How to insert values into database using queries using entity manager, persistence using a java class?

Question

I want to insert data into a table using the following code

    public User registerUser(String usr, String pwd) {

    u=em.find(User.class,usr);
    if(u!=null)
    {
        return null;
    }
    String query1 = "insert into users values('" + usr + "','" + pwd +"')";
    Query q = em.createQuery(query1);
    u=em.find(User.class,usr);
    return u;

}

here 'u' is the object of User class and em is EntityManager.

I get this following exception:

Servlet.service() for servlet action threw exception org.hibernate.hql.ast.QuerySyntaxException: expecting OPEN, found 'values' near line 1, column 19 [insert into users values('pawan','am')]

Renan · Accepted Answer · 2011-02-11 12:06:36Z

16

Try

public User registerUser(String usr, String pwd) {

    u=em.find(User.class,usr);
    if(u!=null)
    {
        return null;
    }

    //Now saving...
    em.getTransaction().begin();
    em.persist(u); //em.merge(u); for updates
    em.getTransaction().commit();
    em.close();

    return u;
}

If the PK is Identity, it will be set automatically in your persisted class, if you are using auto generation strategy (thanks to David Victor).

Edit to @aman_novice comment: set it in your class

//Do this BEFORE getTransaction/persist/commit
//Set names are just a example, change it to your class setters
u.setUsr(usr);
u.setPwd(pwd);

//Now you can persist or merge it, as i said in the first example
em.getTransaction().begin();
(...)

About @David Victor, sorry I forgot about that.

edited Feb 11, 2011 at 12:06

answered Feb 11, 2011 at 10:50

Renan

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2 Comments

aman_novice Over a year ago

how will this insert pwd in the User table, along with usr??

David Victor Over a year ago

The PK will only be set depending on the generation strategy. i.e. an @GeneratedValue(strategy = GenerationType.AUTO) annotation or similar. It is also better to rely on declarative txn boundaries generally.

Michael Borgwardt · Accepted Answer · 2015-05-07 15:27:18Z

5

You're not using SQL but JPAQL, there is no field-based insert. You persist object rather than inserting rows.

You should do something like this:

public User registerUser(String usr, String pwd) {
    u=em.find(User.class,usr);
    if(u!=null)
    {
        return u;
    }
    u = new User(usr, pwd);
    em.persist(u);
    return u;
}

edited May 7, 2015 at 15:27

answered Feb 11, 2011 at 10:28

Michael Borgwardt

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4 Comments

aman_novice Over a year ago

ya true...my bad..wrote it as SQL. however, i would like to point out that User() is a no-parameter constructor. So i cannot create that object!!! :(

Michael Borgwardt Over a year ago

@aman_novice: well, then use the setter methods to set the username and password.

David Victor Over a year ago

I think there will be a problem here with the type of the identifier for the User class. Need to ensure the id type is String on the entity. Which would work - but is unusual. :-)

rockstar Over a year ago

thanks i was looking for something along these lines ! +1 :-)

David Victor · Accepted Answer · 2011-02-11 11:02:13Z

0

This isn't really the way to go. You are trying to insert a row in a table but have no associated attached entity. If you're using the JPA entity manager - then create a new instance - set the properties & persist the entity.

E.g.

User u = new User();
u.setXXX(xx);

em.persist(u);
// em.flush()   <<-- Not required, useful for seeing what is happening

// etc..

If you enable SQL loggging in Hibernate & flush the entity then you'll see what is sent to the database.

E.g. in persistence.xml:

    <property name="hibernate.format_sql" value="true" />

edited Feb 11, 2011 at 11:02

answered Feb 11, 2011 at 10:25

David Victor

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