Question

我在spring-boot上有spring + jpa + rest应用程序（尝试实现）

spring-boot申请

import org.springframework.boot.SpringApplication;
import org.springframework.boot.autoconfigure.EnableAutoConfiguration;
import org.springframework.boot.autoconfigure.SpringBootApplication;
import org.springframework.boot.autoconfigure.domain.EntityScan;
import org.springframework.context.annotation.ComponentScan;
import org.springframework.context.annotation.ImportResource;
import org.springframework.data.jpa.repository.config.EnableJpaRepositories;

import javax.persistence.PersistenceContext;

@SpringBootApplication
@ComponentScan("package")
@EnableJpaRepositories(basePackages = "package.dao", entityManagerFactoryRef = "emf")
@ImportResource("applicationContext.xml")
public class Application {
    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }
}

我休息了-控制器

import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.web.bind.annotation.*;
import package.myService;

@RestController
public class TestController {
    @Autowired
    private MyService myService;

    @PostMapping("create")
    public void test1(){
    }

    @PostMapping("{id}")
    public void test2(@RequestBody Object object){
    }

    @GetMapping("{id}")
    public void test3(@PathVariable Long id){
    }
}

实体：

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import java.math.BigDecimal;

@Entity
public class MyEntity {
    @Id
    @GeneratedValue
    private Long id;
    private String name;

    public MyEntity() {
    }

    public MyEntity(Long id, String name) {
        this.id = id;
        this.name = name;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }
}

存储库为：

MyEntityRepository：

import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.query.Param;
import org.springframework.stereotype.Repository;

@Repository
public interface MyEntityRepositoryextends JpaRepository<myEntity, Long>{
    public void findNameById(@Param("id") Long id);
}

资源中的应用上下文：

<?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:task="http://www.springframework.org/schema/task"
       xmlns:util="http://www.springframework.org/schema/util"
       xmlns:context="http://www.springframework.org/schema/context" xmlns:tx="http://www.springframework.org/schema/tx"
       xsi:schemaLocation="http://www.springframework.org/schema/beans 
       http://www.springframework.org/schema/beans/spring-beans-4.3.xsd
       http://www.springframework.org/schema/task 
       http://www.springframework.org/schema/task/spring-task-4.3.xsd
       http://www.springframework.org/schema/context 
       http://www.springframework.org/schema/context/spring-context.xsd
       http://www.springframework.org/schema/util
       http://www.springframework.org/schema/util/spring-util-4.3.xsd
       http://www.springframework.org/schema/security http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx.xsd">

    <context:annotation-config />
    <bean id="riskMetricService" class="ru.sbrf.risks.services.data.service.RiskMetricServiceImpl"/>
</beans>

如果需要，我可以添加pom.xml

尝试分层时出现错误消息：应用程序启动失败


说明：

构造函数的参数0   org.springframework.data.rest.webmvc.RepositorySearchController   需要一个找不到的名为“ emf”的bean。

操作：

在哪里需要使用dbconnection配置定位文件？如何解决此错误消息？

Answer 1

在您的应用程序中，您引用的是一个名为emf的实体管理器

entityManagerFactoryRef =“ emf”

您应该定义它。您可以在应用程序上下文中或在 @Configuration类。

例如，您可以这样定义它：

    <bean id="emf"
            class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
            ......
    </bean>

尝试在spring-boot上启动jpa时如何解决错误消息？

1 个答案: