Java 使用EntityManagerFactory时出现问题

我正在试图弄清楚如何使用

@PersistenceUnit

，因为我读到它比

@PersistenceContext

好得多。问题是。。。。我不知道如何让它正常工作

@Controller
public class Content {
    @PersistenceUnit(unitName = "CMTPU")
    public EntityManagerFactory emf;
    public EntityManager em = emf.createEntityManager();

    @RequestMapping(value={"/content/edit*"}, method=RequestMethod.GET)
    public ModelAndView edit(Model model) {
        ModelAndView mv = new ModelAndView();
        mv.setViewName("content/edit");

        //get symbols
        List<Symbol> symbols = em.createNamedQuery("Symbol.findAll").getResultList();
        mv.addObject(symbols);

        return mv;
    }
}

我很难确定自己做错了什么

更新

我的模型定义了数据库和所有这些，如下所示。我甚至需要persistence.xml吗

package com.fettergroup.cmt.models;

import java.io.Serializable;
import javax.persistence.Basic;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.Table;
import javax.validation.constraints.NotNull;
import javax.validation.constraints.Size;


    @Entity
    @Table(name = "symbol", catalog = "DATABASE1", schema = "dbo")
    @NamedQueries({
        @NamedQuery(name = "Symbol.findAll", query = "SELECT s FROM Symbol s"),
        @NamedQuery(name = "Symbol.findById", query = "SELECT s FROM Symbol s WHERE s.id = :id"),
        @NamedQuery(name = "Symbol.findBySymbol", query = "SELECT s FROM Symbol s WHERE s.symbol = :symbol"),
        @NamedQuery(name = "Symbol.findByHtmlNumber", query = "SELECT s FROM Symbol s WHERE s.htmlNumber = :htmlNumber"),
        @NamedQuery(name = "Symbol.findByHtmlName", query = "SELECT s FROM Symbol s WHERE s.htmlName = :htmlName"),
        @NamedQuery(name = "Symbol.findByAsciiDec", query = "SELECT s FROM Symbol s WHERE s.asciiDec = :asciiDec"),
        @NamedQuery(name = "Symbol.findByAsciiHex", query = "SELECT s FROM Symbol s WHERE s.asciiHex = :asciiHex")})
    public class Symbol implements Serializable {
        private static final long serialVersionUID = 1L;
        @Id
        @Basic(optional = false)
        @NotNull
        @Column(name = "id")
        private Short id;
        @Size(max = 10)
        @Column(name = "symbol")
        private String symbol;
        @Size(max = 10)
        @Column(name = "html_number")
        private String htmlNumber;
        @Size(max = 10)
        @Column(name = "html_name")
        private String htmlName;
        @Size(max = 10)
        @Column(name = "ascii_dec")
        private String asciiDec;
        @Size(max = 10)
        @Column(name = "ascii_hex")
        private String asciiHex;

        public Symbol() {
        }

        public Symbol(Short id) {
            this.id = id;
        }

        public Short getId() {
            return id;
        }

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

        public String getSymbol() {
            return symbol;
        }

        public void setSymbol(String symbol) {
            this.symbol = symbol;
        }

        public String getHtmlNumber() {
            return htmlNumber;
        }

        public void setHtmlNumber(String htmlNumber) {
            this.htmlNumber = htmlNumber;
        }

        public String getHtmlName() {
            return htmlName;
        }

        public void setHtmlName(String htmlName) {
            this.htmlName = htmlName;
        }

        public String getAsciiDec() {
            return asciiDec;
        }

        public void setAsciiDec(String asciiDec) {
            this.asciiDec = asciiDec;
        }

        public String getAsciiHex() {
            return asciiHex;
        }

        public void setAsciiHex(String asciiHex) {
            this.asciiHex = asciiHex;
        }

        @Override
        public int hashCode() {
            int hash = 0;
            hash += (id != null ? id.hashCode() : 0);
            return hash;
        }

        @Override
        public boolean equals(Object object) {
            // TODO: Warning - this method won't work in the case the id fields are not set
            if (!(object instanceof Symbol)) {
                return false;
            }
            Symbol other = (Symbol) object;
            if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
                return false;
            }
            return true;
        }

        @Override
        public String toString() {
            return "com.project1.models.Symbol[ id=" + id + " ]";
        }

    }

仅使用

@Controller
public class Content {

   @PersistenceContext(unitName = "CMTPU")
   public EntityManager em;

实体管理器应该由spring控制

这是一个例子，它使用Hibernate作为持久性提供者，但我认为您可以调整它

<tx:annotation-driven transaction-manager="transactionManager" />

<bean class="org.springframework.orm.jpa.JpaTransactionManager" id="transactionManager">
    <property name="entityManagerFactory" ref="entityManagerFactory" />

</bean>
<bean
    class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean"
    id="entityManagerFactory">
    <property name="persistenceUnitName" value="myPersistenceUnit" />
    <property name="jpaVendorAdapter">
        <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
            <property name="showSql" value="false" />
        </bean>
    </property>
</bean>

示例持久性单元，用于该配置

<persistence-unit name="myPersistenceUnit" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <properties>
        <property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect" />
        <property name="hibernate.hbm2ddl.auto" value="validate" />
        <property name="hibernate.connection.charSet" value="UTF-8" />
    </properties>
</persistence-unit>

org.hibernate.ejb.HibernatePersistence

我是否在applicationContext.xml中定义bean？这是我第一次尝试Java，如果我在适应我的需要方面遇到困难，请耐心等待。。。。当我尝试复制您拥有的内容并将

myPersistenceUnit

更改为

CMTPU

时，它会抛出错误

创建名为“entityManagerFactory”的bean，该bean在ServletContext资源[/WEB-INF/applicationContext.xml]中定义：在设置bean属性“jpavendorapter”时，无法创建[org.springframework.orm.jpa.vendor.hibernatejbavendorapter]类型为[org.springframework.orm.jpa.vendor.hibernatejbavendorapter]的内部bean“org.springframework.orm.jpa.vendor.hibernatejbavendorapter”查看我以前的评论。。。我没有包括您提到的
持久化单元
更改。在实现该更改后，我发现
以元素“properties”开头的内容无效。此时不需要任何子元素
这个问题在哪里，persistence.xml还是spring config？听起来像是xml问题，请检查xml。（我复制了代码并对其进行了一些修改，我是否可以删除一些结束xml元素）或者您的persistence.xml中是否有2个
properties
元素？
<persistence-unit name="myPersistenceUnit" transaction-type="RESOURCE_LOCAL">
    <provider>org.hibernate.ejb.HibernatePersistence</provider>
    <properties>
        <property name="hibernate.dialect" value="org.hibernate.dialect.MySQL5InnoDBDialect" />
        <property name="hibernate.hbm2ddl.auto" value="validate" />
        <property name="hibernate.connection.charSet" value="UTF-8" />
    </properties>
</persistence-unit>