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• not able to see In app purchase test user on Clicking of Manage user

  - by Gani

  hi everyone, i want to set up a test account to test in app purchase on sandbox, i am logging into intunes connect and following the same procedure as prescribed in the itunes connect developer guide. i am clinking on manage user, but i am not able to see the window where i can select test in app purchase user. do i need to do any change in my profile to make it visible.

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• SQLAuthority News – Wireless Router Security and Attached Devices – Complex Password

  - by pinaldave

  In the last four days (April 21-24), I have received calls from friends who told me that they have got strange emails from me. To my surprise, I did not send them any emails. I was not worried until my wife complained that she was not able to find one of the very important folders containing our daughter’s photo that is located in our shared drive. This was alarming in my par, so I started a search around my computer’s folders. Again, please note that I am by no means a security expert. I checked my entire computer with virus and spyware, and strangely, there I found nothing. I tried to think what can cause this happening. I suddenly realized that there was a power outage in my area for about two hours during the days I have mentioned. Back then, my wireless router needed to be reset, and so I did. I had set up my WPA-PSK [TKIP] + WPA2-PSK [AES] very well. My key was very simple ( ‘SQLAuthority1′), and I never thought of changing it. (It is now replaced with a very complex one). While checking the Attached Devices, I found out that there was another very strange computer name and IP attached to my network. And so as soon as I found out that there is strange device attached to my computer, I shutdown my local network. Afterwards, I reconfigured my wireless router with a more complex security key. Since I created the complex password, I noticed that the user is no more connecting to my machine. Subsequently, I figured out that I can also set up Access Control List. I added my networked computer to that list as well. When I tried to connect from an external laptop which was not in the list but with a valid security key, I was not able to access the network, neither able to connect to it. I wasn’t also able to connect using a remote desktop, so I think it was good. If you have received any nasty emails from me (from my gmail account) during the afore-mentioned days, I want to apologize. I am already paying for my negligence of not putting a complex password; by way of losing the important photos of my daughter. I have already checked with my client, whose password I saved in SSMS, so there was no issue at all. In fact, I have decided to never leave any saved password of production server in my SSMS. Here is the tip SQL SERVER – Clear Drop Down List of Recent Connection From SQL Server Management Studio to clean them. I think after doing all this, I am feeling safe right now. However, I believe that safety is an illusion of many times. I need your help and advice if there is anymore I can do to stop unauthorized access. I am seeking advice and help through your comments. Reference : Pinal Dave (http://www.SQLAuthority.com) Filed under: SQL, SQL Authority, SQL Query, SQL Security, SQL Server, SQL Tips and Tricks, SQLAuthority News, T SQL, Technology

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• Windows 7 Login User Doesn't Exist [closed]

  - by dcolumbus

  I have another interesting issue... because of some issue with a lost password, I had to manually change the password to one of the accounts via and DOS hack. However, somehow in the process I now have a phantom username that I am asked to logon to when Windows first starts... This username doesn't exit. In order to login, I have to "change user" and manually type in the correct username. Is there a way that I can edit which username it prompts me for? I'd like to repair this without having to reinstall just yet.

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• How can I cache a Subversion password on a server, without storing it in unencrypted form?

  - by Zilk

  My Subversion server only provides access via HTTPS; support for svn+ssh has been dropped because we wanted to avoid creating system users on that machine just for SVN access. Now I'm trying to provide a way for users to cache their passwords for a while, without leaving them stored on the filesystem in unencrypted form. This is no problem for Gnome or KDE users, because they can use gnome-keyring and kwallet, respectively. IIRC, TortoiseSVN has a similar caching mechanism, too. But what about users on a non-GUI system? Some context: in this case, we have a development/testing server where one project has been checked out into the Apache htdocs directory. Development for this project is almost complete, and only minor text/layout changes are performed directly on this server. Nevertheless, the changes should be checked into the repository. There's no kwallet and no gnome-keyring on this system, and the ssh-agent can't help because the repository is accessed via https instead of svn+ssh. As far as I know, that leaves them the choice of entering the password every time they talk to the SVN server, or storing it in an insecure way. Is there any way to get something like what gnome-keyring and kwallet provide in a non-GUI environment?

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• User Interface design books/resources for programmers

  - by mmacaulay

  Hi, I'm going to make my monthly trip to the bookstore soon and I'm kind of interested in learning some user interface and/or design stuff - mostly web related, what are some good books I should look at? One that I've seen come up frequently in the past is Don't Make Me Think, which looks promising. I'm aware of the fact that programmers often don't make great designers, and as such this is more of a potential hobby thing than a move to be a professional designer. I'm also looking for any good web resources on this topic. I subscribed to Jakob Nielsen's Alertbox newsletter, for instance, although it seems to come only once a month or so. Thanks! Somewhat related questions: http://stackoverflow.com/questions/75863/what-are-the-best-resources-for-designing-user-interfaces http://stackoverflow.com/questions/7973/user-interface-design

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• Is there an online user agent database?

  - by Gary Richardson

  How do you parse your user agent strings? I'm looking to get: Browser Browser Version OS OS Version from a user agent string. My app is written in perl and was previously using HTTP::BrowserDetect. It's a bit dated and is no longer maintained. I'm in no way tied to using perl for the actual lookup. I've come to the conclusion that automagic parsing is a lost cause. I was thinking of writing a crud type app to show me a list of unclassified UA's and manually keep them up to date. Does such an resource already exist that I can tap into? It would be awesome if I could make an HTTP call to look up the user agent info. Thanks!

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• linux user login/logout log for computer restriction

  - by Cedric

  Hi ! I would like to know how to log the login and logout of a user. I know it's possible to use the command "last". But this command is based on a file that has a r/w permission for the user, hence the possibility to change these data. I would like to log these data over two months. Why would I like to do that ? In fact, I would like to prevent a normal user to use a computer more than an hour a day - except week-ends, and 10 hours in total a week. Cedric System used : kubuntu, Programming language : bash script

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• Unable to resolve user environment variable correctly

  - by Junaid

  I am trying to resolve %USERPROFILE% using WScript.Shell. When I create a vbs file and run directly from Windows, I get the correct path for the logged-in user C:\Documents and Settings\Administrator but it gets resolved to C:\Documents and Settings\Default User instead of logged-in user when I used it inside my classic ASP webapp running on the local machine on IIS. The code I used is as below var oShell = new ActiveXObject("Wscript.Shell"); var userPath = oShell.ExpandEnvironmentStrings("%USERPROFILE%"); Is there a permission/setting which I need to check to get correct value of USERPROFILE when retrieving value from the webapp? PS: I am using javascript to code.

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• Applet User-agent

  - by Jonathan Barbero

  Hello! This is a simple question, but I didn´t found any documentation about this. When an applet makes a request, how is the user agent of the request. I want to know the applet user-agent expression to detect if a request comes from an applet. I make two test, with IE7 and Firefox 3.0.5 with JDK 1.6.0_03 and the user agent was "Mozilla/4.0 (Windows 2003 5.2) Java/1.6.0_03" in both, but I can´t generalize from two test. Thanks in advance, Jonathan.

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• ASP.NET - Exception logging approach for concurrent user scenario

  - by Whiskey-Tango-Foxtrot

  I am involved in designing a asp.net webforms application using .NET 3.5. I have a requirement where we need to log exceptions. What is the best approach for exception handling, given that there would be concurrent users for this application? Is there a need or possibility to log in exceptions at a user level? My support team in-charge wants to have a feature where the support team can get user specific log files. To give you a background, this application is currently on VB 6.0 and we are migrating it along with some enhancements. So, today the support personnel have a provision to get user specific log files.

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• C++ Detecting ENTER key pressed by user

  - by user69514

  I have a loop where I ask the user to enter a name. I need to stop when the user presses the ENTER key..... or when 20 names have been entered. However my method doesn't stop when the user presses the ENTER key //loop until ENTER key is entered or 20 elements have been added bool stop = false; int ind = 0; while( !stop || ind >= 20 ){ cout << "Enter name #" << (ind+1) << ":"; string temp; getline(cin, temp); int enterKey = atoi(temp.c_str()); if(enterKey == '\n'){ stop = true; } else{ names[ind] = temp; } ind++; }

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• How do I create a user history?

  - by ggfan

  I want to create a user history function that allows shows users what they done. ex: commented on an ad, posted an ad, voted on an ad, etc. How exactly do I do this? I was thinking about... in my site, when they log in it stores their user_id ($_SESSION['user_id']) so I guess whenever an user posts an ad(postad.php), comments(comment.php), I would just store in a database table "userhistory" what they did based on whenever or not their user_id was activate. When they comment, I store the user_id in the comment dbc table, so I'll also store it in the "userhistory" table. And then I would just queries all the rows in the dbc for the user to show it Any steps/improvements I can make? :)

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• Storing user settings in table - how?

  - by Mdillion

  I have settings for the user about 200 settings, these include notice settings and tracing settings from user activities on objects. The problem is how to store it in the DB? Should each setting be a row or a column? If colunm then table will have 200 colunms. If row then about 3 colunms but 200 rows per user x even 10 million users = not good. So how else can i store all these settings? NOTE: these settings are a mix of text entry and FK lookups to other tables. Thanks.

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• Django admin panel doesn't work after modify default user model.

  - by damienix

  I was trying to extend user profile. I founded a few solutions, but the most recommended was to create new user class containing foreign key to original django.contrib.auth.models.User class. I did it with this so i have in models.py: class UserProfile(models.Model): user = models.ForeignKey(User, unique=True) website_url = models.URLField(verify_exists=False) and in my admin.py from django.contrib import admin from someapp.models import * from django.contrib.auth.admin import UserAdmin # Define an inline admin descriptor for UserProfile model class UserProfileInline(admin.TabularInline): model = UserProfile fk_name = 'user' max_num = 1 # Define a new UserAdmin class class MyUserAdmin(UserAdmin): inlines = [UserProfileInline, ] # Re-register UserAdmin admin.site.unregister(User) admin.site.register(User, MyUserAdmin) And now when I'm trying to create/edit user in admin panel i have an error: "Unknown column 'content_userprofile.id' in 'field list'" where content is my appname. I was trying to add line AUTH_PROFILE_MODULE = 'content.UserProfile' to my settings.py but with no effect. How to tell panel admin to know how to correctly display fields in user form?

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• How do I reinstate my admin user privileges to global read/write

  - by Matt

  I am running Ubuntu 12.04 LTS. I only have the one user which I created when I installed Ubuntu. Everything has been fine - love it - until I updated a software package recently from the command line using sudo (not gksudo). I was having a little bother which did not make sense to me and in a fluff changed my user read/write privileges through the GUI (not even clear how I got there!). After restart I was stuck in a login loop - using the right login password but kept getting looped back to the login and could only login as Guest. I could still login with my user/password via ctrl + alt + f1 Eventually I was able to login again at start up. Not sure exactly what it was I changed that worked but it was one of/or a combination of installing latest security updates, changing login manager from LightDM to DGM and back again, removing the ICE/Xauthority and chown user. Current dilemma is my primary admin user privileges were read only. In the command line ls -ls /home/user returned this value: drwx------ 48 username username 20480 I have since changed this using sudo chmod 0755 /home/username (from my limited understanding 755 should return my user privileges to their original read/write glory). ls -ld /home/user currently shows my user privileges as: drwxr-xr-x 48 username username 20480 I still seem to have only read access permissions. I've been through lots of threads (and the help file) that talk about creating new users/groups permissions etc. but specific info on returning my existing global/admin/primary users privileges to what they were when I first created that user - baffling me. I feel this is something really simple I'm just not getting it. Please help! sudo mount /dev/sda1 on / type ext4 (rw,errors=remount-ro) proc on /proc type proc (rw,noexec,nosuid,nodev) sysfs on /proc type sysfs (rw,noexec,nosuid,nodev) none on /sys/fs/fuse/connections type fusect1 (rw) none on /sys/kernel/debug type debugfs (rw) none on /sys/kernel/security type securityfs (rw) udev on /dev type devtmpfs (rw,mode=07pe tmpfs55) devpts on /dev/pts type devpts (rw,noexec,nosuid,gid=5,mode=0620) tmpfs on /run type tmpfs (rw,noexec,nosuid,size=10%,mode=0755) none on /run/lock type tmpfs (rw, ,nosuid,nodev,size=5242880 none on /run/shm type tmpfs (rw,nosuid,nodev) gvfs-fuse-daemon on /home/meng/.gvfs type fuse.gvfs-fuse-daemon (rw,nosuid,nodev,user=meng) none on /tmp/guest-1R2Fi5 type tmpsf (rw,mode=700)

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• OS X: Copy a user account from one computer to another?

  - by user18169

  I'm traveling for a week and will be borrowing my roommate's laptop. We both have Macs running Snow Leopard. How can I copy my entire user account (files, settings, etc.) over to his computer? Is it as simple as creating a new user account and copying over my home folder? Do I have to do more?

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• How do you manage the user groups that users are assoicated to in Mac OS X?

  - by Erik Vold

  How do I add/remove users to user groups on Mac OS X 10.5? Like 'bonic_master' for instance.. and out of curiosity how do I create a user group?

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• Which is the simplest way to exclude an user from a launchd agent?

  - by compriots

  I have some launchd plists that are now placed at the administrator per-user level, particularly inside: /Library/LaunchAgents /Library/LaunchDaemons I'd like to avoid that some of those services could start only for one user (possibly avoiding to move all the plists to each ~/Library/LaunchAgents), which is the simplest way? Thanks in advance.

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• How can I transfer a user state of a win7 machine that won't boot?

  - by askvictor

  I have a windows 7 machine that won't boot completely, even in safe mode. I want to re-image the machine using a generic software image, but would like to keep the user data (including settings etc) that are on there ala Windows Easy Transfer. I can mount the hard disk on another machine - can I use Easy Transfer to transfer the user state of an account on the non-booted OS? Or do I need explore USMT?

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• Sql Query - Selecting rows where user can be both friend and user

  - by Gublooo

  Hey Sorry the title is not very clear. This is a follow up to my earlier question where one of the members helped me with a query. I have a following friends Table Friend friend_id - primary key user_id user_id_friend status The way the table is populated is - when I send a friend request to John - my userID appears in user_id and Johns userID appears in user_id_friend. Now another scenario is say Mike sends me a friend request - in this case mike's userID will appear in user_id and my userID will appear in user_id_friend So to find all my friends - I need to run a query to find where my userID appears in both user_id column as well as user_id_friend column What I am trying to do now is - when I search for user say John - I want all users Johns listed on my site to show up along with the status of whether they are my friend or not and if they are not - then show a "Add Friend" button. Based on the previous post - I got this query which does part of the job - My example user_id is 1: SELECT u.user_id, f.status FROM user u LEFT OUTER JOIN friend f ON f.user_id = u.user_id and f.user_id_friend = 1 where u.name like '%' So this only shows users with whom I am friends where they have sent me request ie my userID appears in user_id_friend. Although I am friends with others (where my userID appears in user_id column) - this query will return that as null To get those I need another query like this SELECT u.user_id, f.status FROM user u LEFT OUTER JOIN friend f ON f.user_id_friend = u.user_id and f.user_id = 1 where u.name like '%' So how do I combine these queries to return 1 set of users and what my friendship status with them is. I hope my question is clear Thanks

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• Why do I get Detached Entity exception when upgrading Spring Boot 1.1.4 to 1.1.5

  - by mmeany

  On updating Spring Boot from 1.1.4 to 1.1.5 a simple web application started generating detached entity exceptions. Specifically, a post authentication inteceptor that bumped number of visits was causing the problem. A quick check of loaded dependencies showed that Spring Data has been updated from 1.6.1 to 1.6.2 and a further check of the change log shows a couple of issues relating to optimistic locking, version fields and JPA issues that have been fixed. Well I am using a version field and it starts out as Null following recommendation to not set in the specification. I have produced a very simple test scenario where I get detached entity exceptions if the version field starts as null or zero. If I create an entity with version 1 however then I do not get these exceptions. Is this expected behaviour or is there still something amiss? Below is the test scenario I have for this condition. In the scenario the service layer that has been annotated @Transactional. Each test case makes multiple calls to the service layer - the tests are working with detached entities as this is the scenario I am working with in the full blown application. The test case comprises four tests: Test 1 - versionNullCausesAnExceptionOnUpdate() In this test the version field in the detached object is Null. This is how I would usually create the object prior to passing to the service. This test fails with a Detached Entity exception. I would have expected this test to pass. If there is a flaw in the test then the rest of the scenario is probably moot. Test 2 - versionZeroCausesExceptionOnUpdate() In this test I have set the version to value Long(0L). This is an edge case test and included because I found reference to Zero values being used for version field in the Spring Data change log. This test fails with a Detached Entity exception. Of interest simply because the following two tests pass leaving this as an anomaly. Test 3 - versionOneDoesNotCausesExceptionOnUpdate() In this test the version field is set to value Long(1L). Not something I would usually do, but considering the notes in the Spring Data change log I decided to give it a go. This test passes. Would not usually set the version field, but this looks like a work-around until I figure out why the first test is failing. Test 4 - versionOneDoesNotCausesExceptionWithMultipleUpdates() Encouraged by the result of test 3 I pushed the scenario a step further and perform multiple updates on the entity that started life with a version of Long(1L). This test passes. Reinforcement that this may be a useable work-around. The entity: package com.mvmlabs.domain; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.GenerationType; import javax.persistence.Id; import javax.persistence.Table; import javax.persistence.Version; @Entity @Table(name="user_details") public class User { @Id @GeneratedValue(strategy=GenerationType.AUTO) private Long id; @Version private Long version; @Column(nullable = false, unique = true) private String username; @Column(nullable = false) private Integer numberOfVisits; public Long getId() { return id; } public void setId(Long id) { this.id = id; } public Long getVersion() { return version; } public void setVersion(Long version) { this.version = version; } public Integer getNumberOfVisits() { return numberOfVisits == null ? 0 : numberOfVisits; } public void setNumberOfVisits(Integer numberOfVisits) { this.numberOfVisits = numberOfVisits; } public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } } The repository: package com.mvmlabs.dao; import org.springframework.data.repository.CrudRepository; import com.mvmlabs.domain.User; public interface UserDao extends CrudRepository<User, Long>{ } The service interface: package com.mvmlabs.service; import com.mvmlabs.domain.User; public interface UserService { User save(User user); User loadUser(Long id); User registerVisit(User user); } The service implementation: package com.mvmlabs.service; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.stereotype.Service; import org.springframework.transaction.annotation.Propagation; import org.springframework.transaction.annotation.Transactional; import org.springframework.transaction.support.TransactionSynchronizationManager; import com.mvmlabs.dao.UserDao; import com.mvmlabs.domain.User; @Service @Transactional(propagation=Propagation.REQUIRED, readOnly=false) public class UserServiceJpaImpl implements UserService { @Autowired private UserDao userDao; @Transactional(readOnly=true) @Override public User loadUser(Long id) { return userDao.findOne(id); } @Override public User registerVisit(User user) { user.setNumberOfVisits(user.getNumberOfVisits() + 1); return userDao.save(user); } @Override public User save(User user) { return userDao.save(user); } } The application class: package com.mvmlabs; import org.springframework.boot.SpringApplication; import org.springframework.boot.autoconfigure.EnableAutoConfiguration; import org.springframework.context.annotation.ComponentScan; import org.springframework.context.annotation.Configuration; @Configuration @ComponentScan @EnableAutoConfiguration public class Application { public static void main(String[] args) { SpringApplication.run(Application.class, args); } } The POM: <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <groupId>com.mvmlabs</groupId> <artifactId>jpa-issue</artifactId> <version>0.0.1-SNAPSHOT</version> <packaging>jar</packaging> <name>spring-boot-jpa-issue</name> <description>JPA Issue between spring boot 1.1.4 and 1.1.5</description> <parent> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-parent</artifactId> <version>1.1.5.RELEASE</version> <relativePath /> <!-- lookup parent from repository --> </parent> <dependencies> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-data-jpa</artifactId> </dependency> <dependency> <groupId>org.hsqldb</groupId> <artifactId>hsqldb</artifactId> <scope>runtime</scope> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-test</artifactId> <scope>test</scope> </dependency> </dependencies> <properties> <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> <start-class>com.mvmlabs.Application</start-class> <java.version>1.7</java.version> </properties> <build> <plugins> <plugin> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-maven-plugin</artifactId> </plugin> </plugins> </build> </project> The application properties: spring.jpa.hibernate.ddl-auto: create spring.jpa.hibernate.naming_strategy: org.hibernate.cfg.ImprovedNamingStrategy spring.jpa.database: HSQL spring.jpa.show-sql: true spring.datasource.url=jdbc:hsqldb:file:./target/testdb spring.datasource.username=sa spring.datasource.password= spring.datasource.driverClassName=org.hsqldb.jdbcDriver The test case: package com.mvmlabs; import org.junit.Assert; import org.junit.Test; import org.junit.runner.RunWith; import org.springframework.beans.factory.annotation.Autowired; import org.springframework.boot.test.SpringApplicationConfiguration; import org.springframework.test.context.junit4.SpringJUnit4ClassRunner; import com.mvmlabs.domain.User; import com.mvmlabs.service.UserService; @RunWith(SpringJUnit4ClassRunner.class) @SpringApplicationConfiguration(classes = Application.class) public class ApplicationTests { @Autowired UserService userService; @Test public void versionNullCausesAnExceptionOnUpdate() throws Exception { User user = new User(); user.setUsername("Version Null"); user.setNumberOfVisits(0); user.setVersion(null); user = userService.save(user); user = userService.registerVisit(user); Assert.assertEquals(new Integer(1), user.getNumberOfVisits()); Assert.assertEquals(new Long(1L), user.getVersion()); } @Test public void versionZeroCausesExceptionOnUpdate() throws Exception { User user = new User(); user.setUsername("Version Zero"); user.setNumberOfVisits(0); user.setVersion(0L); user = userService.save(user); user = userService.registerVisit(user); Assert.assertEquals(new Integer(1), user.getNumberOfVisits()); Assert.assertEquals(new Long(1L), user.getVersion()); } @Test public void versionOneDoesNotCausesExceptionOnUpdate() throws Exception { User user = new User(); user.setUsername("Version One"); user.setNumberOfVisits(0); user.setVersion(1L); user = userService.save(user); user = userService.registerVisit(user); Assert.assertEquals(new Integer(1), user.getNumberOfVisits()); Assert.assertEquals(new Long(2L), user.getVersion()); } @Test public void versionOneDoesNotCausesExceptionWithMultipleUpdates() throws Exception { User user = new User(); user.setUsername("Version One Multiple"); user.setNumberOfVisits(0); user.setVersion(1L); user = userService.save(user); user = userService.registerVisit(user); user = userService.registerVisit(user); user = userService.registerVisit(user); Assert.assertEquals(new Integer(3), user.getNumberOfVisits()); Assert.assertEquals(new Long(4L), user.getVersion()); } } The first two tests fail with detached entity exception. The last two tests pass as expected. Now change Spring Boot version to 1.1.4 and rerun, all tests pass. Are my expectations wrong? Edit: This code saved to GitHub at https://github.com/mmeany/spring-boot-detached-entity-issue

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• Struts2 Hibernate Login with User table and group table

  - by J2ME NewBiew

  My problem is, i have a table User and Table Group (this table use to authorization for user - it mean when user belong to a group like admin, they can login into admincp and other user belong to group member, they just only read and write and can not login into admincp) each user maybe belong to many groups and each group has been contain many users and they have relationship are many to many I use hibernate for persistence storage. and struts 2 to handle business logic. When i want to implement login action from Struts2 how can i get value of group member belong to ? to compare with value i want to know? Example I get user from username and password then get group from user class but i dont know how to get value of group user belong to it mean if user belong to Groupid is 1 and in group table , at column adminpermission is 1, that user can login into admincp, otherwise he can't my code: User.java /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.model; import java.io.Serializable; import java.util.Date; import java.util.HashSet; import java.util.Set; import javax.persistence.CascadeType; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.FetchType; import javax.persistence.GeneratedValue; import javax.persistence.Id; import javax.persistence.JoinColumn; import javax.persistence.JoinTable; import javax.persistence.ManyToMany; import javax.persistence.Table; import javax.persistence.Temporal; /** * * @author Administrator */ @Entity @Table(name="User") public class User implements Serializable{ private static final long serialVersionUID = 2575677114183358003L; private Long userId; private String username; private String password; private String email; private Date DOB; private String address; private String city; private String country; private String avatar; private Set<Group> groups = new HashSet<Group>(0); @Column(name="dob") @Temporal(javax.persistence.TemporalType.DATE) public Date getDOB() { return DOB; } public void setDOB(Date DOB) { this.DOB = DOB; } @Column(name="address") public String getAddress() { return address; } public void setAddress(String address) { this.address = address; } @Column(name="city") public String getCity() { return city; } public void setCity(String city) { this.city = city; } @Column(name="country") public String getCountry() { return country; } public void setCountry(String country) { this.country = country; } @Column(name="email") public String getEmail() { return email; } public void setEmail(String email) { this.email = email; } @ManyToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL) @JoinTable(name="usergroup",joinColumns={@JoinColumn(name="userid")},inverseJoinColumns={@JoinColumn( name="groupid")}) public Set<Group> getGroups() { return groups; } public void setGroups(Set<Group> groups) { this.groups = groups; } @Column(name="password") public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } @Id @GeneratedValue @Column(name="iduser") public Long getUserId() { return userId; } public void setUserId(Long userId) { this.userId = userId; } @Column(name="username") public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } @Column(name="avatar") public String getAvatar() { return avatar; } public void setAvatar(String avatar) { this.avatar = avatar; } } Group.java /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.model; import java.io.Serializable; import javax.persistence.Column; import javax.persistence.Entity; import javax.persistence.GeneratedValue; import javax.persistence.Id; import javax.persistence.Table; /** * * @author Administrator */ @Entity @Table(name="Group") public class Group implements Serializable{ private static final long serialVersionUID = -2722005617166945195L; private Long idgroup; private String groupname; private String adminpermission; private String editpermission; private String modpermission; @Column(name="adminpermission") public String getAdminpermission() { return adminpermission; } public void setAdminpermission(String adminpermission) { this.adminpermission = adminpermission; } @Column(name="editpermission") public String getEditpermission() { return editpermission; } public void setEditpermission(String editpermission) { this.editpermission = editpermission; } @Column(name="groupname") public String getGroupname() { return groupname; } public void setGroupname(String groupname) { this.groupname = groupname; } @Id @GeneratedValue @Column (name="idgroup") public Long getIdgroup() { return idgroup; } public void setIdgroup(Long idgroup) { this.idgroup = idgroup; } @Column(name="modpermission") public String getModpermission() { return modpermission; } public void setModpermission(String modpermission) { this.modpermission = modpermission; } } UserDAO /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.dao; import java.util.List; import org.dejavu.software.model.User; import org.dejavu.software.util.HibernateUtil; import org.hibernate.Query; import org.hibernate.Session; /** * * @author Administrator */ public class UserDAO extends HibernateUtil{ public User addUser(User user){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); session.save(user); session.getTransaction().commit(); return user; } public List<User> getAllUser(){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); List<User> user = null; try { user = session.createQuery("from User").list(); } catch (Exception e) { e.printStackTrace(); session.getTransaction().rollback(); } session.getTransaction().commit(); return user; } public User checkUsernamePassword(String username, String password){ Session session = HibernateUtil.getSessionFactory().getCurrentSession(); session.beginTransaction(); User user = null; try { Query query = session.createQuery("from User where username = :name and password = :password"); query.setString("username", username); query.setString("password", password); user = (User) query.uniqueResult(); } catch (Exception e) { e.printStackTrace(); session.getTransaction().rollback(); } session.getTransaction().commit(); return user; } } AdminLoginAction /* * To change this template, choose Tools | Templates * and open the template in the editor. */ package org.dejavu.software.view; import com.opensymphony.xwork2.ActionSupport; import org.dejavu.software.dao.UserDAO; import org.dejavu.software.model.User; /** * * @author Administrator */ public class AdminLoginAction extends ActionSupport{ private User user; private String username,password; private String role; private UserDAO userDAO; public AdminLoginAction(){ userDAO = new UserDAO(); } @Override public String execute(){ return SUCCESS; } @Override public void validate(){ if(getUsername().length() == 0){ addFieldError("username", "Username is required"); }if(getPassword().length()==0){ addFieldError("password", getText("Password is required")); } } public String getPassword() { return password; } public void setPassword(String password) { this.password = password; } public String getRole() { return role; } public void setRole(String role) { this.role = role; } public User getUser() { return user; } public void setUser(User user) { this.user = user; } public String getUsername() { return username; } public void setUsername(String username) { this.username = username; } } other question. i saw some example about Login, i saw some developers use interceptor, im cant understand why they use it, and what benefit "Interceptor" will be taken for us? Thank You Very Much!

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• Couldn't drop privileges: User is missing UID (see mail_uid setting)

  - by drecute

  I'm hoping I can use some help. I'm configuring dovecot_ldap, but I can't seem to be able to get dovecot to authenticate the ldap user. Below is my config and log info: hosts = 192.168.128.45:3268 dn = cn=Administrator,cn=Users,dc=company,dc=example,dc=com dnpass = "passwd" auth_bind = yes ldap_version = 3 base = dc=company, dc=example, dc=com user_attrs = sAMAccountName=home=/var/vmail/example.com/%$,uid=1001,gid=1001 user_filter = (&(sAMAccountName=%Ln)) pass_filter = (&(ObjectClass=person)(sAMAccountName=%u)) dovecot.conf # 2.0.19: /etc/dovecot/dovecot.conf # OS: Linux 3.2.0-33-generic x86_64 Ubuntu 12.04 LTS auth_mechanisms = plain login auth_realms = example.com auth_verbose = yes disable_plaintext_auth = no mail_access_groups = mail mail_location = mbox:~/mail:INBOX=/var/mail/%u mail_privileged_group = mail passdb { driver = pam } passdb { driver = passwd } passdb { args = /etc/dovecot/dovecot-ldap.conf.ext driver = ldap } passdb { args = scheme=CRYPT username_format=%u /etc/dovecot/users driver = passwd-file } protocols = " imap pop3" service auth { unix_listener /var/spool/postfix/private/auth { group = postfix mode = 0660 user = postfix } } service imap-login { inet_listener imap { port = 143 } inet_listener imaps { port = 993 ssl = yes } } ssl_cert = </etc/ssl/certs/dovecot.pem ssl_key = </etc/ssl/private/dovecot.pem userdb { driver = passwd } userdb { args = /etc/dovecot/dovecot-ldap.conf.ext driver = ldap } userdb { args = username_format=%u /etc/dovecot/users driver = passwd-file } protocol imap { imap_client_workarounds = tb-extra-mailbox-sep imap_logout_format = bytes=%i/%o mail_plugins = } mail.log Nov 29 10:51:44 mail dovecot: auth-worker: pam(charyorde,10.10.1.28): pam_authenticate() failed: Authentication failure (password mismatch?) Nov 29 10:51:44 mail dovecot: auth-worker: passwd(charyorde,10.10.1.28): unknown user Nov 29 10:51:44 mail dovecot: auth: passwd(charyorde,10.10.1.28): unknown user Nov 29 10:51:44 mail dovecot: imap-login: Login: user=<charyorde>, method=PLAIN, rip=10.10.1.28, lip=10.10.1.30, mpid=1892, TLS Nov 29 10:51:44 mail dovecot: imap(charyorde): Error: user charyorde: Couldn't drop privileges: User is missing UID (see mail_uid setting) Nov 29 10:51:44 mail dovecot: imap(charyorde): Error: Internal error occurred. Refer to server log for more information. Nov 29 10:51:46 mail dovecot: auth-worker: pam(charyorde,10.10.1.28): pam_authenticate() failed: Authentication failure (password mismatch?) Nov 29 10:51:46 mail dovecot: auth-worker: passwd(charyorde,10.10.1.28): unknown user Nov 29 10:51:46 mail dovecot: auth: passwd(charyorde,10.10.1.28): unknown user Nov 29 10:51:46 mail dovecot: imap-login: Login: user=<charyorde>, method=PLAIN, rip=10.10.1.28, lip=10.10.1.30, mpid=1894, TLS Nov 29 10:51:46 mail dovecot: imap(charyorde): Error: user charyorde: Couldn't drop privileges: User is missing UID (see mail_uid setting) Nov 29 10:51:46 mail dovecot: imap(charyorde): Error: Internal error occurred. Refer to server log for more information. Nov 29 10:51:48 mail dovecot: auth-worker: pam([email protected],10.10.1.28): pam_authenticate() failed: Authentication failure (password mismatch?) Nov 29 10:51:48 mail dovecot: auth-worker: passwd([email protected],10.10.1.28): unknown user Nov 29 10:51:48 mail dovecot: auth: ldap([email protected],10.10.1.28): unknown user Nov 29 10:51:48 mail dovecot: auth: passwd-file([email protected],10.10.1.28): unknown user Nov 29 10:51:54 mail postfix/smtpd[1880]: idle timeout -- exiting Nov 29 10:51:54 mail postfix/smtpd[1879]: idle timeout -- exiting Nov 29 10:51:54 mail postfix/smtpd[1886]: proxymap stream disconnect Nov 29 10:51:54 mail postfix/smtpd[1887]: proxymap stream disconnect Nov 29 10:51:54 mail postfix/smtpd[1886]: auto_clnt_close: disconnect private/tlsmgr stream Nov 29 10:51:54 mail postfix/smtpd[1887]: auto_clnt_close: disconnect private/tlsmgr stream Nov 29 10:51:54 mail postfix/smtpd[1887]: idle timeout -- exiting Nov 29 10:51:54 mail postfix/smtpd[1886]: idle timeout -- exiting Nov 29 10:51:56 mail dovecot: auth-worker: pam([email protected],10.10.1.28): pam_authenticate() failed: Authentication failure (password mismatch?) Nov 29 10:51:56 mail dovecot: auth-worker: passwd([email protected],10.10.1.28): unknown user Nov 29 10:51:56 mail dovecot: auth: ldap([email protected],10.10.1.28): unknown user Nov 29 10:51:56 mail dovecot: auth: passwd-file([email protected],10.10.1.28): unknown user Nov 29 10:52:04 mail dovecot: auth-worker: pam([email protected],10.10.1.28): pam_authenticate() failed: Authentication failure (password mismatch?) Nov 29 10:52:04 mail dovecot: auth-worker: passwd([email protected],10.10.1.28): unknown user Nov 29 10:52:04 mail dovecot: auth: ldap([email protected],10.10.1.28): unknown user Nov 29 10:52:04 mail dovecot: auth: passwd-file([email protected],10.10.1.28): unknown user Nov 29 10:52:06 mail dovecot: imap-login: Disconnected (auth failed, 3 attempts): user=<[email protected]>, method=PLAIN, rip=10.10.1.28, lip=10.10.1.30, TLS Thank you for looking into this.

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• How to migrate user settings and data to new machine?

  - by torbengb

  I'm new to Ubuntu and recently started using it on my PC. I'm going to replace that PC with a new machine. I want to transfer my data and settings to the nettop. What aspects should I consider? Obviously I want to move my data over. What things am I missing if I only copy the entire home folder? This is a home pc (not corporate) so user rights and other security issues are not a concern, except that the files should be accessible on the new machine! Please take into account that the new machine is a nettop that doesn't have an optical drive and doesn't allow me to hook the old SATA disk into it, so any data transfer must be handled via home network (I can have both the old and the new machine turned on and connected to the home LAN) and I have an USB thumbdrive with limited capacity (2GB). This sounds like it might limit the general applicability, but it would in fact make it more general. I'll make this a wiki topic because there could be several "right" answers. Update: Or so I thought. I don't see a choice for that.

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• Run Flyff without elevating user to Admin or requiring Admin Password

  - by AnonJr

  Bottom Line: I need to set up one game on my little sister's laptop to run without requiring an admin password/account. Its the only game that seems to insist on it... so far. Detailed Version: I set up my 14-year-old sister as a regular user on her Windows 7 Home Premium laptop, and almost everything has been fine - until she found a new game (Flyff) that doesn't seem to want to run without an Admin Password (or being logged in as an Admin). For what should be obvious reasons, I'm not going to make her an Admin. or give her the Admin password (which she swears she'll only use to run this game... anyone else buying that? Bueller?) Also, the parents aren't admins on her laptop (they are on their own, but that's another discussion for another day) and I'm not going to set them up as one as I know from past experience that the 3rd time my sister asks them to put in their password, they'll just tell her what it is - at which point I might as well as have just set her up as an admin from the outset. This is a Win7 Home Premium (64-bit, but I doubt that makes a difference) laptop, so using GPEdit is out. I also tried an answer provided in a related (but less specific) question. The app has read/write permissions for its folder in Program Files (x86), yet that doesn't seem to make a difference. I have not yet dug through the registry as mentioned in another answer to the aforementioned question. Just to be thorough, I have checked the "Run as Admin" option on the shortcut's properties to no avail. Am I missing something? Addendum 2010-11-11: Re-Checked permissions as per Joel's answer, and it didn't make a difference. Followed Jane T's suggestion (and Aeo's second) and created a "Games" folder outside Program Files, installing the game there - and making sure regular users had all the permissions they would need. No joy. After the latter of the above two changes, it occurred to me that it may be a UAC issue, so for kicks I turned off UAC - still the damn message. Last item noted: could it be a result of the publisher not being specified/verified? I've been taking a closer look at the error message and it occurred to me that the missing/unverified publisher info could have been the problem all along... Correct me if I'm wrong, but if that's the case, that means there's nothing I can do short of giving her some sort of Admin privileges (i.e. elevating her account, or giving her the password to a separate Admin account) or giving Mom an Admin account.

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