spring-4

问题 I use Spring framework 4.3.2.RELEASE I saw on a tutorial site that both Spring and Hibernate must match their versions in order to integrate Hibernate with Spring . So what is the appropriate version of Hibernate for Spring 4.3.2.RELEASE ? 回答1: Read the link https://spring.io/blog/2016/06/10/spring-framework-4-3-goes-ga Seems to me Hibernate 5.2 is the way to go. 回答2: I belive that correct answer is Hibernate 4.* try with Hibernate Core 4.3.6.Final. 回答3: From the Spring point of view with

问题 I have configured composite primary key for my entity Employee as follows Employee.java: @Entity @Table(name="employee") @Proxy(lazy=false) @IdClass(EmployeeId.class) public class Employee implements Serializable { private static final long serialVersionUID = 1L; private EmployeeId employeeId; private Person person; private Branch branch; private boolean isActive; public Employee() { } @EmbeddedId @AttributeOverrides({ @AttributeOverride(name="person", column = @Column(name="person_id")),

问题 I want to develop a small chat with springs new websocket/stomp support. I guess i cannot use something like this: @MessageMapping("/connect/{roomId}") @SendTo("/topic/newMessage") public String connectToChatRoom(@PathVariable String roomId, Principal p) { return getTimestamp() + " " + p.getName() + " connected to the room."; } What are my options for dynamic mapping here? As a client i want to subscribe only to the room I'm in. Thanks in advance! 回答1: Figured it out, you need to use

问题 Using spring version:4.3.1,Spring data:1.10.2,hibernate core: 5.2.1. using postgres Database pg connector version :9.4.1208.jre7 After upgrading from hibernate 5.1.0 to 5.2.1 when i am trying to create an entity getting below exception at java.lang.Thread.run(Thread.java:745) Caused by: org.springframework.transaction.CannotCreateTransactionException: Could not open JPA EntityManager for transaction; nested exception is java.lang.NoSuchMethodError: org.hibernate.Session.getFlushMode()Lorg

问题 I am working on Spring 4 application with SpringBoot. In com.test.tm package, Application class: @SpringBootApplication @EnableJpaRepositories( repositoryFactoryBeanClass = GenericRepositoryFactoryBean.class ) @Import( { HikariDataSourceConfig.class } ) public class Application { public static void main( String[] args ) { SpringApplication.run(Application.class, args); } } In com.test.tm.entities package, User Class: @Table( name = "test.user" ) @Entity public class User implements

问题 I am working on Spring 4 application with SpringBoot. In com.test.tm package, Application class: @SpringBootApplication @EnableJpaRepositories( repositoryFactoryBeanClass = GenericRepositoryFactoryBean.class ) @Import( { HikariDataSourceConfig.class } ) public class Application { public static void main( String[] args ) { SpringApplication.run(Application.class, args); } } In com.test.tm.entities package, User Class: @Table( name = "test.user" ) @Entity public class User implements

问题 I am using Spring 4.3.0.RELEASE together with Hibernate 5.0.9.Final in my app and I keep getting java.lang.NoSuchMethodError: org.hibernate.engine.spi.SessionFactoryImplementor.getProperties()Ljava/util/Map; If I update the Hibernate 5.2.0.Final, everything works fine. It seems that Spring 4.3.0 is not compatible with Hibernate 5.0, is that correct ? I was not able to find anything about not supporting Hibernate 5.0. In Hibernate 5.2, such method exists, in Hibernate 5.0 the method doesn't

问题 Since starting, I was always confuse of how to deal with InterruptedException and how to properly cancel the http request if they are taking too much time. I have a library in which I have provided two methods, sync and async for our customer. They can call whichever method they feel is right for their purpose. executeSync() - waits until I have a result, returns the result. executeAsync() - returns a Future immediately which can be processed after other things are done, if needed. They will

问题 I use xml config spring4 is work well.like this： <bean id="multipartResolver" class="org.springframework.web.multipart.commons.CommonsMultipartResolver" p:defaultEncoding="utf-8" > <!-- one of the properties available; the maximum file size in bytes --> <!-- <property name="maxUploadSize" value="40000000" /> --> </bean> when I config spring4 without xml.like this: @Configuration @ComponentScan({ "common.user", "service" }) @EnableWebMvc public class SpringMVCConfig { @Bean public ViewResolver

问题 I am working with Spring Core 4.0.7 and Spring Web Flow 2.4.0 Using Spring Tool Suite I am able to create a Spring Web Flow Definition File For the <flow> tag it generates <?xml version="1.0" encoding="UTF-8"?> <flow xmlns="http://www.springframework.org/schema/webflow" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/webflow http://www.springframework.org/schema/webflow/spring-webflow-2.0.xsd"> Observe spring-webflow-2.0.xsd When