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  • How to use JTA support in Tomcat 6 for Hibernate ?

    - by EugeneP
    They recommend using JTA transaction support in JEE environment. But how to configure JTA in Tomcat6 so that Hibernate Session could use it ? Starting with version 3.0.1, Hibernate added the SessionFactory.getCurrentSession() method. Initially, this assumed usage of JTA transactions, where the JTA transaction defined both the scope and context of a current session. Given the maturity of the numerous stand-alone JTA TransactionManager implementations, most, if not all, applications should be using JTA transaction management, whether or not they are deployed into a J2EE container. Based on that, the JTA-based contextual sessions are all you need to use.

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  • "SELECT ... FOR UPDATE" not working for Hibernate and MySQL

    - by Andres Rodriguez
    Hi, We have a system in which we must use pessimistic locking in one entity. We are using hibernate, so we use LockMode.UPGRADE. However, it does not lock. The tables are InnoDB We have checked that locking works correctly in the database (5.0.32), so this bug http://bugs.mysql.com/bug.php?id=18184 seems to be no problem. We have checked that datasource includes the autoCommit = false parameter. We have checked that the SQL hibernate (version 3.2) generates includes the " FOR UPDATE". Thanks,

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  • How to user Hibernate @Valid constraint with Spring 3.x?

    - by Burak Dede
    I am working on simple form to validate fields like this one. public class Contact { @NotNull @Max(64) @Size(max=64) private String name; @NotNull @Email @Size(min=4) private String mail; @NotNull @Size(max=300) private String text; } I provide getter and setters hibernate dependencies on my classpath also.But i still do not get the how to validate simple form there is actually not so much documentation for spring hibernate combination. @RequestMapping(value = "/contact", method = RequestMethod.POST) public String add(@Valid Contact contact, BindingResult result) { .... } Could you explain it or give some tutorial , except original spring 3.x documentation

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  • Hibernate: How do I link a subclass to its superclass?

    - by Markus
    Hey there! I'm having a little problem setting up my webshop project. Thing is, I have a User() superclass and two subclasses, PrivateUser and BusinessUser. Now, I'm not quite sure how to get my head around storing this relationship via hibernate. For the purpose of this question, the User() class contains only one field: String address; the PrivateUser contains: String firstName; and the BusinessUser contains: String CompanyName; Each field has its getter and setter. As is right now, I would only store and be able to get firstName and companyName. When I fetch a user from my DB using Hibernate I would get a PrivateUser/BusinessUser with a null address. Bottom line is, could someone point me towards a useful tutorial or better yet show a similar example code? Thanks!

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  • How do I share a Hibernate SessionFactory across web applications?

    - by Jer
    I have two web applications that are running on a single Tomcat server and are connected to the same database with Hibernate. I am concerned that having two SessionFactory instances running around might cause some issues. Also, since both web applications share much of the same application logic, I thought it would be a good idea to centralize as much as I could. And since I use Spring for DI and Hibernate configuration it would make sense to have a single ApplicationContext as well. How would I go about doing something like this? Do I need to deploy a headless WAR that creates an ApplicationContext and thus a SessionFactory and allow each application access to it? Is this even a good idea?

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  • How to detect open database connection with Hibernate / JPA?

    - by John K
    I am learning JPA w/Hibernate using a Java SE 6 project. I'd simply like to be able to detect if the connection between Hibernate and my database (MS SQL Server) is open. For example, I'd like to be able to detect this, log it, and try reconnecting again in 60 seconds. This is what I thought would work but isOpen() doesn't appear to be what I want (always is true): EntityManagerFactory emf = Persistence.createEntityManagerFactory("rcc", props); if (emf != null && emf.isOpen()) { EntityManager em = emf.createEntityManager(); if (em == null || !emf.isOpen()) // error connecting to database else ... This seems to me to be a simple problem, but I cannot find an answer!

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  • Hibernate constraint ConstraintViolationException. Is there an easy way to ignore duplicate entries?

    - by vincent
    Basically I've got the below schema and I'm inserting records if they don't exists. However when it comes to inserting a duplicate it throws and error as I would expect. My question is whether there is an easy way to make Hibernate to just ignore inserts which would in effect insert duplicates? CREATE TABLE IF NOT EXISTS `method` ( `id` bigint(20) NOT NULL AUTO_INCREMENT, `name` varchar(10) DEFAULT NULL, PRIMARY KEY (`id`), UNIQUE KEY `name` (`name`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=2 ; SEVERE: Duplicate entry 'GET' for key 'name' Exception in thread "pool-11-thread-4" org.hibernate.exception.ConstraintViolationException: could not insert:

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  • Combine Hibernate class with @Bindable for SwingBuilder without Griffon?

    - by Misha Koshelev
    Dear All: I have implemented a back-end for my application in Groovy/Gradle, and am now trying to implement a GUI. I am using Hibernate for my data storage (with HSQLDB) per http://groovy.codehaus.org/Using+Hibernate+with+Groovy (with Jasypt for encryption) and it is working quite well. I was wondering if there are any good tips for using @Bindable with, e.g., an @Entity class such as @Entity class Book { @Id @GeneratedValue(strategy = GenerationType.AUTO) public Long id @OneToMany(cascade=CascadeType.ALL) public Set<Author> authors public String title String toString() { "$title by ${authors.name.join(', ')}" } } or if I am: (i) asking for Griffon (ii) completely on the wrong track? Thank you! Misha

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  • Is there a way to combine streaming data retrieval with hibernate?

    - by Steve B.
    For the purposes of handling very large collections (and by very large I just mean "likely to throw OutOfMemory exception"), it seems problematic to use Hibernate because normally collection retrieval is done in a block, i.e. List values=session.createQuery("from X").list(), where you monolithically grab all N-million values and then process them. What I'd prefer to do is to retrieve the values as an iterator so that I grab 1000 or so (or whatever's a reasonable page size) at a time. Apart from writing my own iteration (which seems like it's likely to be re-inventing the wheel) is there a hibernate-native way to handle this?

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  • How to map a property for HQL usage only (in Hibernate)?

    - by ManBugra
    i have a table like this one: id | name | score mapped to a POJO via XML with Hibernate. The score column i only need in oder by - clauses in HQL. The value for the score column is calculated by an algorithm and updated every 24 hours via SQL batch process (JDBC). So i dont wanna pollute my POJO with properties i dont need at runtime. For a single column that may be not a problem, but i have several different score columns. Is there a way to map a property for HQL use only? For example like this: <property name="score" type="double" ignore="true"/> so that i still can do this: from Pojo p order by p.score but my POJO implementation can look like this: public class Pojo { private long id; private String name; // ... } No Setter for score provided or property added to implementation. using the latest Hibernate version for Java.

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  • What should I do in order to be able to work with maven + eclipse + wicket + hibernate + spring in M

    - by spderosso
    I want to create a web app that will use wicket, hibernate and spring frameworks. My IDE of choice is Eclipse, I am using maven for the .war generation and I am running Mac OS. What steps should I follow to correctly install and configure all the tools so as to have a project running that relies on these 3 frameworks. I was able to successfully set up wicket but I am having trouble for setting up hibernate and spring. I went through multiple tutorials but I still couldn't find the solution. Thanks!

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  • How does hibernate use an empty string for an equality restriction?

    - by Stephen
    I have a column that potentially has some bad data and I can't clean it up, so I need to check for either null or empty string. I'm doing a Hibernate Criteria query so I've got the following that returns incorrectly right now: Session session = getSessionFactory().openSession(); Transaction tx = session.beginTransaction(); Criteria myCriteria = session.createCriteria(Object); ... myCriteria.add(Restrictions.or(Restrictions.isNull("stringColumn"), Restrictions.eq("stringColumn", ""))); List<Objects> list = myCriteria.list(); I can't get it to properly return the results I'd expect. So as an experiment I changed the second restriction to read: Restrictions.eq("stringColumn", "''") And it started returning the expected results, so is hibernate incorrectly translating my empty string (e.g. "") into a SQL empty string (e.g. ''), or am I just doing this wrong?

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  • How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way?

    - by EugeneP
    How to deal with the Hibernate hql multi-join query result in an Object-Oriented Way? As I see it returns a list of Objects. yes, it is tricky and only you who write the query know what should the query return (what objects). But are there ways to simplify things, so that it returned specific objects with no need in casting Object to a specific class according to its position in the query ? Maybe Spring can simplify things here? It has the similar functionality for JDBC, but I don't see if it can help in a similar way with Hibernate.

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  • out-of-the-box way to get an idmap from hibernate for a given entity?

    - by Geert-Jan
    Over and over again I notive myself getting a list from hibernate, and the first thing next is put it in an idmap like: List<House> entities = s.createCriteria(House.class).list(); Map<String,House> entitymap = new HashMap<String,House>(); for(TA_entity e:entities){ entitymap.put(e.getId(), e); } Is there a way to get this directly out of hibenerate? afterall Hibernate is familiar with the id.

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  • A typical lifecycle of a Hibernate object in a web app - ?

    - by EugeneP
    Describe please a typical lifecycle of a Hibernate object (that maps to a db table) in a web app. Suppose, you create a new instance of an object and persist in the db. But during the app lifetime you'll be working on a detached object and finally you need to update it in the database, for example on exit. How does it look like with hibernate and spring? p.s. Can transactions and sessions live between servlet transitions? So that we opened 1 session and use it in all servlets without a need to reopen it?

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  • Windows 7 unidentified network (or limited access) after hibernate.

    - by null
    My windows 7 network will show limited access or unidentified network after coming up from hibernation. In the office I normally use LAN connection, I turn-off my wireless card (DELL Latitude has on/off switch for the wireless card). When I back at home I will turn on the wireless card, but it will take about 15 seconds to detect my home WIFI and then show limited access. I will have to restart the notebook and it will be able to connect to my WIFI and internet. The problem will be solved if I restart the notebook, but that defeats the purpose of hibernation doesn't it? I have tried uninstalling the wireless card driver but still does not solve it. I also tried updating my network card driver but windows says I am using the latest driver. On support.dell.com also showing I am using the latest driver.

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  • Delete all previous records and insert new ones

    - by carlos
    When updating an employee with id = 1 for example, what is the best way to delete all previous records in the table certificate for this employee_id and insert the new ones?. create table EMPLOYEE ( id INT NOT NULL auto_increment, first_name VARCHAR(20) default NULL, last_name VARCHAR(20) default NULL, salary INT default NULL, PRIMARY KEY (id) ); create table CERTIFICATE ( id INT NOT NULL auto_increment, certificate_name VARCHAR(30) default NULL, employee_id INT default NULL, PRIMARY KEY (id) ); Hibernate mapping <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD//EN" "http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"> <hibernate-mapping> <class name="Employee" table="EMPLOYEE"> <id name="id" type="int" column="id"> <generator class="sequence"> <param name="sequence">employee_seq</param> </generator> </id> <set name="certificates" lazy="false" cascade="all"> <key column="employee_id" not-null="true"/> <one-to-many class="Certificate"/> </set> <property name="firstName" column="first_name"/> <property name="lastName" column="last_name"/> <property name="salary" column="salary"/> </class> <class name="Certificate" table="CERTIFICATE"> <id name="id" type="int" column="id"> <param name="sequence">certificate_seq</param> </id> <property name="employee_id" column="employee_id" insert="false" update="false"/> <property name="name" column="certificate_name"/> </class> </hibernate-mapping>

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  • How to convert an ORM to its subclass using Hibernate ?

    - by Gaaston
    Hi everybody, For example, I have two classes : Person and Employee (Employee is a subclass of Person). Person : has a lastname and a firstname. Employee : has also a salary. On the client-side, I have a single HTML form where i can fill the person informations (like lastname and firstname). I also have a "switch" between "Person" and "Employee", and if the switch is on Employee I can fill the salary field. On the server-side, Servlets receive informations from the client and use the Hibernate framework to create/update data to/from the database. The mapping i'm using is a single table for persons and employee, with a discriminator. I don't know how to convert a Person in an Employee. I firstly tried to : load the Person p from the database create an empty Employee e object copy values from p into e set the salary value save e into the database But i couldn't, as I also copy the ID, and so Hibernate told me they where two instanciated ORM with the same id. And I can't cast a Person into an Employee directly, as Person is Employee's superclass. There seems to be a dirty way : delete the person, and create an employee with the same informations, but I don't really like it.. So I'd appreciate any help on that :) Some precisions : The person class : public class Person { protected int id; protected String firstName; protected String lastName; // usual getters and setters } The employee class : public class Employee extends Person { // string for now protected String salary; // usual getters and setters } And in the servlet : // type is the "switch" if(request.getParameter("type").equals("Employee")) { Employee employee = daoPerson.getEmployee(Integer.valueOf(request.getParameter("ID"))); modifyPerson(employee, request); employee.setSalary(request.getParameter("salary")); daoPerson.save(employee ); } else { Person person = daoPerson.getPerson(Integer.valueOf(request.getParameter("ID"))); modifyPerson(employee, request); daoPerson.save(person); } And finally, the loading (in the dao) : public Contact getPerson(int ID){ Session session = HibernateSessionFactory.getSession(); Person p = (Person) session.load(Person.class, new Integer(ID)); return p; } public Contact getEmployee(int ID){ Session session = HibernateSessionFactory.getSession(); Employee = (Employee) session.load(Employee.class, new Integer(ID)); return p; } With this, i'm getting a ClassCastException when trying to load a Person using getEmployee. XML Hibernate mapping : <class name="domain.Person" table="PERSON" discriminator-value="P"> <id name="id" type="int"> <column name="ID" /> <generator class="native" /> </id> <discriminator column="type" type="character"/> <property name="firstName" type="java.lang.String"> <column name="FIRSTNAME" /> </property> <property name="lastName" type="java.lang.String"> <column name="LASTNAME" /> </property> <subclass name="domain.Employee" discriminator-value="E"> <property name="salary" column="SALARY" type="java.lang.String" /> </subclass> </class> Is it clear enough ? :-/

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  • How to handle large dataset with JPA (or at least with Hibernate)?

    - by Roman
    I need to make my web-app work with really huge datasets. At the moment I get either OutOfMemoryException or output which is being generated 1-2 minutes. Let's put it simple and suppose that we have 2 tables in DB: Worker and WorkLog with about 1000 rows in the first one and 10 000 000 rows in the second one. Latter table has several fields including 'workerId' and 'hoursWorked' fields among others. What we need is: count total hours worked by each user; list of work periods for each user. The most straightforward approach (IMO) for each task in plain SQL is: 1) select Worker.name, sum(hoursWorked) from Worker, WorkLog where Worker.id = WorkLog.workerId group by Worker.name; //results of this query should be transformed to Multimap<Worker, Long> 2) select Worker.name, WorkLog.start, WorkLog.hoursWorked from Worker, WorkLog where Worker.id = WorkLog.workerId; //results of this query should be transformed to Multimap<Worker, Period> //if it was JDBC then it would be vitally //to set resultSet.setFetchSize (someSmallNumber), ~100 So, I have two questions: how to implement each of my approaches with JPA (or at least with Hibernate); how would you handle this problem (with JPA or Hibernate of course)?

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  • How can I go through a Set in JSP? (Hibernate Associations)

    - by Parris
    Hi All, So I am pretty new to JSP. I have tried this a few ways. Ways that would make sense in PHP or automagicy frameworks... I am probably thinking too much in fact... I have a hibernate one to many association. That is class x has many of class y. In class x's view.jsp. I would like to grab all of class y, where the foreign key of y matches the primary key of x and display them. It seems that hibernate properly puts this stuff into a set. Now, the question is how can I iterate through this set and then output it's contents... I am kind of stumped here. I tried to write a scriptlet, <% java.util.Iterator iter = aBean.getYs().iter(); // aBeans is the bean name // getYs would return the set and iter would return an iterator for the set while(iter.hasNext) { model.X a = new iter.next() %> <h1><%=a.getTitle()%></h1> <% } %> It would seem that that sort of thing should work? Hmmmmmm

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  • JSP/Struts2/Hibernate: loop through a self-referencing table.

    - by TBW
    Hello everyone, Let's say we have a self-referencing table called PERSON, with the following columns: ID, PARENT, where PARENT is a foreign key to the ID column of another element in the PERSON table. Of course, many persons can have the same parent. I use Hibernate 3 in lazy fetching mode to deal with the database. Hibernate fetches a person element from the database, which is then put in the ValueStack by the Struts2 action, to be used on the result JSP page. Now the question is : In JSP, how can I do to display all the child (and the child's child, and so on, like a family tree) of this person element? Of course, for the n+1 children I can use the < s:iterator tag over the person.person. I can also nest another < s:iterator tag over person.person.person to get the n+2 children. But what if I want to do this in an automated manner, up to the last n+p child, displaying in the process all the children of all the n+1..n+p elements? I hope I have been clear enough. Thank you all for your time. -- TBW.

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  • Hibernate saveOrUpdate fails when I execute it on empty table.

    - by Vladimir
    I'm try to insert or update db record with following code: Category category = new Category(); category.setName('catName'); category.setId(1L); categoryDao.saveOrUpdate(category); When there is a category with id=1 already in database everything works. But if there is no record with id=1 I got following exception: org.hibernate.StaleStateException: Batch update returned unexpected row count from update [0]; actual row count: 0; expected: 1: Here is my Category class setters, getters and constructors ommited for clarity: @Entity public class Category { @Id @GeneratedValue(strategy = GenerationType.AUTO) private Long id; private String name; @ManyToOne private Category parent; @OneToMany(fetch = FetchType.LAZY, mappedBy = "parent") private List<Category> categories = new ArrayList<Category>(); } In the console I see this hibernate query: update Category set name=?, parent_id=? where id=? So looks like hibernates tryis to update record instead of inserting new. What am I doing wrong here?

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  • Ordering the results of a Hibernate Criteria query by using information of the child entities of the

    - by pkainulainen
    I have got two entities Person and Book. Only one instance of a specific book is stored to the system (When a book is added, application checks if that book is already found before adding a new row to the database). Relevant source code of the entities is can be found below: @Entity @Table(name="persons") @SequenceGenerator(name="id_sequence", sequenceName="hibernate_sequence") public class Person extends BaseModel { @Id @Column(name = "id") @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_sequence") private Long id = null; @ManyToMany(targetEntity=Book.class) @JoinTable(name="persons_books", joinColumns = @JoinColumn( name="person_id"), inverseJoinColumns = @JoinColumn( name="book_id")) private List<Book> ownedBooks = new ArrayList<Book>(); } @Entity @Table(name="books") @SequenceGenerator(name="id_sequence", sequenceName="hibernate_sequence") public class Book extends BaseModel { @Id @Column(name = "id") @GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "id_sequence") private Long id = null; @Column(name="name") private String name = null; } My problem is that I want to find persons, which are owning some of the books owned by a specific persons. The returned list of persons should be ordered by using following logic: The person owning most of the same books should be at the first of the list, second person of the the list does not own as many books as the first person, but more than the third person. The code of the method performing this query is added below: @Override public List<Person> searchPersonsWithSimilarBooks(Long[] bookIds) { Criteria similarPersonCriteria = this.getSession().createCriteria(Person.class); similarPersonCriteria.add(Restrictions.in("ownedBooks.id", bookIds)); //How to set the ordering? similarPersonCriteria.addOrder(null); return similarPersonCriteria.list(); } My question is that can this be done by using Hibernate? And if so, how it can be done? I know I could implement a Comparator, but I would prefer using Hibernate to solve this problem.

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