Search Results

Search found 94030 results on 3762 pages for 'user tips'.

Page 27/3762 | < Previous Page | 23 24 25 26 27 28 29 30 31 32 33 34  | Next Page >

• 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

  Read the article

• 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

  Read the article

• 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!

  Read the article

• 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.

  Read the article

• 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.

  Read the article

• "log on as a batch job" user rights removed by what GPO?

  - by LarsH

  I am not much of a server administrator, but get my feet wet when I have to. Right now I'm running some COTS software on a Windows 2008 Server machine. The software installer creates a few user accounts for running its processes, and then gives those users the right to "log on as a batch job". Every so often (e.g. yesterday at 2:52pm and this morning at 7:50am), those rights disappear. The software then stops working. I can verify that the user rights are gone by using secedit /export /cfg e:\temp\uraExp.inf /areas USER_RIGHTS and I have a script that does this every 30 seconds and logs the results with a timestamp, so I know when the rights disappear. What I see from the export is that in the "good" state, i.e. after I install the software and it's working correctly, the line for SeBatchLogonRight from the secedit export includes the user accounts created by the software. But every few hours (sometimes more), those user accounts are removed from that line. The same thing can be seen by using the GUI tool Local Security Policy > Security Settings > Local Policies > User Rights Assignment > Log on as a batch job: in the "good" state, that policy includes the needed user accounts, and in the bad state, the policy does not. Based on the above-mentioned logging script and the timestamps at which the user rights are being removed, I can see clearly that some GPOs are causing the change. The GPO Operational log shows GPOs being processed at exactly the right times. E.g.: Starting Registry Extension Processing. List of applicable GPOs: (Changes were detected.) Local Group Policy I have run GPOs on demand using gpupdate /force, and was able to verify that this caused the User Rights to be removed. We have looked over local group policies till our eyes are crossed, trying to figure out which one might be stripping these User Rights to "log on as a batch job." We have not configured any local group policies on this machine, that we know of; so is there a default local group policy that might typically do such a thing? Are there typical domain policies that would do this? I have been working with our IT staff colleagues to troubleshoot the problem, but none of them are really GPO experts... They wear many hats, and they do what they need to do in order to keep most things running. Any suggestions would be greatly appreciated!

  Read the article

• Degrading administrative privilege to standard with single admin user account

  - by Vivek S Panicker

  I recently met with a severe issue with user accounts. In my system, there is only administrator user named vivek. I added another user with name vivi and changed its privilege to administrator. After clicked on my username, vivek,and changed its privilege to standard. Since vivek is being the current user, I dropped with all administrator privileges. No password was set for the new administrator user vivi and hence it was disabled by default. I no longer access to any administrative activities. Later I corrected this by editing etc/group file. Isn't this a severe bug? Being the current administrator user, how could I degrade myself to a standard user and got out from administrator's seat? I did not get any warning messages indicating no other administrators exists to manage my system. I suggest this warning should be included there in user accounts when an administrator user changes his privilege without any enabled administrators. Your thoughts?

  Read the article

• Oracle Linux Tips and Tricks: Using SSH

  - by Robert Chase

  Out of all of the utilities available to systems administrators ssh is probably the most useful of them all. Not only does it allow you to log into systems securely, but it can also be used to copy files, tunnel IP traffic and run remote commands on distant servers. It’s truly the Swiss army knife of systems administration. Secure Shell, also known as ssh, was developed in 1995 by Tau Ylonen after the University of Technology in Finland suffered a password sniffing attack. Back then it was common to use tools like rcp, rsh, ftp and telnet to connect to systems and move files across the network. The main problem with these tools is they provide no security and transmitted data in plain text including sensitive login credentials. SSH provides this security by encrypting all traffic transmitted over the wire to protect from password sniffing attacks. One of the more common use cases involving SSH is found when using scp. Secure Copy (scp) transmits data between hosts using SSH and allows you to easily copy all types of files. The syntax for the scp command is: scp /pathlocal/filenamelocal remoteuser@remotehost:/pathremote/filenameremote In the following simple example, I move a file named myfile from the system test1 to the system test2. I am prompted to provide valid user credentials for the remote host before the transfer will proceed.  If I were only using ftp, this information would be unencrypted as it went across the wire.  However, because scp uses SSH, my user credentials and the file and its contents are confidential and remain secure throughout the transfer.  [user1@test1 ~]# scp /home/user1/myfile user1@test2:/home/user1user1@test2's password: myfile                                    100%    0     0.0KB/s   00:00 You can also use ssh to send network traffic and utilize the encryption built into ssh to protect traffic over the wire. This is known as an ssh tunnel. In order to utilize this feature, the server that you intend to connect to (the remote system) must have TCP forwarding enabled within the sshd configuraton. To enable TCP forwarding on the remote system, make sure AllowTCPForwarding is set to yes and enabled in the /etc/ssh/sshd_conf file: AllowTcpForwarding yes Once you have this configured, you can connect to the server and setup a local port which you can direct traffic to that will go over the secure tunnel. The following command will setup a tunnel on port 8989 on your local system. You can then redirect a web browser to use this local port, allowing the traffic to go through the encrypted tunnel to the remote system. It is important to select a local port that is not being used by a service and is not restricted by firewall rules.  In the following example the -D specifies a local dynamic application level port forwarding and the -N specifies not to execute a remote command.   ssh –D 8989 [email protected] -N You can also forward specific ports on both the local and remote host. The following example will setup a port forward on port 8080 and forward it to port 80 on the remote machine. ssh -L 8080:farwebserver.com:80 [email protected] You can even run remote commands via ssh which is quite useful for scripting or remote system administration tasks. The following example shows how to  log in remotely and execute the command ls –la in the home directory of the machine. Because ssh encrypts the traffic, the login credentials and output of the command are completely protected while they travel over the wire. [rchase@test1 ~]$ ssh rchase@test2 'ls -la'rchase@test2's password: total 24drwx------  2 rchase rchase 4096 Sep  6 15:17 .drwxr-xr-x. 3 root   root   4096 Sep  6 15:16 ..-rw-------  1 rchase rchase   12 Sep  6 15:17 .bash_history-rw-r--r--  1 rchase rchase   18 Dec 20  2012 .bash_logout-rw-r--r--  1 rchase rchase  176 Dec 20  2012 .bash_profile-rw-r--r--  1 rchase rchase  124 Dec 20  2012 .bashrc You can execute any command contained in the quotations marks as long as you have permission with the user account that you are using to log in. This can be very powerful and useful for collecting information for reports, remote controlling systems and performing systems administration tasks using shell scripts. To make your shell scripts even more useful and to automate logins you can use ssh keys for running commands remotely and securely without the need to enter a password. You can accomplish this with key based authentication. The first step in setting up key based authentication is to generate a public key for the system that you wish to log in from. In the following example you are generating a ssh key on a test system. In case you are wondering, this key was generated on a test VM that was destroyed after this article. [rchase@test1 .ssh]$ ssh-keygen -t rsaGenerating public/private rsa key pair.Enter file in which to save the key (/home/rchase/.ssh/id_rsa): Enter passphrase (empty for no passphrase): Enter same passphrase again: Your identification has been saved in /home/rchase/.ssh/id_rsa.Your public key has been saved in /home/rchase/.ssh/id_rsa.pub.The key fingerprint is:7a:8e:86:ef:59:70:ef:43:b7:ee:33:03:6e:6f:69:e8 rchase@test1The key's randomart image is:+--[ RSA 2048]----+|                 ||  . .            ||   o .           ||    . o o        ||   o o oS+       ||  +   o.= =      ||   o ..o.+ =     ||    . .+. =      ||     ...Eo       |+-----------------+ Now that you have the key generated on the local system you should to copy it to the target server into a temporary location. The user’s home directory is fine for this. [rchase@test1 .ssh]$ scp id_rsa.pub rchase@test2:/home/rchaserchase@test2's password: id_rsa.pub                  Now that the file has been copied to the server, you need to append it to the authorized_keys file. This should be appended to the end of the file in the event that there are other authorized keys on the system. [rchase@test2 ~]$ cat id_rsa.pub >> .ssh/authorized_keys Once the process is complete you are ready to login. Since you are using key based authentication you are not prompted for a password when logging into the system.   [rchase@test1 ~]$ ssh test2Last login: Fri Sep  6 17:42:02 2013 from test1 This makes it much easier to run remote commands. Here’s an example of the remote command from earlier. With no password it’s almost as if the command ran locally. [rchase@test1 ~]$ ssh test2 'ls -la'total 32drwx------  3 rchase rchase 4096 Sep  6 17:40 .drwxr-xr-x. 3 root   root   4096 Sep  6 15:16 ..-rw-------  1 rchase rchase   12 Sep  6 15:17 .bash_history-rw-r--r--  1 rchase rchase   18 Dec 20  2012 .bash_logout-rw-r--r--  1 rchase rchase  176 Dec 20  2012 .bash_profile-rw-r--r--  1 rchase rchase  124 Dec 20  2012 .bashrc As a security consideration it's important to note the permissions of .ssh and the authorized_keys file.  .ssh should be 700 and authorized_keys should be set to 600.  This prevents unauthorized access to ssh keys from other users on the system.   An even easier way to move keys back and forth is to use ssh-copy-id. Instead of copying the file and appending it manually to the authorized_keys file, ssh-copy-id does both steps at once for you.  Here’s an example of moving the same key using ssh-copy-id.The –i in the example is so that we can specify the path to the id file, which in this case is /home/rchase/.ssh/id_rsa.pub [rchase@test1]$ ssh-copy-id -i /home/rchase/.ssh/id_rsa.pub rchase@test2 One of the last tips that I will cover is the ssh config file. By using the ssh config file you can setup host aliases to make logins to hosts with odd ports or long hostnames much easier and simpler to remember. Here’s an example entry in our .ssh/config file. Host dev1 Hostname somereallylonghostname.somereallylongdomain.com Port 28372 User somereallylongusername12345678 Let’s compare the login process between the two. Which would you want to type and remember? ssh somereallylongusername12345678@ somereallylonghostname.somereallylongdomain.com –p 28372 ssh dev1 I hope you find these tips useful.  There are a number of tools used by system administrators to streamline processes and simplify workflows and whether you are new to Linux or a longtime user, I'm sure you will agree that SSH offers useful features that can be used every day.  Send me your comments and let us know the ways you  use SSH with Linux.  If you have other tools you would like to see covered in a similar post, send in your suggestions.

  Read the article

• Simulating user activities on a GMail page

  - by Vitaliy

  I create a program that simulates me browsing to gmail, entering the user name and password and clicking the submit button. All this with C#. I would appreciate two kinds of answers: One that tells how to do this programaticaly. Since I may be interested in automating more sophisticated user activities. On that tells me about a program that already does that. Thanks!!

  Read the article

• How to detect user agent in WCF web service

  - by Kangkan

  How can I detect the user agent in a web service? My web service is implemented using a WCF webservice with basicHTTPBinding. It will be a post from some SOAP clients. I wish to know the user-agent from the clients. I shall like to see some sample code for this.

  Read the article

• Django: How to get current user in admin forms

  - by lazerscience

  In Django's ModelAdmin I need to display forms customized according to the permissions an user has. Is there a way of getting the current user object into the form class, so that i can customize the form in its __init__ method? I think saving the current request in a thread local would be a possibility but this would be my last resort think I'm thinking it is a bad design approach....

  Read the article

• Best practices for custom http user-agent strings?

  - by Noufal Ibrahim

  I'm developing an application that communicates with an internal web service using HTTP. Are there any "best practices" for custom user-agent strings so that I can put a nice one in my app? It's a Python library and the lower transport is Python's own httplib. Should the user-agent string say that or something else?

  Read the article

• How do you test the usability of your user interfaces

  - by Martin

  How do you test the usability of the user interfaces of your applications - be they web or desktop? Do you just throw it all together and then tweak it based on user experience once the application is live? Or do you pass it to a specific usability team for testing prior to release? We are a small software house, but I am interested in the best practices of how to measure usability. Any help appreciated.

  Read the article

• Django microblog showing a logged in user only his posts

  - by dana

  i have a miniblog application, with a class named New(refering to a new post), having a foreign key to an user(who has posted the entry). above i have a method that displays all the posts from all the users. I'd like to show to the logged in user, only his posts How can i do it? Thanks in advance! def paginate(request): paginator = New.objects.all() return render_to_response('news/newform.html', { 'object_list': paginator, }, context_instance=RequestContext(request))

  Read the article

• Writing user stories for internal technical tasks

  - by John Nolan

  I am attempting to manage my projects a little better so I am looking at attempting to apply some of (eventually all) the features of scrum. Looking at user stories specifically the high level format seems to be: As a User I can Feature Description or Artifact is Doing Something How would I write "Upgrade the Database"? Is it simply Upgrade the Database? I think I am being thrown off as there is no specific actor/customer and that the customer is the IT department.

  Read the article

• how to get the 'user' object in django..

  - by zjm1126

  i want to get the user who login now .. how to do this .. thanks this is my code : ps=Position.objects.filter(name=User.username)

  Read the article

• User accounts in Symfony?

  - by gruner

  I'm new to Symfony. Is my understanding correct that the User class is actually for controlling sessions? But is there built-in login and account creation? I'm not finding it. But if there's an admin backend generator, how can it function without user logins?

  Read the article

• problem in batch script read user input

  - by JCH

  hi, i use set /p below to read user input it seems to work outside the if block but the one inside if block doesn't work. When i run the script second time the user input in the if block prints the previous user input. test script: @echo off set cond=true echo %cond% if %cond%==true ( echo "cond is true" REM the below input doesn't work set /p name1="enter your name" echo name is: %name1% ) REM it works here set /p name2="enter your name" echo name is: %name2% thank you

  Read the article

• drupal user login redirect

  - by Arun

  hi i wrote a module for user page generation. After login the path goes to users/[user-name] and i want to customize it using my module. how to handle it?

  Read the article

• JavaScript Trigger window.open When a User Clicks Anywhere on Page

  - by Joe Hammonds

  I have a problem that I have been trying to figure out and haven't been able to get past it because Chrome/FireFox/IE do not publicly publish their "rules" for pop up blocking when it comes to JavaScript, Flash, etc. I am trying to trigger a window.open() when a user clicks anywhere on page. I've tried this: document.onclick = window.open("http://msn.com"); But all 3 browsers are blocking the popup, even though it is a user interaction.

  Read the article

• User input without waiting for enter

  - by Hermann Ingjaldsson

  I am trying to make an interactive shell script in perl. The only user input I can find is the following: $name = <STDIN>; print STDOUT "Hello $name\n"; But in this the user must always press enter for the changes to take effect. How can I get the program to proceed immediately after a button has been pressed?

  Read the article

• Is it possible to have a local group for an LDAP user

  - by fakedrake

  I have an LDAP server to which i do not have full privileges and an ubuntu system with LDAP authentication to which i am root. Is it possible to add an LDAP user to a local group? (i dont know if i phrase this correctly but all i want is to have a user in LDAP in a group without edititing the actual database)

  Read the article

• ASP.NET/AJAX - Web user control update panel triggering from outside

  - by Julian

  Hi, I've got an web user control with an updatepanel. On my mainpage I got like 3 of those controls. Now I want to have an timer in the mainpage which would trigger the updatepanels in the web user controls. How can I manage this? Thanks in advance.

  Read the article

• Using a user-defined type as a primary key

  - by Chris Kaminski

  Suppose I have a system where I have metadata such as: table: ====== key name address ... Then suppose I have a user-defined type described as so: datasource datasource-key A) are there systems where it's possible to have keys based on user-defined types? B) if so, how do you decompose the keys into a form suitable for querying? C) is this a case where I'm just better off with a composite primary key?

  Read the article

• programatically add user permission to a list in sharepoint

  - by Adyt

  how do i programatically add user permission to a list in sharepoint. i want to add the permission "Contribute" to a user or group for a certain list. im using c# Thanks..

  Read the article

< Previous Page | 23 24 25 26 27 28 29 30 31 32 33 34  | Next Page >