我是JPA的新手,我试着从书中做一个简单的例子。但无论我做什么,我都会收到以下错误:
Exception in thread "main" javax.persistence.PersistenceException: No Persistence provider for EntityManager named EmployeeService
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:89)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:60)
at com.mycompany.simpleentity.EmployeeTest.main(EmployeeTest.java:18)我用谷歌搜索了很多,我做了所有我读到的关于JPA的事情。
以下是我的项目的目录树:
.
|-- pom.xml
`-- src
|-- main
| |-- java
| | `-- com
| | `-- mycompany
| | `-- simpleentity
| | |-- Employee.java
| | |-- EmployeeService.java
| | `-- EmployeeTest.java
| `-- resources
| `-- META-INF
| `-- persistence.xml
`-- test这是我的pom.xml:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd">
<modelVersion>4.0.0</modelVersion>
<groupId>com.mycompany</groupId>
<artifactId>SimpleEntity</artifactId>
<packaging>jar</packaging>
<version>1.0-SNAPSHOT</version>
<name>SimpleEntity</name>
<url>http://maven.apache.org</url>
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>3.8.1</version>
<scope>test</scope>
</dependency>
<dependency>
<groupId>javax.persistence</groupId>
<artifactId>persistence-api</artifactId>
<version>1.0</version>
</dependency>
<dependency>
<groupId>postgresql</groupId>
<artifactId>postgresql</artifactId>
<version>9.0-801.jdbc4</version>
</dependency>
</dependencies>
<build>
<plugins>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-compiler-plugin</artifactId>
<configuration>
<source>1.5</source>
<target>1.5</target>
</configuration>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<configuration>
<archive>
<manifest>
<addClasspath>true</addClasspath>
<mainClass>com.mycompany.simpleentity.EmployeeTest</mainClass>
<!-- <classpathLayoutType>repository</classpathLayoutType> -->
<classpathMavenRepositoryLayout>true</classpathMavenRepositoryLayout>
<classpathPrefix>${env.HOME}/.m2/repository</classpathPrefix>
</manifest>
</archive>
</configuration>
</plugin>
</plugins>
</build>
</project>这是我的源代码: EmployeeTest.java:
package com.mycompany.simpleentity;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;
public class EmployeeTest {
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("EmployeeService");
EntityManager em = emf.createEntityManager();
}
}这是我的persistance.xml
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence
http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"
version="1.0">
<persistence-unit name="EmployeeService" transaction-type="RESOURCE_LOCAL">
<class>com.mycompany.simpleentity.Employee</class>
<properties>
<property name="toplink.jdbc.driver"
value="org.postgresql.Driver"/>
<property name="toplink.jdbc.url"
value="jdbc:postgresql://localhost:5432/testdb;create=true"/>
<property name="toplink.jdbc.user" value="postgres"/>
<property name="toplink.jdbc.password" value="111"/>
</properties>
</persistence-unit>
</persistence>我做错了什么?提前谢谢你。
发布于 2010-11-24 06:22:59
JPA是由多个JPA提供者(Hibernate、EclipseLink、OpenJPA、Toplink)实现的规范。
您需要选择要使用的提供程序,并将适当的依赖项添加到pom.xml。此外,您还需要在persistence.xml中指定您的提供者。
例如,如果您使用OpenJPA (我在本例中选择它,因为它的最新版本可以在Maven Central Repo中找到,所以不需要配置特定于供应商的存储库):
<dependencies>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>3.8.1</version>
<scope>test</scope>
</dependency>
<!-- Note that you don't need persistence-api dependency - it's transitive -->
<dependency>
<groupId>org.apache.openjpa</groupId>
<artifactId>openjpa-all</artifactId>
<version>2.0.1</version>
</dependency>
<dependency>
<groupId>postgresql</groupId>
<artifactId>postgresql</artifactId>
<version>9.0-801.jdbc4</version>
</dependency>
</dependencies> 。
<persistence xmlns="http://java.sun.com/xml/ns/persistence"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/persistence
http://java.sun.com/xml/ns/persistence/persistence_1_0.xsd"
version="1.0">
<persistence-unit name="EmployeeService" transaction-type="RESOURCE_LOCAL">
<!-- Provider specification -->
<provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider>
<class>com.mycompany.simpleentity.Employee</class>
<properties>
<property name="javax.persistence.jdbc.driver"
value="org.postgresql.Driver"/>
<property name="javax.persistence.jdbc.url"
value="jdbc:postgresql://localhost:5432/testdb;create=true"/>
<property name="javax.persistence.jdbc.user" value="postgres"/>
<property name="javax.persistence.jdbc.password" value="111"/>
</properties>
</persistence-unit>
</persistence> 发布于 2012-09-14 21:04:31
如果您使用JPA + eclipselink提供程序,则使用以下代码
<properties>
<property name="javax.persistence.jdbc.url" value="jdbc:postgresql://localhost/Database name" />
<property name="javax.persistence.jdbc.user" value="" />
<property name="javax.persistence.jdbc.password" value="" />
<property name="javax.persistence.jdbc.driver" value="org.postgresql.Driver" />
</properties>发布于 2012-12-03 23:20:45
1)确保您已经定义了持久性提供者(对于任何提供者):Ex For openjpa:<persistence-unit ...> <provider>org.apache.openjpa.persistence.PersistenceProviderImpl</provider> ... ... </persistence-unit>
2)如果您使用的是自定义构建/编译过程(maven等),请确保将Meta-INF/persistance.xml复制到已编译的/classes文件夹中。
https://stackoverflow.com/questions/4261435
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