applications Cloud User Experience Partner & Sales Briefing This event is for Oracle Applications partners and Oracle sales who want to find out what’s up with release 9 user experience highlights for: Oracle Sales Cloud, Oracle HCM Cloud, cloud extensibility, and Paas4SaaS. It will be held the day before Oracle OpenWorld kicks off. All attendees must be approved to attend. Register HERE.Date and time: 10:30 a.m. - 12:30 p.m. Sunday, Sept. 28, 2014Location: Intercontinental Hotel, 888 Howard Street, San Francisco, Calif. , in the Telegraph Hill room. Oracle OpenWorld 2014 OAUX Applications Cloud Exchange.This daylong, demo-intensive event is for Oracle customers, partners, and sales representatives who want to see what the future of Oracle’s cloud user experiences will look like. Attendees will also see what’s cooking in Oracle’s research and development kitchen – concepts that aren’t products … yet.All attendees must be approved to attend and have signed Oracle’s non-disclosure agreement. Register HERE.Date and time: 1 - 4 p.m. and 6 - 8:00 p.m. Monday, Sept. 29, 2014 Location: Intercontinental Hotel, 888 Howard Street, San Francisco, Calif., on the Spa Terrace.

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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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session management: verifying a user's log-in state

- by good_computer

I am storing sessions in my database. Everytime a user logs in, I create a new row corresponding to the new session, generate a new session id and send it as a cookie to the browser. My session data looks something like this: { 'user_id': 1234 'user_name': 'Sam' ... } When a request comes, I check whether a cookie with a session id is sent. If it is, I fetch session data from my database (or memcache) corresponding to that session id. When the user logs out, I remove the session data from my database (and memcache), and delete the cookie from the user's browser too. Notice that in my session data, I don't have something like logged_in: true. This is because if I find a session record in the database (or memcache) I deduce that the user is logged in, and if there is no session record found, the user is not logged in. My question is: is this the right approach? Should I have a logged_in key in my session data? Is there any possibility that a session record may be present on the server where the corresponding user is actually NOT logged in? Are there any security implications in having or not having such a key?

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Storing User-uploaded Images

- by Nyxynyx

What is the usual practice for handling user uploaded photos and storing them on the database and server? For a user profile image: After receiving the image file from user, rename file to <image_id>_<username> Move image to /images/userprofile Add img filename to a table users containing their profile details like first_name, last_name, age, gender, birthday For a image for a review done by user: After receiving the image file from user, rename file to <image_id>_<review_id> Move image to /images/reviews Add img filename to a table reviews containing their profile details like review_id, review_content, user_id, score. Question 1: How should I go about storing the image filenames if the user can upload multiple photos for a particular review? Serialize? Question 2: Or have another table review_images with columns review_id, image_id, image_filename just for tracking images? Will doing a JOIN when retriving the image_filename from this table slow down performance noticeably? Question 3: Should all the images be stored in a single folder? Will there be a problem when we have 100K photos in the same folder? Is there a more efficient way to go about doing this?

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SQL DB design to support user feeds (in application like facebook)

- by Yoav

I have a social network server with a MySql DB. I want to show the users feeds like done in Facebook. Example - UserX now Friend with userY, userX did like on postX etc. Currently I have table: C1 : UserId C2 : LogType (now friend, did like etc) C3 : ObjectId (Can be userId or postId) - set depending on the LogType. Currently to get all related logs to show to the user I do the following queries: 1. Get All user Friends userIds 2. Query all rows which C1 is in userIds (I query completed) 3. Scan the DB and see - if LogType equals DidLike, check if post's OwnerId is the userId - if yes add it to logs. And so on. Obvious this is not efficient at all. I am looking for a better way. I thought I had in mind: Create a new table (in addition to the Log table) C1 : UserId C2 : LogId (from Log table) C3 : UserID of the one who did the action When querying logs - look in the table and get related Logs (by LogId) from LogTable. Updating the table: Whenever user doing action that should be in the log: 1. Add the Log entry to LogTable. 2. Scan the DB and see which users are interested with the Log (Who my friends are, Who is the owner of the post) and add related entries to the new table. (must be done in BG). 3. If user UNFRIEND another user - then look in the logs for all rows where C3 == UNFRIENDED user id and delete them. Any opinions? Other suggestions?

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When overriding initWithCoder is it always necessary to call [super initWithCoder: coder]

- by Evan

In this code I am loading a View Controller (and associated View) from a .xib: -(id)initWithCoder:(NSCoder *)coder { [super initWithCoder:coder]; return self; } This successfully works, but I do not really understand what the line [super initWithCoder:coder] is accomplishing. Is that initializing my View Controller after my View has been initialized? Please be as explicit as possible when explaining. Thanks.

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What is Super Object in CodeIgniter?

- by Morgan Cheng

I saw the term "Super Object" in CodeIgniter manual, but the term is not explained in details. So, what is exactly "super object" in CodeIgnter?

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super function doesn't work inside a maya python module

- by sfjedi

Somehow, this works fine in the Maya/Python script editor, but fails when it's inside of my module code. Anyone have any ideas? class ControlShape(object): def __init__(self, *args, **kwargs): print 'Inside ControlShape...' class Cross(ControlShape): def __init__(self, *args, **kwargs): print 'Entering Cross...' super(Cross, self).__init__(*args, **kwargs) print 'Leaving Cross...' x = Cross() This gives me a TypeError: super(type, obj): obj must be an instance or subtype of type.

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why Cannot invoke super constructor from enum constructor ?

- by hilal

public enum A { A(1); private A(int i){ } private A(){ super(); // compile - error // Cannot invoke super constructor from enum constructor A() } } and here is the hierarchy of enum A extends from abstract java.lang.Enum extends java.lang.Object Class c = Class.forName("/*path*/.A"); System.out.println(c.getSuperclass().getName()); System.out.println(Modifier.toString(c.getSuperclass().getModifiers()).contains("abstract")); System.out.println(c.getSuperclass().getSuperclass().getName());

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Python, invoke super constructor

- by Mike

class A: def __init__(self): print "world" class B(A): def __init__(self): print "hello" B() hello In all other languages I've worked with the super constructor is invoked implicitly. How does one invoke it in Python? I would expect super(self) but this doesn't work

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Calling super()

- by Mike

When do you call super() in Java? I see it in some constructors of the derived class, but isn't the constructors for each of the parent class called automatically? Why would you need to use super?

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SQL SERVER – Solution – User Not Able to See Any User Created Object in Tables – Security and Permissions Issue

- by pinaldave

There is an old quote “A Picture is Worth a Thousand Words”. I believe this quote immensely. Quite often I get phone calls that something is not working if I can help. My reaction is in most of the cases, I need to know more, send me exact error or a screenshot. Until and unless I see the error or reproduce the scenario myself I prefer not to comment. Yesterday I got a similar phone call from an old friend, where he was not sure what is going on. Here is what he said. “When I try to connect to SQL Server, it lets me connect just fine as well let me open and explore the database. I noticed that I do not see any user created instances but when my colleague attempts to connect to the server, he is able to explore the database as well see all the user created tables and other objects. Can you help me fix it? “ My immediate reaction was he was facing security and permission issue. However, to make the same recommendation I suggested that he send me a screenshot of his own SSMS and his friend’s SSMS. After carefully looking at both the screenshots, I was very confident about the issue and we were able to resolve the issue. Let us reproduce the same scenario and many there is some learning for us. Issue: User not able to see user created objects First let us see the image of my friend’s SSMS screen. (Recreated on my machine) Now let us see my friend’s colleague SSMS screen. (Recreated on my machine) You can see that my friend could not see the user tables but his colleague was able to do the same for sure. Now I believed it was a permissions issue. Further to this I asked him to send me another image where I can see the various permissions of the user in the database. My friends screen My friends colleagues screen This indeed proved that my friend did not have access to the AdventureWorks database and because of the same he was not able to access the database. He did have public access which means he will have similar rights as guest access. However, their SQL Server had followed my earlier advise on having limited access for guest access, which means he was not able to see any user created objects. My next question was to validate what kind of access my friend’s colleague had. He replied that the colleague is the admin of the server. I suggested that if my friend was suppose to have admin access to the database, he should request of having admin access to his colleague. My friend promptly asked for the same to his colleague and on following screen he added him as an admin. You can do the same using following T-SQL script as well. USE [AdventureWorks2012] GO ALTER ROLE [db_owner] ADD MEMBER [testguest] GO Once my friend was admin he was able to access all the user objects just like he was expecting. Please note, this complete exercise was done on a development server. One should not play around with security on live or production server. Security is such an issue, which should be left with only senior administrator of the server. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, T SQL, Technology

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OAuth 2.0: Can a user-agent client avoid forwarding fragments?

- by Bosh

In the OAuth 2.0 draft specification, user-agent clients receive authorization in the form of a bearer token via redirection (from an authentication server) to a URL such as HTTP/1.1 302 Found Location: http://example.com/rd#access_token=FJQbwq9&expires_in=3600 According to Section 3.5.2 it is then the user-agent's job to GET the URL in question, but "The user-agent SHALL NOT include the fragment component with the request." In other words, as a result of the example redirection above, the user-agent should GET /rd HTTP/1.1 Host: example.com without passing #access_token to the server. My question: what user agents behave this way? I thought redirection in Firefox, for example, would (logically) include the fragment in the GET request. Am I just wrong about this, or does the OAuth 2.0 specification rely on non-standard user-agent behavior?

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What are the default groups assigned to the first user in Ubuntu Server?

- by Wayne Koorts

I just made a silly mistake on my Ubuntu Server box: I added myself to a group using usermod -G, after which I discovered the -a option... The result is that I am now out of the admin group, and lost my sudo rights. I can sort that out, but I want to know what other groups I may been removed from? My user was the first one so what I'm looking for is a list of groups that the first user gets added to at installation time.

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multiple ssh aliases is selecting wrong user when forwarding

- by Chris Beck

I'm following the dual identity procedure for bitbucket: I have 2 bitbucket accounts ccmcbeck and chrisbeck. The former is personal, the latter is work. On my local Mac, I have this in my ~/.ssh/config Host *.work.com User chris ForwardAgent yes IdentityFile ~/.ssh/work_dsa Host bitbucket-personal HostName bitbucket.org User ccmcbeck ForwardAgent no IdentityFile ~/.ssh/bitbucket_ccmcbeck_rsa Host bitbucket-work HostName bitbucket.org User chrisbeck ForwardAgent no IdentityFile ~/.ssh/bitbucket_chrisbeck_rsa On my local Mac I ssh -T all is good, I get: $ ssh -T git@bitbucket-personal logged in as ccmcbeck. $ ssh -T git@bitbucket-work logged in as chrisbeck. On my local Mac, the ssh version is OpenSSH_6.2p2, OSSLShim 0.9.8r 8 Dec 2011 When I ssh foo.work.com to my Linux box, I get: $ ssh-add -l 1024 ... /Users/chris/.ssh/work_dsa (DSA) 2048 ... /Users/chris/.ssh/bitbucket_ccmcbeck_rsa (RSA) 2048 ... /Users/chris/.ssh/bitbucket_chrisbeck_rsa (RSA) On foo.work.com, I also have this in my ~/.ssh/config Host bitbucket-personal HostName bitbucket.org User ccmcbeck ForwardAgent no IdentityFile ~/.ssh/bitbucket_ccmcbeck_rsa Host bitbucket-work HostName bitbucket.org User chrisbeck ForwardAgent no IdentityFile ~/.ssh/bitbucket_chrisbeck_rsa However, on foo.work.com when I ssh -T, it references the wrong User for git@bitbucket-work $ ssh -T git@bitbucket-personal logged in as ccmcbeck. $ ssh -T git@bitbucket-work logged in as ccmcbeck. On foo.work.com, the ssh version is OpenSSH_4.3p2, OpenSSL 0.9.8e-fips-rhel5 01 Jul 2008 Why is my configuration causing foo.work.com to reference the wrong User?

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Finding the groups of a user in WLS with OPSS

- by user12587121

How to find the group memberships for a user from a web application running in Weblogic server ? This is useful for building up the profile of the user for security purposes for example. WLS as a container offers an identity store service which applications can access to query and manage identities known to the container. This article for example shows how to recover the groups of the current user, but how can we find the same information for an arbitrary user ? It is the Oracle Platform for Securtiy Services (OPSS) that looks after the identity store in WLS and so it is in the OPSS APIs that we can find the way to recover this information. This is explained in the following documents. Starting from the FMW 11.1.1.5 book list, with the Security Overview document we can see how WLS uses OPSS: Proceeding to the more detailed Application Security document, we find this list of useful references for security in FMW. We can follow on into the User/Role API javadoc. The Application Security document explains how to ensure that the identity store is configured appropriately to allow the OPSS APIs to work. We must verify that the jps-config.xml file where the application is deployed has it's identity store configured--look for the following elements in that file: <serviceProvider type="IDENTITY_STORE" name="idstore.ldap.provider" class="oracle.security.jps.internal.idstore.ldap.LdapIdentityStoreProvider"> <description>LDAP-based IdentityStore Provider</description> </serviceProvider> <serviceInstance name="idstore.ldap" provider="idstore.ldap.provider"> <property name="idstore.config.provider" value="oracle.security.jps.wls.internal.idstore.WlsLdapIdStoreConfigProvider"/> <property name="CONNECTION_POOL_CLASS" value="oracle.security.idm.providers.stdldap.JNDIPool"/></serviceInstance> <serviceInstanceRef ref="idstore.ldap"/> The document contains a code sample for using the identity store here. Once we have the identity store reference we can recover the user's group memberships using the RoleManager interface: RoleManager roleManager = idStore.getRoleManager(); SearchResponse grantedRoles = null; try{ System.out.println("Retrieving granted WLS roles for user " + userPrincipal.getName()); grantedRoles = roleManager.getGrantedRoles(userPrincipal, false); while( grantedRoles.hasNext()){ Identity id = grantedRoles.next(); System.out.println(" disp name=" + id.getDisplayName() + " Name=" + id.getName() + " Principal=" + id.getPrincipal() + "Unique Name=" + id.getUniqueName()); // Here, we must use WLSGroupImpl() to build the Principal otherwise // OES does not recognize it. retSubject.getPrincipals().add(new WLSGroupImpl(id.getPrincipal().getName())); } }catch(Exception ex) { System.out.println("Error getting roles for user " + ex.getMessage()); ex.printStackTrace(); } }catch(Exception ex) { System.out.println("OESGateway: Got exception instantiating idstore reference"); } This small JDeveloper project has a simple servlet that executes a request for the user weblogic's roles on executing a get on the default URL. The full code to recover a user's goups is in the getSubjectWithRoles() method in the project.

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php-cgi cpu usage is super high

- by Ryan Thompson

I am getting constantly high and wildly fluctuating CPU usage % for php-cgi commands as seen via "top" on my Centos server.. I have a server density account and it seems that this is a common trend: User - PID - CPU % - MEM % - VSZ - RSS - TT - Stat - Started - Time - Command 500 - 6389 - 22.4 - 3 - 271136 - 32380 - ? - S - 20:26 - 0:40 - /usr/bin/php-cgi Seems there are about 6 or so of those records in my processes list at any given check-in. Any ideas what's causing this? I have fast_cgi installed and the module is loading.. Not sure why it isn't handling this though. Any help would be greatly appreciated! Ryan

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Run Active Directory Admin Center as another user

- by dunxd

I am trying to run Active Directory Admin Center (dsac.exe) on Windows 7 as another user by means of creating a shortcut, rather than having to Shift+Right click and specify the user. On Windows XP I could create a runas shortcut like this (forget for a moment that dsac.exe does not exist in Windows XP): runas /user:DOMAIN\user dsac.exe When I run this on Windows 7, the cmd style windows pops up and asks for the password for DOMAIN\user, but I get the following message: Attempting to start dsac.exe as user "DOMAIN\user" ... RUNAS ERROR: Unable to run - dsac.exe 740: The requested operation requires elevation. How do I get Windows 7 to automatically run dsac.exe as a specified user? I'm happy to fill in a password prompt for the specified user, but would be even happier if there was a solution that cached the password, so I didn't have to enter it more than once a day.

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Access denied for user 'diduser'@'localhost' to database 'diddata' (1044, 42000)

- by Arlen Beiler

I am trying to setup a MySQL server and when I went to create a second user it wouldn't give it permissions for the database. I can connect fine as long as I don't specify a database. Access denied for user 'user'@'localhost' to database 'diddata' The connection details are: { 'host' : 'localhost', 'user' : 'user', 'password' : 'password' , 'database': 'diddata' }; And to create the DB and table I did: CREATE DATABASE IF NOT exists diddata; CREATE USER 'user'@'localhost' IDENTIFIED BY 'password'; GRANT ALL ON user.* TO 'user'@'localhost'; Note that I've changed the username and password in this question. I've already checked the privileges in MySQL workbench and they are there.

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transparently set up Windows 7 as remote workstation

- by Áxel

Maybe is a very basic question, but I can't find the exact terms to Google for it and find the concrete answer to my doubt. Suppose we have several PCs in which individual employees work. One of them has an extremely powerful CPU, and it's very useful to use that computer to perform heavy computations, but go there and set up your task means its user has to stop working for a while. Is it possible to allow a secondary user account to remotly log in, for example via Remote Desktop, and work with a full user environment, while the main user keeps working under his user session? I've used remote desktop many times in the past, but it always blocked current user session, or even terminated it. Lots of thanks in advance guys.

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Losing file permissions after rebooting Windows 7

- by SMTF

I have a User directory full of files that are not accessible permission wise for the user who's home directory it is. Said user can't run Explorer. For example, it provides an error complaining that permission is not available for required files. I tried various ways to give said user permission of his home directory and things are fine until after rebooting the machine; the permissions reset to the previous state and the problem persists. I followed the solution outlined here. And again things worked until I reboot the machine. I'm in this mess because I replaced a corrupted user profile as outlined here. The original user and the new replacement on are/where both admin accounts. In case it is relevant I will mention that the Users directory is not on the C volume but a D volume on the same machine. Any insight is appreciated.

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Java inheritance and super() isn't working as expected

- by dwwilson66

For a homework assignment, I'm working with the following. It's an assigned class structure, I know it's not the best design by a long shot. Class | Extends | Variables -------------------------------------------------------- Person | None | firstName, lastName, streetAddress, zipCode, phone CollegeEmployee | Person | ssn, salary,deptName Faculty | CollegeEmployee | tenure(boolean) Student | person | GPA,major So in the Faculty class... public class Faculty extends CollegeEmployee { protected String booleanFlag; protected boolean tenured; public Faculty(String firstName, String lastName, String streetAddress, String zipCode, String phoneNumber,String ssn, String department,double salary) { super(firstName,lastName,streetAddress,zipCode,phoneNumber, ssn,department,salary); String booleanFlag = JOptionPane.showInputDialog (null, "Tenured (Y/N)?"); if(booleanFlag.equals("Y")) tenured = true; else tenured = false; } } It was my understanding that super() in Faculty would allow access to the variables in CollegeEmployee as well as Person. With the code above, it compiles fine when I ONLY include the Person variables. As soon as I try to use ssn, department, or salary I get the following compile errors. Faculty.java:15: error: constructor CollegeEmployee in class CollegeEmployee can not be applied to the given types: super(firstName,lastName,streetAddress,zipCode,phoneNumber,ssn,department,salary); ^ Required: String,String,String,String,String Found: String,String,String,String,String,String,String,String reason: actual and formal argument lists differ in length I'm completely confused by this error...which is the actual and formal? Person has five arguments, CollegeEmployee has 3, so my guess is that something's funky with how the parameters are being passed...but I'm not quite sure where to begin fixing it. What am I missing?

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what web based tool, to allow a non-technical user to manage authorized keys files on a Linux (fedora/centos/ubuntu/debian) server

- by Tom H

(Edit: clarification below) We have a number of groups of developers that change frequently, and a security policy to require individual logins to servers using rsa or dsa public keys, which is achieved via the standard method of adding id_dsa.pub to their authorized keys file. I am using chef to sync the user accounts across machines, however our previous method of using webmin to manage the user passwords is not designed for key based auth, and hence is not easy to use for non-technical users. The developers are logging in from the WAN using ssh, they can either provide their own key, or an administrator will send them a private key. The development machines are located in the cloud and we have a single server available to host the master set of accounts. Obviously I could deploy ldap or other centralised authentication system, but that seems a bit over blown when webmin worked well for the simple case. It is easy to achieve synchronised users, groups and passwords across a bunch of low security development boxes using webmin clustered users and groups. However looking at the currently installed webmin it is not so easy to create the authorized keys as it is to create user accounts and passwords. (its possible, but its not easy - some functionality is in the usermin module, or would required some tedious steps) Ideally I'd like a web interface that is pretty much dedicated to creating users and groups, and can generate key pairs on the fly, and can accepted pasted in public keys to add to the users authorized keys file. If the tool sync'ed the users and keys as well, that would be great, but I can use chef to do that part if the accounts are created correctly on the "master" server.

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How to change Windows admin password from guest user

- by John Smiith

How to gain access of admin account of Windows, I activated a guest user and I want to change the admin password from the command line. When I type: net user administrator password the response is System error 5 has occurred. Access is denied I am using winxp pro sp2 I am running this command from cdm.exe and I am running this command from guest user. I actually want to change my admin password from guest user.

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Hide user login in Windows XP

- by Tony Borf

I added a user account to my Windows XP box. Now this user is only accessing the pc remotely. My question is how can I remove that user from the login welcome screen? In fact how can I eliminate the welcome screen alltogeather and just log into the box automatically from the main user account? Thanks

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