CODEWITHSUNDEEP
javasqlhibernate
Mike Minaev
Mike Minaev

Reputation: 2062

Select by field hibernate

i have such structure 

package logic;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Id;
import javax.persistence.Table;

import org.hibernate.annotations.GenericGenerator;

@Entity
@Table(name="users")
public class user {
    private Long id;
    private String name;
    private String pass;

    public user(){
        name = null;
    }

    public user(user u){
        name = u.getName();
    }

    @Id
    @GeneratedValue(generator = "increment")
    @GenericGenerator(name="increment", strategy = "increment")
    @Column(name="id")
    public Long getId(){
        return id;
    }

    @Column(name = "username", unique = true)
    public String getName() {
        return name;
    }

    @Column(name = "password")
    public String getPass(){
        return pass;
    }

    public void setId(Long i){
        id = i;
    }

    public void setName(String n){
        name = n;
    }

    public void setPass(String p){
        pass = p;
    }
}

i want to select from this table by username, in SQL smth like this select * from users where username = "abc";, but how i can do it with hibernate?

Upvotes: 1

Views: 3982

Answers (5)

sakura
sakura

Reputation: 2279

you can use something like that

  public user getUser(String username) {
    Session session = SessionUtil.sessionFactory.getCurrentSession();
    Transaction tx = session.beginTransaction();
    Query query = session.createQuery("From user where name=:name");
            query.setString("name", username);
            user result = (user) query.uniqueResult();
    tx.commit();
    return result;
}

if you want to use pattern then you can use something like below:

    Query query = session.createQuery("From user where name like :name");
    query.setString("name", username);
    List result = query.list();

Above will use the pattern, and return list of users with matching username pattern.

Upvotes: 1

Rahul Chudasama
Rahul Chudasama

Reputation: 38

 Session session = HibernateUtil.getSessionFactory().openSession();

        Query query = session.createQuery("from users where username = :name ");
        query.setParameter("name", "abc");

        // You can replace the above to commands with this one
        // Query query = session.createQuery("from Student where studentId = 1 ");
        List<?> list = query.list();

        User user = (User)list.get(0);

with this you can easily get idea what you want with any table.

Upvotes: 0

Gundamaiah
Gundamaiah

Reputation: 786

  Query query=session.createQuery("FROM users where name=:name");
  query.setParameter("name","abc");
   List list=query.list();

Upvotes: 0

Rahul
Rahul

Reputation: 45060

You can write a query in HQL which looks like this:

Query query = session.createQuery("from user where name = :name");
query.setParameter("name", "abc");
List list = query.list(); // List of users
user usr = (user) query.uniqueResult(); // Single user

Upvotes: 0

Prabhakaran Ramaswamy
Prabhakaran Ramaswamy

Reputation: 26094

Do like this

Query query = getSession().createQuery("from user where name =:name ")
            .setParameter("name ", "abc");

Single user with name abc

 user u = (user) query.uniqueResult();

All users with name abc

 List list = query.list();

Upvotes: 2

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