JPA criteria query load entire table

Greg picture Greg · Jul 26, 2012 · Viewed 20.3k times · Source

I feel like this is a silly question, but I can't find the answer. I have a class as follows:

import java.io.Serializable;

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

@Entity
@Table(name="DEMO_VARIABLES")
public class Variable implements Serializable
{
    private static final long serialVersionUID = -1734898766626582592L;

    @Id
    @SequenceGenerator(name="VARIABLE_ID_GENERATOR", sequenceName="DEMO_VARIABLE_ID_SEQ", allocationSize=1)
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="VARIABLE_ID_GENERATOR")
    @Column(name="VARIABLE_ID", unique=true, nullable=false, precision=22)
    private long variableId;

    @Column(name="VARIABLE_NAME", nullable=false, length=50)
    private String variableName;

    @Column(name="VARIABLE_VALUE", nullable=false, length=500)
    private String variableValue;

    public Variable()
    {

    }

    public long getVariableId()
    {
        return variableId;
    }

    public void setVariableId(long variableId)
    {
        this.variableId = variableId;
    }

    public String getVariableName()
    {
        return variableName;
    }

    public void setVariableName(String variableName)
    {
        this.variableName = variableName;
    }

    public String getVariableValue()
    {
        return variableValue;
    }

    public void setVariableValue(String variableValue)
    {
        this.variableValue = variableValue;
    }
}

Now I want to use a criteria query to load the entire table (ie "select * from variables"). I'd like to use a criteria query more for code consistency than anything else. I get this exception though:

java.lang.IllegalStateException: No criteria query roots were specified
    at org.hibernate.ejb.criteria.CriteriaQueryImpl.validate(CriteriaQueryImpl.java:303)
    at org.hibernate.ejb.criteria.CriteriaQueryCompiler.compile(CriteriaQueryCompiler.java:145)
    at org.hibernate.ejb.AbstractEntityManagerImpl.createQuery(AbstractEntityManagerImpl.java:437

The query I am using is:

public List<Variable> loadAllVariables()
{
    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Variable> query = builder.createQuery(Variable.class);

    return em.createQuery(query).getResultList();
} 

I know that the exception means it wants this:

Root<Variable> variableRoot = query.from(Variable.class);

But without a Predicate I don't see how to get the Root object into the query?

Answer

Mikko Maunu picture Mikko Maunu · Jul 26, 2012

I am not sure, did I understood you right, but if goal is to choose fetch list of all Variable entities, then following is way to go:

public List<Variable> loadAllVariables() {
    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Variable> query = builder.createQuery(Variable.class);
    Root<Variable> variableRoot = query.from(Variable.class);
    query.select(variableRoot);

    return em.createQuery(query).getResultList();
} 

Difference is that select is used. All implementations do not implicitly use last call of from in place of select. In JPA 2.0 specification this is told as follows:

Portable applications should use the select or multiselect method to specify the query’s selection list. Applications that do not use one of these methods will not be portable.