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• LINQ to XML Query Help

  - by cw

  Hello, I am trying to get a "diff" of 2 xml documents and end up with a list of Elements that are different. Below is the XML, I was wondering if anyone can assist. In the case below, I want the list to contain the "file2.xml" element and the "file3.xml" element because they are both different or new than the first xml document. Thanks in advance! <?xml version="1.0" encoding="utf-8" ?> <versioninfo> <files> <file version="1.0">file1.xml</file> <file version="1.0">file2.xml</file> </files> </versioninfo> <?xml version="1.0" encoding="utf-8" ?> <versioninfo> <files> <file version="1.0">file1.xml</file> <file version="1.1">file2.xml</file> <file version="1.0">file3.xml</file> </files> </versioninfo>

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• Why are transactions not rolling back when using SpringJUnit4ClassRunner/MySQL/Spring/Hibernate

  - by Trevor

  I am doing unit testing and I expect that all data committed to the MySQL database will be rolled back... but this isn't the case. The data is being committed, even though my log was showing that the rollback was happening. I've been wrestling with this for a couple days so my setup has changed quite a bit, here's my current setup. LoginDAOTest.java: @RunWith(SpringJUnit4ClassRunner.class) @ContextConfiguration(locations={"file:web/WEB-INF/applicationContext-test.xml", "file:web/WEB-INF/dispatcher-servlet-test.xml"}) @TransactionConfiguration(transactionManager = "transactionManager", defaultRollback = true) public class UserServiceTest { private UserService userService; @Test public void should_return_true_when_user_is_logged_in () throws Exception { String[] usernames = {"a","b","c","d"}; for (String username : usernames) { userService.logUserIn(username); assertThat(userService.isUserLoggedIn(username), is(equalTo(true))); } } ApplicationContext-Text.xml: <?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:p="http://www.springframework.org/schema/p" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:tx="http://www.springframework.org/schema/tx" xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-2.5.xsd http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-2.5.xsd http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-2.5.xsd"> <bean id="dataSource" class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close"> <property name="driverClassName" value="com.mysql.jdbc.Driver"/> <property name="url" value="jdbc:mysql://localhost:3306/test"/> <property name="username" value="root"/> <property name="password" value="Ecosim07"/> </bean> <tx:annotation-driven transaction-manager="transactionManager"/> <bean id="userService" class="Service.UserService"> <property name="userDAO" ref="userDAO"/> </bean> <bean id="userDAO" class="DAO.UserDAO"> <property name="hibernateTemplate" ref="hibernateTemplate"/> </bean> <bean id="sessionFactory" class="org.springframework.orm.hibernate3.LocalSessionFactoryBean"> <property name="dataSource" ref="dataSource"/> <property name="mappingResources"> <list> <value>/himapping/User.hbm.xml</value> <value>/himapping/setup.hbm.xml</value> <value>/himapping/UserHistory.hbm.xml</value> </list> </property> <property name="hibernateProperties"> <props> <prop key="hibernate.dialect">org.hibernate.dialect.SQLServerDialect</prop> <prop key="hibernate.show_sql">true</prop> </props> </property> </bean> <bean id="transactionManager" class="org.springframework.orm.hibernate3.HibernateTransactionManager" p:sessionFactory-ref="sessionFactory"/> <bean id="hibernateTemplate" class="org.springframework.orm.hibernate3.HibernateTemplate"> <property name="sessionFactory"> <ref bean="sessionFactory"/> </property> </bean> </beans> I have been reading about the issue, and I've already checked to ensure that the MySQL database tables are setup to use InnoDB. Also I have been able to successfully implement rolling back of transactions outside of my testing suite. So this must be some sort of incorrect setup on my part. Any help would be greatly appreciated :)

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• create dynamic navigation from xml file in flash cs3

  - by mjr

  action script newbie here :-) What I want to do is this, hand an xml file to a swf and have the swf generate a dynamic text box and button for each of the links in the xml file a rudimentary navigation here's the xml <?xml version="1.0" encoding="UTF-8"?> <page> <page name="Page Name 1" url="/page-1/" /> <page name="Page Name 2" url="/page-2/" /> <page name="Page Name 3" url="/page-3/" /> <page name="Page Name 4" url="/page-4/" /> </page> and in my fla I have a button in my library named 'nav_button' there's a layer named actions and in frame 1 I have this var xml:XML; var xmlList:XMLList; var xmlLoader:URLLoader = new URLLoader(); var button:Button = new Button(); xmlLoader.load(new URLRequest("links.xml")); xmlLoader.addEventListener(Event.COMPLETE, xmlLoaded); function xmlLoaded(event:Event):void { xml = XML(event.target.data); xmlList = xml.children(); trace(xml.length()); for(var i:int = 0; i < xmlList.length(); i++) { button = new Button(); button.x = 25; button.y = i * 50 +25; addChild(button); } } the xml imports fine, but when it comes to the for loop and adding the buttons and text boxes to the stage, I'm toast

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• Spring import runs hibernate persistence twice

  - by Jaanus

  I have 2 spring configurations : spring-servlet.xml spring-security.xml needed to add this line to security: <beans:import resource="spring-servlet.xml"/> Now hibernate is ran twice, this is log screenshot : my web.xml: <servlet> <servlet-name>spring</servlet-name> <servlet-class> org.springframework.web.servlet.DispatcherServlet </servlet-class> <load-on-startup>1</load-on-startup> </servlet> <servlet-mapping> <servlet-name>spring</servlet-name> <url-pattern>/</url-pattern> </servlet-mapping> <listener> <listener-class> org.springframework.web.context.ContextLoaderListener </listener-class> </listener> <context-param> <param-name>contextConfigLocation</param-name> <param-value> /WEB-INF/spring-security.xml </param-value> </context-param>

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• JPA 2.0 Provider Hibernate 3.6 for DB2 v9.5 type 2 driver is throwing exception in configuration prepration

  - by Deep Saurabh

  The JPA 2.0 Provider Hibernate is throwing exception while preparing configuration for entity manager factory, I am using DB2 v9.5 database and DB2 v9.5 JDBC type 2 driver . java.sql.SQLException: [IBM][JDBC Driver] CLI0626E getDatabaseMajorVersion is not supported in this version of DB2 JDBC 2.0 driver. at COM.ibm.db2.jdbc.app.SQLExceptionGenerator.throwNotSupportedByDB2(Unknown Source) at COM.ibm.db2.jdbc.app.DB2DatabaseMetaData.getDatabaseMajorVersion(Unknown Source) at org.hibernate.cfg.SettingsFactory.buildSettings(SettingsFactory.java:117) at org.hibernate.cfg.Configuration.buildSettingsInternal(Configuration.java:2833) at org.hibernate.cfg.Configuration.buildSettings(Configuration.java:2829) at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1840) at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:902) at org.hibernate.ejb.HibernatePersistence.createEntityManagerFactory(HibernatePersistence.java:57) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:48) at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:32)

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• How to populate Java (web) application with initial data using Spring/JPA/Hibernate

  - by Tuukka Mustonen

  I want to setup my database with initial data programmatically. I want to populate my database for development runs, not for testing runs (it's easy). The product is built on top of Spring and JPA/Hibernate. Developer checks out the project Developer runs command/script to setup database with initial data Developer starts application (server) and begins developing/testing then: Developer runs command/script to flush the database and set it up with new initial data because database structures or the initial data bundle were changed What I want is to setup my environment by required parts in order to call my DAOs and insert new objects into database. I do not want to create initial data sets in raw SQL, XML, take dumps of database or whatever. I want to programmatically create objects and persist them in database as I would in normal application logic. One way to accomplish this would be to start up my application normally and run a special servlet that does the initialization. But is that really the way to go? I would love to execute the initial data setup as Maven task and I don't know how to do that if I take the servlet approach. There is somewhat similar question. I took a quick glance at the suggested DBUnit and Unitils. But they seem to be heavily focused in setting up testing environments, which is not what I want here. DBUnit does initial data population, but only using xml/csv fixtures, which is not what I'm after here. Then, Maven has SQL plugin, but I don't want to handle raw SQL. Maven also has Hibernate plugin, but it seems to help only in Hibernate configuration and table schema creation (not in populating db with data). How to do this?

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• Duplicate a collection of entities and persist in Hibernate/JPA

  - by Michael Bavin

  Hi, I want to duplicate a collection of entities in my database. I retreive the collection with: CategoryHistory chNew = new CategoryHistory(); CategoryHistory chLast = (CategoryHistory)em.createQuery("SELECT ch from CategoryHistory ch WHERE ch.date = MAX(date)").getSingleResult; List<Category> categories = chLast.getCategories(); chNew.addCategories(categories)// Should be a copy of the categories: OneToMany Now i want to duplicate a list of 'categories' and persist it with EntityManager. I'm using JPA/Hibernate. UPDATE After knowing how to detach my entities, i need to know what to detach: current code: CategoryHistory chLast = (CategoryHistory)em.createQuery("SELECT ch from CategoryHistory ch WHERE ch.date=(SELECT MAX(date) from CategoryHistory)").getSingleResult(); Set<Category> categories =chLast.getCategories(); //detach org.hibernate.Session session = ((org.hibernate.ejb.EntityManagerImpl) em.getDelegate()).getSession(); session.evict(chLast);//detaches also its child-entities? //set the realations chNew.setCategories(categories); for (Category category : categories) { category.setCategoryHistory(chNew); } //set now create date chNew.setDate(Calendar.getInstance().getTime()); //persist em.persist(chNew); This throws a failed to lazily initialize a collection of role: entities.CategoryHistory.categories, no session or session was closed exception. I think he wants to lazy load the categories again, as i have them detached. What should i do now?

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• Using Hibernate's ScrollableResults to slowly read 90 million records

  - by at

  I simply need to read each row in a table in my MySQL database using Hibernate and write a file based on it. But there are 90 million rows and they are pretty big. So it seemed like the following would be appropriate: ScrollableResults results = session.createQuery("SELECT person FROM Person person") .setReadOnly(true).setCacheable(false).scroll(ScrollMode.FORWARD_ONLY); while (results.next()) storeInFile(results.get()[0]); The problem is the above will try and load all 90 million rows into RAM before moving on to the while loop... and that will kill my memory with OutOfMemoryError: Java heap space exceptions :(. So I guess ScrollableResults isn't what I was looking for? What is the proper way to handle this? I don't mind if this while loop takes days (well I'd love it to not). I guess the only other way to handle this is to use setFirstResult and setMaxResults to iterate through the results and just use regular Hibernate results instead of ScrollableResults. That feels like it will be inefficient though and will start taking a ridiculously long time when I'm calling setFirstResult on the 89 millionth row... UPDATE: setFirstResult/setMaxResults doesn't work, it turns out to take an unusably long time to get to the offsets like I feared. There must be a solution here! Isn't this a pretty standard procedure?? I'm willing to forgo Hibernate and use JDBC or whatever it takes. UPDATE 2: the solution I've come up with which works ok, not great, is basically of the form: select * from person where id > <offset> and <other_conditions> limit 1 Since I have other conditions, even all in an index, it's still not as fast as I'd like it to be... so still open for other suggestions..

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• Hibernate limitations on using variables in queries

  - by sammichy

  I had asked the following question I have the following table structure for a table Player Table Player { Long playerID; Long points; Long rank; } Assuming that the playerID and the points have valid values, can I update the rank for all the players based on the number of points in a single query? If two people have the same number of points, they should tie for the rank. And received the answer from Daniel Vassalo (thank you). UPDATE player JOIN (SELECT p.playerID, IF(@lastPoint <> p.points, @curRank := @curRank + 1, @curRank) AS rank, IF(@lastPoint = p.points, @curRank := @curRank + 1, @curRank), @lastPoint := p.points FROM player p JOIN (SELECT @curRank := 0, @lastPoint := 0) r ORDER BY p.points DESC ) ranks ON (ranks.playerID = player.playerID) SET player.rank = ranks.rank; When I try to execute this as a native query in Hibernate, the following exception is thrown. java.lang.IllegalArgumentException: org.hibernate.QueryException: Space is not allowed after parameter prefix ':' Apparently this has been an open issue for the last couple of years, I want to know if the ranking query can be made to work either Without using any variables in the SQL query OR Using any workaround for Hibernate.

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• Ternary (and n-ary) relationships in Hibernate

  - by Bytecode Ninja

  Q 1) How can we model a ternary relationship using Hibernate? For example, how can we model the ternary relationship presented here using Hibernate (or JPA)? Ideally I prefer my model to be like this: class SaleAssistant { Long id; //... } class Customer { Long id; //... } class Product { Long id; //... } class Sale { SalesAssistant soldBy; Customer buyer; Product product; //... } Q 1.1) How can we model this variation, in which each Sale item might have many Products? class SaleAssistant { Long id; //... } class Customer { Long id; //... } class Product { Long id; //... } class Sale { SalesAssistant soldBy; Customer buyer; Set<Product> products; //... } Q 2) In general, how can we model n-ary, n = 3 relationships with Hibernate? Thanks in advance.

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• How does Linq-to-Xml convert objects to strings?

  - by Eamon Nerbonne

  Linq-to-Xml contains lots of methods that allow you to add arbitrary objects to an xml tree. These objects are converted to strings by some means, but I can't seem to find the specification of how this occurs. The conversion I'm referring to is mentioned (but not specified) in MSDN. I happen to need this for javascript interop, but that doesn't much matter to the question. Linq to Xml isn't just calling .ToString(). Firstly, it'll accept null elements, and secondly, it's doing things no .ToString() implementation does: For example: new XElement("elem",true).ToString() == "<elem>true</elem>" //but... true.ToString() == "True" //IIRC, this is culture invariant, but in any case... true.ToString(CultureInfo.InvariantCulture) == "True" Other basic data types are similarly specially treated. So, does anybody know what it's doing and where that's described?

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• executing stored procedure from Spring-Hibernate using Annotations

  - by HanuAthena

  I'm trying to execute a simple stored procedure with Spring/Hibernate using Annotations. Here are my code snippets: DAO class: public class UserDAO extends HibernateDaoSupport { public List selectUsers(final String eid){ return (List) getHibernateTemplate().execute(new HibernateCallback() { public Object doInHibernate(Session session) throws HibernateException, SQLException { Query q = session.getNamedQuery("SP_APPL_USER"); System.out.println(q); q.setString("eid", eid); return q.list(); } }); } } my entity class: @Entity @Table(name = "APPL_USER") @Inheritance(strategy = InheritanceType.SINGLE_TABLE) @DiscriminatorFormula(value = "SUBSCRIBER_IND") @DiscriminatorValue("N") @NamedQuery(name = "req.all", query = "select n from Requestor n") @org.hibernate.annotations.NamedNativeQuery(name = "SP_APPL_USER", query = "call SP_APPL_USER(?, :eid)", callable = true, readOnly = true, resultClass = Requestor.class) public class Requestor { @Id @Column(name = "EMPL_ID") public String getEmpid() { return empid; } public void setEmpid(String empid) { this.empid = empid; } @Column(name = "EMPL_FRST_NM") public String getFirstname() { return firstname; } ... } public class Test { public static void main(String[] args) { ApplicationContext ctx = new ClassPathXmlApplicationContext( "applicationContext.xml"); APFUser user = (APFUser)ctx.getBean("apfUser"); List selectUsers = user.getUserDAO().selectUsers("EMP456"); System.out.println(selectUsers); } } and the stored procedure: create or replace PROCEDURE SP_APPL_USER (p_cursor out sys_refcursor, eid in varchar2) as empId varchar2(8); fname varchar2(50); lname varchar2(50); begin empId := null; fname := null; lname := null; open p_cursor for select l.EMPL_ID, l.EMPL_FRST_NM, l.EMPL_LST_NM into empId, fname, lname from APPL_USER l where l.EMPL_ID = eid; end; If i enter invalid EID, its returning empty list which is OK. But when record is there, following exception is thrown: Exception in thread "main" org.springframework.jdbc.BadSqlGrammarException: Hibernate operation: could not execute query; bad SQL grammar [call SP_APPL_USER(?, ?)]; nested exception is java.sql.SQLException: Invalid column name Do I need to modify the entity(Requestor.class) ? How will the REFCURSOR be converted to the List? The stored procedure is expected to return more than one record.

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• How to read nested XML using xDocument in Silver light?

  - by Subhen

  Hi currently I have a nested XMl , having the following Structure : <?xml version="1.0" encoding="utf-8" ?> <Response><Result><item id="something" /><price na="something" /> <?xml version="1.0" encoding="UTF-8" ?><DIDL-Lite xmlns="urn:schemas-upnp-org:metadata-1-0/DIDL-Lite/" xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:upnp="urn:schemas-upnp-org:metadata-1-0/upnp/" xmlns:dlna="urn:schemas-dlna-org:metadata-1-0/"></Result><NumberReturned>10</NumberReturned><TotalMatches>10</TotalMatches></Response> Any help on how to read this using Xdocument or XMLReader will be really helpfull. Thanks, Subhendu

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• Setting database-agnostic default column timestamp using Hibernate

  - by unsquared

  I'm working on a java project full of Hibernate (3.3.1) mapping files that have the following sort of declaration for most domain objects. <property name="dateCreated" generated="insert"> <column name="date_created" default="getdate()" /> </property> The problem here is that getdate() is an MSSQL specific function, and when I'm using something like H2 to test subsections of the project, H2 screams that getdate() isn't a recognized function. It's own timestamping function is current_timestamp(). I'd like to be able to keep working with H2 for testing, and wanted to know whether there was a way of telling Hibernate "use this database's own mechanism for retrieving the current timestamp". With H2, I've come up with the following solution. CREATE ALIAS getdate AS $$ java.util.Date now() { return new java.util.Date(); } $$; CALL getdate(); It works, but is obviously H2 specific. I've tried extending H2Dialect and registering the function getdate(), but that doesn't seem to be invoked when Hibernate is creating tables. Is it possible to abstract the idea of a default timestamp away from the specific database engine?

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• Hibernate triggering constraint violations using orphanRemoval

  - by ptomli

  I'm having trouble with a JPA/Hibernate (3.5.3) setup, where I have an entity, an "Account" class, which has a list of child entities, "Contact" instances. I'm trying to be able to add/remove instances of Contact into a List<Contact> property of Account. Adding a new instance into the set and calling saveOrUpdate(account) persists everything lovely. If I then choose to remove the contact from the list and again call saveOrUpdate, the SQL Hibernate seems to produce involves setting the account_id column to null, which violates a database constraint. What am I doing wrong? The code below is clearly a simplified abstract but I think it covers the problem as I'm seeing the same results in different code, which really is about this simple. SQL: CREATE TABLE account ( INT account_id ); CREATE TABLE contact ( INT contact_id, INT account_id REFERENCES account (account_id) ); Java: @Entity class Account { @Id @Column public Long id; @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true) @JoinColumn(name = "account_id") public List<Contact> contacts; } @Entity class Contact { @Id @Column public Long id; @ManyToOne(optional = false) @JoinColumn(name = "account_id", nullable = false) public Account account; } Account account = new Account(); Contact contact = new Contact(); account.contacts.add(contact); saveOrUpdate(account); // some time later, like another servlet request.... account.contacts.remove(contact); saveOrUpdate(account); Result: UPDATE contact SET account_id = null WHERE contact_id = ? Edit #1: It might be that this is actually a bug http://opensource.atlassian.com/projects/hibernate/browse/HHH-5091

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• can i use javabeans with hibernate ?

  - by Dilllllo

  Hello i'm using a plugin of hibernate2 in my webproject with jsp ,in my project i have a register page. Can i use javabeans to send information from a html <form> using hibernate class's ? with out hibernate i creat class with get and set like that package com.java2s; public class Lang { private String choix; private String comm; public String getChoix() { return choix; } public void setChoix(String choix) { this.choix = choix; //System.out.println(choix); } public String getComm() { return comm; } public void setComm(String comm) { this.comm = comm; // System.out.println(comm); } } but i know that hibernate generate a get and set class ! and recive it with that : <jsp:useBean id='user' class='com.java2s.Lang' type='com.java2s.Lang' scope='session' /> <jsp:setProperty name='user' property='*'/> any idea how to do that ?

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• How to prevent Hibernate from nullifying relationship column during entity removal

  - by Grzegorz

  I have two entities, A and B. I need to easily retrieve entities A, joined with entities B on the condition of equal values of some column (some column from A equal to some column in B). Those columns are not primary or foreign keys, they contain same business data. I just need to have access from each instance of A to the collection of B's with the same value of this column. So I model it like this: class A { @OneToMany @JoinColumn(name="column_in_B", referencedColumnName="column_in_A") Collection<B> bs; This way, I can run queries like "select A join fetch a.bs b where b...." (Actually, the real relationship here is many-to-many. But when I use @ManyToMany, Hibernate forces me to use join table, which doesnt exist here. So I have to use @OneToMany as workaround). So far so good. The main problem is: whenever I delete an instance of A, hibernate calls "Update B set column_in_B = null", becuase it thinks the column_in_B is foreign key pointing at primary key in A (and because row in A is deleted, it tries to clean the foreign key in B). BUT the column_in_B IS NOT a foreign key, and can't be modified, because it causes data lost (and this column is NOT NULL anyway in my case, causing data integerity exception to be thrown). Plese help me with this. How to model such relationships with Hibernate? (I would call it "virtual relationships", or "secondary relationships" or so: as they are not based on foreign keys, they are just some shortcuts which allows for retrieving related objects and quering for them with HQL)

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• Applying the Hibernate filter attribute to a Bag with a many-to-many relationship

  - by David P

  Consider the following Hibernate mapping file: <hibernate-mapping ...> <class name="ContentPackage" table="contentPackages"> <id name="Id" column="id" type="int"><generator class="native" /></id> ... <bag name="Clips" table="contentAudVidLinks"> <key column="fk_contentPackageId"></key> <many-to-many class="Clip" column="fk_AudVidId"></many-to-many> <filter name="effectiveDate" condition=":asOfDate BETWEEN startDate and endDate" /> </bag> </class> </hibernate-mapping> When I run the following command: _session.EnableFilter("effectiveDate").SetParameter("asOfDate", DateTime.Today); IList<ContentPackage> items = _session.CreateCriteria(typeof(ContentPackage)) .Add(Restrictions.Eq("Id", id)) .List<ContentPackage>(); The resulting SQL has the WHERE clause on the intermediate mapping table (contentAudVidLinks), rather than the "Clips" table even though I have added the filter attribute to the Bag of Clips. What am I doing wrong?

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• Eager loading OneToMany in Hibernate with JPA2

  - by pihentagy

  I have a simple @OneToMany between Person and Pet entities: @OneToMany(mappedBy="owner", cascade=CascadeType.ALL, fetch=FetchType.EAGER) public Set<Pet> getPets() { return pets; } I would like to load all Persons with associated Pets. So I came up with this (inside a test class): @RunWith(SpringJUnit4ClassRunner.class) @ContextConfiguration public class AppTest { @Test @Rollback(false) @Transactional(readOnly = false) public void testApp() { CriteriaBuilder qb = em.getCriteriaBuilder(); CriteriaQuery<Person> c = qb.createQuery(Person.class); Root<Person> p1 = c.from(Person.class); SetJoin<Person, Pet> join = p1.join(Person_.pets); TypedQuery<Person> q = em.createQuery(c); List<Person> persons = q.getResultList(); for (Person p : persons) { System.out.println(p.getName()); for (Pet pet : p.getPets()) { System.out.println("\t" + pet.getNick()); } } However, turning the SQL logging on shows, that it executes 3 queries (having 2 Persons in the DB). Hibernate: select person0_.id as id0_, person0_.name as name0_, person0_.sex as sex0_ from Person person0_ inner join Pet pets1_ on person0_.id=pets1_.owner_id Hibernate: select pets0_.owner_id as owner3_0_1_, pets0_.id as id1_, pets0_.id as id1_0_, pets0_.nick as nick1_0_, pets0_.owner_id as owner3_1_0_ from Pet pets0_ where pets0_.owner_id=? Hibernate: select pets0_.owner_id as owner3_0_1_, pets0_.id as id1_, pets0_.id as id1_0_, pets0_.nick as nick1_0_, pets0_.owner_id as owner3_1_0_ from Pet pets0_ where pets0_.owner_id=? Any tips? Thanks Gergo

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• Escaping colons in hibernate createSQLQuery

  - by Stratosgear

  I am confused on how I can create an SQL statement containing colons. I am trying to create a view and I am using (notice the double colons): create view MyView as ( SELECT tableA.colA as colA, tableB.colB as colB, round(tableB.colD / 1024)::numeric, 2) as calcValue, FROM tableA, tableB WHERE tableA.colC = 'someValue' ); This is a postgres query and I am forced to use the double colons (::) in order to correctly run the statement. I then pass the above statement through: s.createSQLQuery(myQuery).executeUpdate(); and I get a: Exception in thread "main" org.hibernate.exception.DataException: \ could not execute native bulk manipulation query at org.hibernate.exception.SQLStateConverter.convert(\ SQLStateConverter.java:102) ... more stacktrace... with an output of my above statement changed as (notice the question mark): create view MyView as ( SELECT tableA.colA as colA, tableB.colB as colB, round(tableB.colD / 1024)?, 2) as calcValue, FROM tableA, tableB WHERE tableA.colC = 'someValue' ); Obviously, hibernate confuses my colons with named parameters. Is there a way to escape the colons (a google suggestion that mentions that a single colon is escaped as a double colon does NOT work) or another way of running this statement? Thanks.

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• Avoiding secondary selects or joins with Hibernate Criteria or HQL query

  - by Ben Benson

  I am having trouble optimizing Hibernate queries to avoid performing joins or secondary selects. When a Hibernate query is performed (criteria or hql), such as the following: return getSession().createQuery(("from GiftCard as card where card.recipientNotificationRequested=1").list(); ... and the where clause examines properties that do not require any joins with other tables... but Hibernate still performs a full join with other tables (or secondary selects depending on how I set the fetchMode). The object in question (GiftCard) has a couple ManyToOne associations that I would prefer to be lazily loaded in this case (but not necessarily all cases). I want a solution that I can control what is lazily loaded when I perform the query. Here's what the GiftCard Entity looks like: @Entity @Table(name = "giftCards") public class GiftCard implements Serializable { private static final long serialVersionUID = 1L; private String id_; private User buyer_; private boolean isRecipientNotificationRequested_; @Id public String getId() { return this.id_; } public void setId(String id) { this.id_ = id; } @ManyToOne @JoinColumn(name = "buyerUserId") @NotFound(action = NotFoundAction.IGNORE) public User getBuyer() { return this.buyer_; } public void setBuyer(User buyer) { this.buyer_ = buyer; } @Column(name="isRecipientNotificationRequested", nullable=false, columnDefinition="tinyint") public boolean isRecipientNotificationRequested() { return this.isRecipientNotificationRequested_; } public void setRecipientNotificationRequested(boolean isRecipientNotificationRequested) { this.isRecipientNotificationRequested_ = isRecipientNotificationRequested; } }

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• Hibernate: Querying objects by attributes of inherited classes

  - by MichaelD

  Hi all, I ran into a problem with Hibernate concerning queries on classes which use inheritance. Basically I've the following class hierarchy: @Entity @Table( name = "recording" ) class Recording { ClassA attributeSet; ... } @Entity @Inheritance( strategy = InheritanceType.JOINED ) @Table( name = "classA" ) public class ClassA { String Id; ... } @Entity @Table( name = "ClassB1" ) @PrimaryKeyJoinColumn( name = "Id" ) public class ClassB1 extends ClassA { private Double P1300; private Double P2000; } @Entity @Table( name = "ClassB2" ) @PrimaryKeyJoinColumn( name = "Id" ) public class ClassB2 extends ClassA { private Double P1300; private Double P3000; } The hierarchy is already given like this and I cannot change it easily. As you can see ClassB1 and ClassB2 inherit from ClassA. Both classes contain a set of attributes which sometimes even have the same names (but I can't move them to ClassA since there are possible more sub-classes which do not use them). The Recording class references one instance of one of this classes. Now my question: What I want to do is selecting all Recording objects in my database which refer to an instance of either ClassB1 or ClassB2 with e.g. the field P1300 == 15.5 (so this could be ClassB1 or ClassB2 instances since the P1300 attribute is declared in both classes). What I tried is something like this: Criteria criteria = session.createCriteria(Recording.class); criteria.add( Restrictions.eq( "attributeSet.P1300", new Double(15.5) ) ); criteria.list(); But since P1300 is not an attribute of ClassA hibernate throws an exception telling me: could not resolve property: P1300 of: ClassA How can I tell hibernate that it should search in all subclasses to find the attribute I want to filter? Thanks MichaelD

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• [hibernate - jpa] @OneToOne annotoation problem (i think...)

  - by blow

  Hi all, im new in hibernate and JPA and i have some problems with annotations. My target is to create this table in db (PERSON_TABLE with personal-details) ID ADDRESS NAME SURNAME MUNICIPALITY_ID First of all, i have a MUNICIPALITY table in db containing all municipality of my country. I mapped this table in this ENTITY: @Entity public class Municipality implements Serializable { @Id @GeneratedValue(strategy=GenerationType.IDENTITY) private Long id; private String country; private String province; private String name; @Column(name="cod_catasto") private String codCatastale; private String cap; public Municipality() { } ... Then i make an EMBEDDABLE class Address containing fields that realize a simple address... @Embeddable public class Address implements Serializable { @OneToOne(cascade=CascadeType.ALL) @JoinColumn(name="id_municipality") private Municipality municipality; @Column(length=45) private String address; public Address() { } ... Finally i embedded this class into Person ENTITY @Entity public class Person implements Serializable { @Id @GeneratedValue(strategy=GenerationType.IDENTITY) private Long id; private String name; private String surname; @Embedded private Address address; public Person() { } ... All works good when i have to save a new Person record, in fact hibernate creates a PERSON_TABLE as i want, but if i try to retrieve a Person record i have an exception. HQL is simply "from Person" The excpetion is (Entities is the package containing all classes above-mentioned): org.hibernate.AnnotationException: @OneToOne or @ManyToOne on Entities.Person.address.municipality references an unknown entity: Entities.Municipality Is the @OneToOne annotation the problem? Thanks.

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• jboss cache as hibernate 2nd level - cluster node doesn't persist replicated data

  - by Sergey Grashchenko

  I'm trying to build an architecture basically described in user guide http://www.jboss.org/file-access/default/members/jbosscache/freezone/docs/3.2.1.GA/userguide_en/html/cache_loaders.html#d0e3090 (Replicated caches with each cache having its own store.) but having jboss cache configured as hibernate second level cache. I've read manual for several days and played with the settings but could not achieve the result - the data in memory (jboss cache) gets replicated across the hosts, but it's not persisted in the datasource/database of the target (not original) cluster host. I had a hope that a node might become persistent at eviction, so I've got a cache listener and attached it to @NoveEvicted event. I found that though I could adjust eviction policy to fully control it, no any persistence takes place. Then I had a though that I could try to modify CacheLoader to set "passivate" to true, but I found that in my case (hibernate 2nd level cache) I don't have a way to access a loader. I wonder if replicated data persistence is possible at all by configuration tuning ? If not, will it work for me to create some manual peristence in CacheListener (I could check whether the eviction event is local, and if not - persist it to hibernate datasource somehow) ? I've used mvcc-entity configuration with the modification of cacheMode - set to REPL_ASYNC. I've also played with the eviction policy configuration. Last thing to mention is that I've tested entty persistence and replication in project that has been generated with Seam. I guess it's not important though.

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• Going "behind Hibernate's back" to update foreign key values without an associated entity

  - by Alex Cruise

  Updated: I wound up "solving" the problem by doing the opposite! I now have the entity reference field set as read-only (insertable=false updatable=false), and the foreign key field read-write. This means I need to take special care when saving new entities, but on querying, the entity properties get resolved for me. I have a bidirectional one-to-many association in my domain model, where I'm using JPA annotations and Hibernate as the persistence provider. It's pretty much your bog-standard parent/child configuration, with one difference being that I want to expose the parent's foreign key as a separate property of the child alongside the reference to a parent instance, like so: @Entity public class Child { @Id @GeneratedValue Long id; @Column(name="parent_id", insertable=false, updatable=false) private Long parentId; @ManyToOne(cascade=CascadeType.ALL) @JoinColumn(name="parent_id") private Parent parent; private long timestamp; } @Entity public class Parent { @Id @GeneratedValue Long id; @OrderBy("timestamp") @OneToMany(mappedBy="parent", cascade=CascadeType.ALL, fetch=FetchType.LAZY) private List<Child> children; } This works just fine most of the time, but there are many (legacy) cases when I'd like to put an invalid value in the parent_id column without having to create a bogus Parent first. Unfortunately, Hibernate won't save values assigned to the parentId field due to insertable=false, updatable=false, which it requires when the same column is mapped to multiple properties. Is there any nice way to "go behind Hibernate's back" and sneak values into that field without having to drop down to JDBC or implement an interceptor? Thanks!

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