hibernate-mapping

问题 I want to pass the join query results from controller to view , but I am getting error "java.lang.NumberFormatException: For input string: "id" I have already go through the duplicated questions and applied the solutions for my scenario but it is not getting successful. Duplicated Question -1 Duplicated Question -2 Duplicated Question -3 DAO code public List<Object [] > showEmployee_Role(){ try { Session session=HibernateUtil.getSessionFactory().openSession(); SQLQuery query = session

问题 As shown in query and result, I want to fetch and add the result in list but how do I do it in Spring data. How to hold the result in a list? Should I create a new entity class? Keeping in mind that I absolutely do not need to map my model class to database. I simple want to fetch and use the list in controller. Thanks to anyone who can help. 回答1: You can use projections. In your case, for example: public interface TotalPerMonth { String getMonth(); Long getTotal(); } Then use it in your

问题 I'm working on an application with spring, i'm encountering the following error and i don't know how to deal with it . Etat HTTP 500 - Handler processing failed; nested exception is java.lang.NoClassDefFoundError: Could not initialize class com.my.util.HibernateUtil type Rapport d''exception message Handler processing failed; nested exception is java.lang.NoClassDefFoundError: Could not initialize class com.my.util.HibernateUtil description Le serveur a rencontré une erreur interne qui l''a

问题 Could someone help me with the possible hibernate exceptions when two threads update the same Object? ex: employee with name "a", age "30" and address "test" thread1 tries to update "a" to "b" and thread2 tries to update "a" to "c" Thanks in advance, Kathir 回答1: If your object is a Hibernate entity, then two threads shouldn't have a reference to the same object in the first place. Each thread will have its own Hibernate session, and each session will have its own copy of the entity. If you

问题 I am getting org.hibernate.MappingException: No Dialect mapping for JDBC type: -1 error while fetching row from database. ERROR details exception org.springframework.web.util.NestedServletException: Request processing failed; nested exception is org.hibernate.MappingException: No Dialect mapping for JDBC type: -1 org.springframework.web.servlet.FrameworkServlet.processRequest(FrameworkServlet.java:659) org.springframework.web.servlet.FrameworkServlet.doPost(FrameworkServlet.java:563) javax

问题 I've got two tables A and B with simple PK's. @Entity public class A { @Id public int idA; } @Entity public class B { @Id public int idB; } I want to map a new association class AB that simply stores the relations between A and B , with composite PK idA + idB . AB doesn't have any extra columns, just the relation between idA and idB . Is it possible to map AB using a single class? I want to avoid having to create a new ABId class just to use it as @IdClass or @EmbeddedId in AB , and I don't

问题 I have mapped a column in Mysql DB to enum in java. But there are invalid entries in the table column (created manually) and which is not in the enum I created in java. When I try to load the entries I get the below exception. Is it possible to suppress this exception i.e. set the enum to 'null' when you have a invalid entry in the column in the Database? java.lang.IllegalArgumentException: Unknown name value [] for enum class [model.enums.PeriodUnit] at org.hibernate.type.EnumType

问题 I am getting this issue and not able to resolve. I want to implement OneToMany/ManyToOne mapping without creating extra table. I have two Entity: (1) Employee.java @Entity(name="EMPLOYEE_DETAILS") public class Employee { private int empId; private String employeeName; private Date joiningDate; private Collection<Project> project = new ArrayList<Project>(); @Id @GeneratedValue (strategy = GenerationType.AUTO) public int getEmpId() { return empId; } public void setEmpId(int employeeId) { this

问题 This question is very similar to the one asked here Could someone shed some light as why hibernate decides to ignore the property, which has a formula, completely while persisting. If so whats the alternative to persist a property which has a formula ? is there any additional config ? Query fired by hibernate : insert into fighterjet (max_speed, country, jet_id) values (?, ?, ?) Note how hibernate ignores the 'name' property in the insert query, which has a formula in the hbm. HBM :

问题 class A{ private List<B> bs; ... } class B{ private Long id; private String name; ... } And I'd like to have this: class A{ // the map should have B.name as key private Map<String,B> bs; ... } class B{ private Long id; private String name; private A a; ... } I don't know if it is clear what I'd like to do, but it is as simple as mapping a one to many relationship to a Map with the name of B as the key of the map. Thanks in advance, Neuquino 回答1: Try the hibernate annotation MapKey @MapKey