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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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  • Stopping an specify Apache instance

    - by user1435991
    I have two Apache instances setup in my server (Solaris 10): Instance 1: /etc/apache2 Instance 2: /etc/apache2-instance2 To start the instance 1, I execute the following command: /usr/apache2/bin/apachectl -f /etc/apache2/httpd.conf And instance 2: /usr/apache2/bin/apachectl -f /etc/apache2-instance2/httpd.conf Both instances run perfectly, however the problem comes when I want to stop the instances. I have not been able to find a parameter to indicate what instance I want to stop. if I execute this command: /usr/apache2/bin/apachectl -k stop It will stop always the Instance 1 (the default one). The only solution that I could find to stop the instance 2 was to do this: kill -TERM 'cat /var/run/apache2-instance2/httpd.pid' Is this the only way to do it? or what is the best solution? I remember that I did something similar in Ubuntu setting a the global variable APACHE_CONFDIR before calling apachectl

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  • Cannot connect to a 2008 sql server named instance hosted in a azure virtual machine

    - by emardini
    When I try to connect to a named instance in a SQL SERVER hosted in a azure VM I get this message: A network-related or instance-specific error occurred while establishing a connection to SQL Server. The server was not found or was not accessible. Verify that the instance name is correct and that SQL Server is configured to allow remote connections. (provider: SQL Network Interfaces, error: 26 - Error Locating Server/Instance Specified) (Microsoft SQL Server, Error: -1) The problem is the sql browser is not working properly, when I start the sql browser service it closes after a few seconds and the event log says "There are no instances of SQL Server or SQL Server Analysis Services." But I do have a named instance, I can connect locally to this instance. I've re-installed sql browser and the instance but ii does not work. The host is a azure virtual machine windows server 2008 datacenter. Please help. Thank you

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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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  • Wisdom of merging 100s of Oracle instances into one instance

    - by hoytster
    Our application runs on the web, is mostly an inquiry tool, does some transactions. We host the Oracle database. The app has always had a different instance of Oracle for each customer. A customer is a company which pays us to provide our service to the company's employees, typically 10,000-25,000 employees per customer. We do a major release every few years, and migrating to that new release is challenging: we might have a team at the customer site for a couple weeks, explaining new functionality and setting up the driving data to suit that customer. We're considering going multi-client, putting all our customers into a single shared Oracle 11g instance on a big honkin' Windows Server 2008 server -- in order to reduce costs. I'm wondering if that's advisable. There are some advantages to having separate instances for each customer. Tell me if these are bogus, please. In my rough guess about decreasing importance: Our customers MyCorp and YourCo can be migrated separately when breaking changes are made to the schema. (With multi-client, we'd be migrating 300+ customers overnight!?!) MyCorp's data can be easily backed up and (!!!) restored, without affecting other customers. MyCorp's data is securely separated from their competitor YourCo's data, without depending on developers to get the code right and/or DBAs getting the configuration right. Performance is better because the database is smaller (5,000 vs 2,000,000 rows in ~50 tables). If MyCorp's offices are (mostly) in just one region, then the MyCorp's instance can be geographically co-located there, so network lag doesn't hurt performance. We can provide better service to global clients, for the same reason. In MyCorp wants to take their database in-house, then we can easily export their instance, to get MyCorp their data. Load-balancing is easier because instances can be placed on different servers (this is with a web farm). When a DEV or QA instance is needed, it's easier to clone the real instance and anonymize the data, because there's much less data. Because they're small enough, developers can have their own instance running locally, so they can work on code while waiting at the airport and while in-flight, without fighting VPN hassles. Q1: What are other advantages of separate instances? We are contemplating changing the database schema and merging all of our customers into one Oracle instance, running on one hefty server. Here are advantages of the multi-client instance approach, most important first (my WAG). Please snipe if these are bogus: Less work for the DBAs, since they only need to maintain one instance instead of hundreds. Less DBA work translates to cheaper, our main motive for this change. With just one instance, the DBAs can do a better job of optimizing performance. They'll have time to add appropriate indexes and review our SQL. It will be easier for developers to debug & enhance the application, because there is only one schema and one app (there might be dozens of schema versions if there are hundreds of instances, with a different version of the app for each version of the schema). This reduces costs too. The alternative is having to start every debug session with (1) What version is this customer running and (2) Let's struggle to recreate the corresponding development environment, code and database. (We need a Virtual Machine that includes the code AND database instance for each patch and release!) Licensing Oracle is cheaper because it's priced per server irrespective of heft (or something -- I don't know anything about the subject). The database becomes a viable persistent store for web session data, because there is just one instance. Some database operations are easier with one multi-client instance, like finding a participant when they're hazy about which customer they (or their spouse, maybe) works for: all the names are in one table. Reporting across customers is straightforward. Q2: What are other advantages of having multiple clients in one instance? Q3: Which approach do you think is better (why)? Instance per customer, or all customers in one instance? I'm concerned that having one multi-client instance makes migration near-impossible, and that's a deal killer... ... unless there is a compromise solution like having two multi-client instances, the old and the new. In that case case, we would design cross-instance solutions for finding participants, reporting, etc. so customers could go from one multi-client instance to the next without anything breaking. THANKS SO MUCH for your collective advice! This issue is beyond me -- but not beyond the collective you. :) Hoytster

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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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  • Single Instance of Child Forms in MDI Applications

    - by Akshay Deep Lamba
    In MDI application we can have multiple forms and can work with multiple forms i.e. MDI childs at a time but while developing applications we don't pay attention to the minute details of memory management. Take this as an example, when we develop application say preferably an MDI application, we have multiple child forms inside one parent form. On MDI parent form we would like to have menu strip and tab strip which in turn calls other forms which build the other parts of the application. This also makes our application looks pretty and eye-catching (not much actually). Now on a first go when a user clicks a menu item or a button on a tab strip an application initialize a new instance of a form and shows it to the user inside the MDI parent, if a user again clicks the same button the application creates another new instance for the form and presents it to the user, this will result in the un-necessary usage of the memory. Therefore, if you wish to have your application to prevent generating new instances of the forms then use the below method which will first check if the the form is visible among the list of all the child forms and then compare their types, if the form types matches with the form we are trying to initialize then the form will get activated or we can say it will be bring to front else it will be initialize and set visible to the user in the MDI parent window. The method we are using: private bool CheckForDuplicateForm(Form newForm) { bool bValue = false; foreach (Form frm in this.MdiChildren) { if (frm.GetType() == newForm.GetType()) { frm.Activate(); bValue = true; } } return bValue; } Usage: First we need to initialize the form using the NEW keyword ReportForm ReportForm = new ReportForm(); We can now check if there is another form present in the MDI parent. Here, we will use the above method to check the presence of the form and set the result in a bool variable as our function return bool value. bool frmPresent = CheckForDuplicateForm(Reportfrm); Once the above check is done then depending on the value received from the method we can set our form. if (frmPresent) return; else if (!frmPresent) { Reportfrm.MdiParent = this; Reportfrm.Show(); } In the end this is the code you will have at you menu item or tab strip click: ReportForm Reportfrm = new ReportForm(); bool frmPresent = CheckForDuplicateForm(Reportfrm); if (frmPresent) return; else if (!frmPresent) { Reportfrm.MdiParent = this; Reportfrm.Show(); }

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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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  • 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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