JPA条件查询加载整个表

Question

我觉得这是一个愚蠢的问题，但我找不到答案。我的课如下:

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;
    }
}

现在，我想使用条件查询来加载整个表(即“从变量中选择*”)。我想更多地使用条件查询来确保代码的一致性，而不是其他任何事情。我得到这个异常:

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

我正在使用的查询是:

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

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

我知道异常意味着它想要这样:

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

但是没有谓词，我看不到如何将Root对象放入查询中？

Answer 1

我不确定，我是否理解正确，但是如果目标是选择所有Variable实体的获取列表，那么可以采取以下方法:

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();
} 

区别在于使用了select。所有实现都不会隐式地使用from的最后一次调用代替select。在JPA 2.0规范中，这被告知如下:

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.