hibernate-mapping

Hibernate 2nd level cache objects that are lazy=false, result in a default fetch=join, is it documented anywhere?

半城伤御伤魂 提交于 2020-01-01 05:27:09
问题 I experience the following apparently undocumented issue, and I want to understand if I did something wrong Did anyone encounter the same issue? Is it really not documented anywhere? or did I miss something? The behavior is this Assume the following mapping <class name="org.sample.Foo" table="foo"> ... <many-to-one name="bar" class="org.sample.Bar"/> </class> <class name="org.sample.Bar" table="bar" lazy="false"> ... </class> First, as a background, Hibernate default value for the fetch

Hibernate CriteriaBuilder to join multiple tables

邮差的信 提交于 2019-12-31 20:15:54
问题 I'm trying to join 4 tables using hibernate criteriabuilder.. Below are the tables respectively.. ` @Entity public class BuildDetails { @Id private long id; @Column private String buildNumber; @Column private String buildDuration; @Column private String projectName; } @Entity public class CodeQualityDetails{ @Id private long id; @Column private String codeHealth; @ManyToOne private BuildDetails build; //columnName=buildNum } @Entity public class DeploymentDetails{ @Id private Long id; @Column

Doesn't work setting default value of property in Hibernate

若如初见. 提交于 2019-12-31 20:10:53
问题 I have a boolean property in my entity. Here's my annotations for it: @Column(name = "IS_ACTIVE", nullable = false, columnDefinition="BIT DEFAULT 1", length = 1) public Boolean getActive() { return isActive; } But columnDefinition="BIT DEFAULT 1" doen't work perfectly. Here's SQL code I get as result for generated table: IS_ACTIVE BIT(1) NOT NULL, What am I doing wrong? And so when I try to save an instance of this class to the database I get the exception: `com.mysql.jdbc.exceptions.jdbc4

How to implement IdentifierGenerator with PREFIX and separate Sequence for each entity

删除回忆录丶 提交于 2019-12-31 01:51:07
问题 I have followed basic Vlad's guide to implement prefixed generator, but haveing trouldles. The goal is to simultaneously have: Separate sequence per each entity Prefix which I define via annotation ( @GenericGenerator I guess) All sequences are dropped and created on each app restart (redeploy, whatever..) with hibernate.hbm2ddl.auto=create . Being able to set the prefix on PACKAGE level not only on class / id-field How to implement those conditions simultaneously ? THE (PARTIAL!) SOLUTION I

mapping multiple sets in one table in hibernate

为君一笑 提交于 2019-12-30 17:22:47
问题 I've got a User an a Log class (which I cannot change): class User { private long id; private String name; private Set<Log> accessLogs; private Set<Log> actionLogs; } class Log { private String what; private Date when; } A possible mapping will look like: <class name="com.example.User" table="users"> <id name="id" access="field"> <generator class="native" /> </id> <property name="name" length="256" /> <set name="accessLogs" table="user_access_logs" cascade="all-delete-orphan" order-by="`when`

org.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.entity.PojoEntityTuplizer]

假装没事ソ 提交于 2019-12-30 09:32:49
问题 I am developing a web application using Hibernate framework. I got this error when trying to run webapp. Error Console: Exception caught in Create Account Dataorg.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.entity.PojoEntityTuplizer] java.lang.NullPointerException My mapping file (hbm.xml) : <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net

org.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.entity.PojoEntityTuplizer]

谁都会走 提交于 2019-12-30 09:32:16
问题 I am developing a web application using Hibernate framework. I got this error when trying to run webapp. Error Console: Exception caught in Create Account Dataorg.hibernate.HibernateException: Unable to instantiate default tuplizer [org.hibernate.tuple.entity.PojoEntityTuplizer] java.lang.NullPointerException My mapping file (hbm.xml) : <?xml version="1.0" encoding="utf-8"?> <!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN" "http://hibernate.sourceforge.net

how to create a composite primary key hibernate JPA?

不打扰是莪最后的温柔 提交于 2019-12-29 08:19:27
问题 i try to create composite primary key in table from 2 foreign key using hibernate JPA,but gives me error.I find some solution on net but nothing The relations between tables looks like this: Table Client clientID(PK) FirstName LastName . Table CarService serviceID(PK) DescriptionOfFailure . . Table ServiceDepartment clientID(PK) serviceID(PK) price . implementation of my code looks like this: @Entity public class ServiceDepartmentBean implements ServiceDepartmentI { @EmbeddedId private

java.security.AccessControlException: access denied (java.lang.RuntimePermission modifyThreadGroup)

久未见 提交于 2019-12-28 16:02:06
问题 I have a problem in using of my own Database server with Google web application. Am using eclips (java EE IDE), installed all google plugins init, and develop a sample google web aplication , deployed it into web.Its working correctly. Now i want to use my own Database in my app.(MYSQL installed in localhost). Here am using Hibernate to connect with database. All required jar file are placed in lib directory which is in WEB-INF folder. When i run my application it gives an error like .. com

org.hibernate.MappingException: Could not determine type for: java.util.Set, at table: USERS, for columns: [org.hibernate.mapping.Column(invoices)]

不打扰是莪最后的温柔 提交于 2019-12-28 10:52:00
问题 I have a problem that Hibernate is unable to determine the type for Set at the table USERS. I am trying to create a foreign key of table INVOICES through one-to-many relationship. One User can generate many Invoices. My User.java is given below. @Entity @Table(name="USERS") public class User { @Id @Column(name="User_Id",nullable=false) @GeneratedValue(strategy=GenerationType.AUTO) private Integer user_id; @Column(name="NAME") private String name; @Column(name="Address") private String address