Confused about inserting a single entity as explained in Murach's Java Servlets/JSP 3rd edition

[Lisa Austin]

Hi,
In the book Murach's Java Servlets/JSP 3rd edition pg. 444 - 445 it explains how to insert , update , delete a single entity. I need some clarification please. My instructor has us using examples from the book. The example below is what is on pg 445 except two additional lines which I added based on my instructor's feedback .

I don't understand how the value "user" in the code

em.persist ( user );

knows what column to insert "user" into. I am use to SQL insert into statements so I'm having trouble with this.

The examples are this.

EntityManager em = DBUtil.getEmFactory().createEntityManager(); // I added this.  Not part of example in book
EntityTransation trans = em.getTransation();

		
String user = "root"; // I added this. Not part of example in book 

try {
  trans.begin();
  em.persist ( user );
  trans.commit ();
  
}  catch (Exception ex ) {
  trans.rollback();
} finally {
  em.close();
}

I understand that the line

em.persist ( user );

is to insert a single entity but I don't think I don't understand what exactly is being inserted and what table . I have the below persistence.xml .

Am I correctly understanding :

that the database schema " JPA_Test " will be created.

The table "User" will be created and the columns are based on the values I have in my JPA Entity ( code pasted below the persistence.xml code ).

If I am correct then is the value being inserted by

em.persist ( user );

just " root" ? If it's "root" then how does this specify WHAT column "root" should be inserted into?


Thank You for your help.

<?xml version="1.0" encoding="UTF-8"?>
<persistence 
	version ="2.1" 
	xmlns="http //xmlns.jcp.org/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="myChoicePU"
 transaction-type="RESOURCE_LOCAL">
 <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
 <exclude-unlisted-classes>false</exclude-unlisted-classes>
 <properties>
 <property name="javax.persistence.jdbc.url"
 value="jdbc:mysql://localhost:3306/JPA_Test"/>
 <property name="javax.persistence.jdbc.driver"
 value="com.mysql.jdbc.Driver"/>
 <property name="javax.persistence.jdbc.user" value="root"/>
 <property name="javax.persistence.jdbc.password" value="*****"/>
 <property name="javax.persistence.schema-generation.database.action"
 value="create"/>
 <property name="hibernant.show_sql" value="true"></property>
 </properties>
</persistence-unit>
</persistence>
 

JPA entity

import java.io.Serializable; 
import javax.persistence.Entity; 
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType; 
import javax.persistence.Id;  

@Entity 
public class User implements Serializable {  
    @Id     
    @GeneratedValue(strategy = GenerationType.AUTO)
     private Long userId;
     private String firstName;
     private String lastName;
     private String email;  

    public Long getUserId() {
         return userId;
     }
     public void setUserId(Long userId) {
         this.userId = userId;
     }     // the rest of the get and set methods for the fields 
}

[Dave Tolls]

The value passed into the persist() method must be an Entity.
It's the Entity (via its mappings) that is persisted.

In the case of your User entity, it will use the default values, so the table will be "User", and all the columns will match the other data and will be mapped to a relevant column datatype.

[Lisa Austin]

Dave Tolls wrote:The value passed into the persist() method must be an Entity.
It's the Entity (via its mappings) that is persisted.

In the case of your User entity, it will use the default values, so the table will be "User", and all the columns will match the other data and will be mapped to a relevant column datatype.

Hey Dave,
I'm sorry to be a pain.
I thought entity is a single value? As in if I have a table which has the columns User, Name, UserID. Entity would be one single value in the column "User"? Is that wrong?

[Dave Tolls]

At it's most basic, and Entity is the Java class that represents a table in the database.
The @Entity annotation tells JPA that the class that's been annotated is represented by a table.

It's not a value in a column, it's the whole row (give or take).

So, to take your User class above, JPA will expect there to be a table in the database called User, looking something like (depending on database):
userId IDENTITY
firstName VARCHAR2
lastName VARCHAR2
email VARCHAR2

Hibernate can even generate that for you, though I've never used the auto generation thing.

[Roel De Nijs]

Lisa Austin wrote:In the book Murach's Java Servlets/JSP 3rd edition pg. 444 - 445 it explains how to insert , update , delete a single entity.  I need some clarification please.

As i seems you are an ORM (JPA) greenhorn, it might be useful to have a few other (free) resources besides the Murach book to use as complementary resources when you need some additional clarification.

I really like the Java Persistence API WikiBook. When I'm having an issue with JPA, JPQL, or something related it's the first resource I'll check to find a solution. It's really an excellent resource (in my humble opinion)! It's not a step-by-step tutorial, but you'll find everything required in this wiki book: from basic entity mappings to complex JPQL queries.

If you are looking for a step-by-step tutorial, I would definitely have a look at the Hibernate Getting Started Guide.

Hope it helps!
Kind regards,
Roel