java javax.persistence.PersistenceException：无法定位持久性单元
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/31305334/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
javax.persistence.PersistenceException: Unable to locate persistence units
提问by Java Developer
I had a Maven project, I converted it to JPA project that now works with Maven. My persistence.xmlis as follows:
我有一个 Maven 项目，我将它转换为现在可与 Maven 一起使用的 JPA 项目。我persistence.xml的如下：
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1" xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="my-pu" transaction-type="RESOURCE_LOCAL">
<description>My Persistence Unit</description>
<provider>org.hibernate.ejb.HibernatePersistence</provider>
<class>model.entity.Tweet</class>
<class>model.entity.TweetHashtag</class>
<class>model.entity.TweetUrl</class>
<class>model.entity.User</class>
<exclude-unlisted-classes>true</exclude-unlisted-classes>
<properties>
<property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>
<property name="hibernate.connection.driver_class" value="com.mysql.jdbc.Driver"/>
<property name="hibernate.connection.url" value="jdbc:mysql://localhost:3306/tweetDB"/>
<property name="hibernate.connection.username" value="root"/>
<property name="hibernate.connection.password" value="root"/>
<property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/tweetDB"/>
<property name="javax.persistence.jdbc.user" value="root"/>
<property name="javax.persistence.jdbc.password" value=""/>
<property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver"/>
</properties>
</persistence-unit>
</persistence>
…and I have the following entities:
……我有以下实体：
package model.entity;
@Entity
@Table(name="tweets")
public class Tweet implements Serializable{
private static final long serialVersionUID = -1041037108182045708L;
@Id
@Column(name="tweet_id")
private int tweetId;
@Column(name="tweet_text")
private String tweetText;
@Temporal(TemporalType.TIMESTAMP)
@Column(name = "created_at", nullable = false, length = 19)
private Date createdAt;
@Column(name="lang_code")
private String languageCode;
@ManyToOne(optional=false)
@JoinColumn(name = "user_id",referencedColumnName="user_id")
private User user;
@OneToMany(mappedBy="tweet")
private Set<TweetHashtag> hashtags=new HashSet<TweetHashtag>(0);
@OneToMany(mappedBy="tweet")
private Set<TweetUrl> links=new HashSet<TweetUrl>(0);
public int getTweetId() {
return tweetId;
}
public void setTweetId(int tweetId) {
this.tweetId = tweetId;
}
public String getTweetText() {
return tweetText;
}
public void setTweetText(String tweetText) {
this.tweetText = tweetText;
}
public Date getCreatedAt() {
return createdAt;
}
public void setCreatedAt(Date createdAt) {
this.createdAt = createdAt;
}
public String getLanguageCode() {
return languageCode;
}
public void setLanguageCode(String languageCode) {
this.languageCode = languageCode;
}
public User getUser() {
return user;
}
public void setUser(User user) {
this.user = user;
}
public Set<TweetHashtag> getHashtags() {
return hashtags;
}
public void setHashtags(Set<TweetHashtag> hashtags) {
this.hashtags = hashtags;
}
public Set<TweetUrl> getLinks() {
return links;
}
public void setLinks(Set<TweetUrl> links) {
this.links = links;
}
}
//////////////////////////////////
package model.entity;
@Entity
@Table(name="tweet_hashtag")
public class TweetHashtag implements Serializable{
/**
*
*/
private static final long serialVersionUID = 6302465277669302540L;
@Id @GeneratedValue
@Column(name="hashtag_id")
private int hastagId;
public int getHastagId() {
return hastagId;
}
public void setHastagId(int hastagId) {
this.hastagId = hastagId;
}
@ManyToOne
@JoinColumn(name = "tweet_id",referencedColumnName="tweet_id")
private Tweet tweet;
@Column(name="hashtag")
private String hashtag;
public Tweet getTweet() {
return tweet;
}
public void setTweet(Tweet tweet) {
this.tweet = tweet;
}
public String getHashtag() {
return hashtag;
}
public void setHashtag(String hashtag) {
this.hashtag = hashtag;
}
}
package model.entity;
///////////////////////////////////////
@Entity
@Table(name="tweet_urls")
public class TweetUrl implements Serializable{
/**
*
*/
private static final long serialVersionUID = -606690240421717136L;
@Id @GeneratedValue
@Column(name="url_id")
private int urlId;
public int getUrlId() {
return urlId;
}
public void setUrlId(int urlId) {
this.urlId = urlId;
}
@ManyToOne
@JoinColumn(name="tweet_id",referencedColumnName="tweet_id")
private Tweet tweet;
@Column(name="url")
private String url;
public Tweet getTweet() {
return tweet;
}
public void setTweet(Tweet tweet) {
this.tweet = tweet;
}
public String getUrl() {
return url;
}
public void setUrl(String url) {
this.url = url;
}
}
////////////////////////////
package model.entity;
@Entity
@Table(name="users")
public class User implements Serializable{
@Id
@Column(name="user_id")
private int userId;
@Column(name="screen_name")
private String screenName;
@Column(name="profile_image_url")
private String profileImageUrl;
@Column(name="statuses_count")
private int statusesCount;
@OneToMany(mappedBy="user")
private List<Tweet> tweets;
public int getUserId() {
return userId;
}
public void setUserId(int userId) {
this.userId = userId;
}
public String getScreenName() {
return screenName;
}
public void setScreenName(String screenName) {
this.screenName = screenName;
}
public String getProfileImageUrl() {
return profileImageUrl;
}
public void setProfileImageUrl(String profileImageUrl) {
this.profileImageUrl = profileImageUrl;
}
public int getStatusesCount() {
return statusesCount;
}
public void setStatusesCount(int statusesCount) {
this.statusesCount = statusesCount;
}
public List<Tweet> getTweets() {
return tweets;
}
public void setTweets(List<Tweet> tweets) {
this.tweets = tweets;
}
}
//////////////////
Here, the error occurs:
在这里，发生错误：
package model;
public class TweetPersistent {
private EntityManagerFactory emf;
private EntityManager em;
public void persist(Pair<String, QueryResult> pair) {
List<Status> tweets = pair.getValue().getTweets();
for (Status tweet : tweets) {
this.insertToDB(pair.getKey(), tweet);
// or this.batchInsertToDB(pair.getKey(), tweet);
}
}
private void insertToDB(String key, Status tweet) {
emf = Persistence.createEntityManagerFactory("TwitterQueryApp");
em = emf.createEntityManager();
em.getTransaction().begin();
em.persist(new Pair<String, Status>(key, tweet));
em.getTransaction().commit();
// Insert code here
}
private void batchInsertToDB(String key, Status tweet) {
// insert batch
}
}
///////////
Now the error: "javax.persistence.PersistenceException: Unable to locate persistence units"
现在错误：“ javax.persistence.PersistenceException：无法找到持久性单元”
Picked up JAVA_TOOL_OPTIONS: -javaagent:/usr/share/java/jayatanaag.jar
log4j:WARN No appenders could be found for logger (twitter4j.HttpClientImpl).
log4j:WARN Please initialize the log4j system properly.
log4j:WARN See http://logging.apache.org/log4j/1.2/faq.html#noconfig for more info.
Exception in thread "Thread-1" javax.persistence.PersistenceException: Unable to locate persistence units
at org.hibernate.jpa.HibernatePersistenceProvider.getEntityManagerFactoryBuilderOrNull(HibernatePersistenceProvider.java:101)
at org.hibernate.jpa.HibernatePersistenceProvider.getEntityManagerFactoryBuilderOrNull(HibernatePersistenceProvider.java:88)
at org.hibernate.jpa.HibernatePersistenceProvider.createEntityManagerFactory(HibernatePersistenceProvider.java:69)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:79)
at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:54)
at model.TweetPersistent.insertToDB(TweetPersistent.java:32)
at model.TweetPersistent.persist(TweetPersistent.java:24)
at controller.TwitterStreamConsumer.run(TwitterStreamConsumer.java:25)
at java.lang.Thread.run(Thread.java:745)
Caused by: javax.persistence.PersistenceException: Invalid persistence.xml.
Error parsing XML [line : -1, column : -1] : cvc-elt.1: Deklaration des Elements "persistence" kann nicht gefunden werden.
at org.hibernate.jpa.boot.internal.PersistenceXmlParser.validate(PersistenceXmlParser.java:377)
at org.hibernate.jpa.boot.internal.PersistenceXmlParser.loadUrl(PersistenceXmlParser.java:310)
at org.hibernate.jpa.boot.internal.PersistenceXmlParser.parsePersistenceXml(PersistenceXmlParser.java:114)
at org.hibernate.jpa.boot.internal.PersistenceXmlParser.doResolve(PersistenceXmlParser.java:104)
at org.hibernate.jpa.boot.internal.PersistenceXmlParser.locatePersistenceUnits(PersistenceXmlParser.java:86)
at org.hibernate.jpa.HibernatePersistenceProvider.getEntityManagerFactoryBuilderOrNull(HibernatePersistenceProvider.java:97)
... 8 more
I included different jars: hibernate-validation, java-persistence, hibernate-jpaand hibernate-entityManagerin the build path and also class path of project itself. I also copied persistence.xmlin resource folder.
I did lots of things, but it does not work, getting always the same error.
我在构建路径和项目本身的类路径中包含了不同的 jars: hibernate-validation, java-persistence, hibernate-jpaand hibernate-entityManager。我也复制persistence.xml到资源文件夹中。我做了很多事情，但它不起作用，总是出现相同的错误。
Thank you in advance for your suggestions!
预先感谢您的建议！
采纳答案by Pawe? G?owacz
This looks like a problem with the XML schema for JPA 2.1 version.
这看起来像是 JPA 2.1 版本的 XML 模式的问题。
Try and look at this solution. Hope it works for you.
试着看看这个解决方案。希望对你有效。
回答by Rzubakov
If you use Eclipse, please check into export window 'copy required libraries into a subfolder next to the generated JAR'.
如果您使用 Eclipse，请检查导出窗口“将所需的库复制到生成的 JAR 旁边的子文件夹中”。
回答by Pawel Hawro
You should pass persistence unit name to Persistence.createEntityManagerFactorywhich is present in yours persistence.xml:
您应该将Persistence.createEntityManagerFactory存在于您的 persistence.xml 中的持久性单元名称传递给：
<persistence-unit name="my-pu" transaction-type="RESOURCE_LOCAL">
Try calling:
尝试调用：
Persistence.createEntityManagerFactory("my-pu");
Name given as argument in createEntityManagerFactoryshould exist in one of persistence elements in your xml file.
作为参数给出的名称createEntityManagerFactory应该存在于您的 xml 文件中的持久性元素之一中。
回答by krevelen
When creating an EntityManagerFactorywithin a webapp running on a container managed by the Cargo Maven plugin, for instance an embedded Tomcat8xcontainer, configure Cargo so it copies the javax.transaction:jtaand (implicit) org.hibernate:hibernate-entitymanagerdependencies into Tomcat's Shared Classpath, as per the documentation:
当EntityManagerFactory在由Cargo Maven 插件管理的容器（例如嵌入式Tomcat8x容器）上运行的 web 应用程序中创建一个时，配置 Cargo 使其将javax.transaction:jta和（隐式）org.hibernate:hibernate-entitymanager依赖项复制到 Tomcat 的共享类路径中，根据文档：
<project ...>
:
<dependencies>
:
<dependency>
<groupId>javax.transaction</groupId>
<artifactId>jta</artifactId>
<version>${jta.version}</version>
<scope>provided</scope>
</dependency>
<dependency>
<groupId>org.hibernate.ogm</groupId>
<artifactId>hibernate-ogm-neo4j</artifactId>
<version>${hibernate.version}</version>
</dependency>
</dependencies>
:
<build>
:
<plugins>
:
<plugin>
<groupId>org.codehaus.cargo</groupId>
<artifactId>cargo-maven2-plugin</artifactId>
<version>${cargo.version}</version>
<configuration>
:
<container>
<type>embedded</type>
<artifactInstaller>
<groupId>org.apache.tomcat</groupId>
<artifactId>tomcat</artifactId>
<version>${tomcat.version}</version>
</artifactInstaller>
:
<dependencies>
<dependency>
<groupId>org.hibernate</groupId>
<artifactId>hibernate-entitymanager</artifactId>
<classpath>shared</classpath>
</dependency>
<dependency>
<groupId>javax.transaction</groupId>
<artifactId>jta</artifactId>
<classpath>shared</classpath>
</dependency>
</dependencies>
</container>
</configuration>
</plugin>
</plugins>
</build>
</project>
回答by Ken
Could you please try removing the below from xml ?
您能否尝试从 xml 中删除以下内容？
<?xml version="1.0" encoding="UTF-8"?>
