javax.persistence.PersistenceException: Unable to locate persistence units

Question

I had a Maven project, I converted it to JPA project that now works with Maven. My persistence.xml is as follows:

<?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:

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:

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"

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-jpa and hibernate-entityManager in the build path and also class path of project itself. I also copied persistence.xml in resource folder. I did lots of things, but it does not work, getting always the same error.

Thank you in advance for your suggestions!

Solution 1

This looks like a problem with the XML schema for JPA 2.1 version.

Try and look at this solution. Hope it works for you.

Solution 2

If you use Eclipse, please check into export window 'copy required libraries into a subfolder next to the generated JAR'.

Solution 3

You should pass persistence unit name to Persistence.createEntityManagerFactory which is present in yours persistence.xml:

<persistence-unit name="my-pu" transaction-type="RESOURCE_LOCAL"> 

Try calling:

Persistence.createEntityManagerFactory("my-pu");

Name given as argument in createEntityManagerFactory should exist in one of persistence elements in your xml file.

Comments

[Java Developer]

now it says :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.

[Java Developer]

Exception in thread "Thread-1" javax.persistence.PersistenceException: Unable to locate persistence units

[Java Developer]

Tnx now it is better: Exception in thread "Thread-1" java.lang.IllegalArgumentException: Unknown entity: javafx.util.Pair at org.hibernate.jpa.spi.AbstractEntityManagerImpl.persist(Abst‌​ractEntityManagerImp‌​l.java:1184) at model.TweetPersistent.insertToDB(TweetPersistent.java:35) at model.TweetPersistent.persist(TweetPersistent.java:24) at controller.TwitterStreamConsumer.run(TwitterStreamConsumer.j‌​ava:25) at java.lang.Thread.run(Thread.java:745)

[Paweł Głowacz]

I have no idea what Pair is. EntityManager works only if you define @Entity annotation for your class that need to be persisted and you need to define this class in persistence.xml. Otherwise you will see above exception.