使用外键在 2 个表上建立关联时出现 Criteria 问题

Question

我正在尝试了解休眠和标准。 我有 2 张桌子 RATS 和 SICKNESS。 我在 RATS 中设置了一个外键：RATS.Sickness_Id = SICKNESS.ID。

我正在尝试使用 Criteria 获得等效的 SQL：

select * from RATS r, SICKNESS s where s.id = r.sickness_id

我以为是这种关联：

session
.createCriteria(Rats.class)
.createCriteria(Sickness.class)
.toList()

不幸的是，结果如下： org.hibernate.QueryException：无法解析属性：实体：entities.Rats

奇怪的是两者都有：

session.createCriteria(Rats.class).toList()

和

    session.createCriteria(Sickness.class).toList()

work fine....

I'm a bit puzzled.
Here are my entities classes code:

@Entity
@Table(name = "RATS")
public class Rats implements java.io.Serializable {

    private int id;
    private Sickness sickness;
    private String name;
    private int age;

    public Rats() {
    }

    public Rats(int id, String name, int age) {
        this.id = id;
        this.name = name;
        this.age = age;
    }

    public Rats(int id, Sickness sickness, String name, int age) {
        this.id = id;
        this.sickness = sickness;
        this.name = name;
        this.age = age;
    }

    @Id
    @Column(name = "ID", unique = true, nullable = false)
    public int getId() {
        return this.id;
    }

    public void setId(int id) {
        this.id = id;
    }

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "Sickness_Id")
    public Sickness getSickness() {
        return this.sickness;
    }

    public void setSickness(Sickness sickness) {
        this.sickness = sickness;
    }

    @Column(name = "Name", nullable = false, length = 50)
    public String getName() {
        return this.name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @Column(name = "Age", nullable = false)
    public int getAge() {
        return this.age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public String toString() {
        String returnString =   "My name is "   + getName() + ", I am " + getAge()+ ". ";
        returnString += getSickness() == null ? "I am healthy like hell! :)" :  "I suffer from " + getSickness().getNom();
        return returnString;
    }




}

和

@Entity
@Table(name = "SICKNESS")
public class Sickness implements java.io.Serializable {

    private int id;
    private String nom;
    private Set<Rats> ratses = new HashSet<Rats>(0);

    public Sickness() {
    }

    public Sickness(int id) {
        this.id = id;
    }

    public Sickness(int id, String nom, Set<Rats> ratses) {
        this.id = id;
        this.nom = nom;
        this.ratses = ratses;
    }

    @Id
    @Column(name = "Id", unique = true, nullable = false)
    public int getId() {
        return this.id;
    }

    public void setId(int id) {
        this.id = id;
    }

    @Column(name = "Nom", length = 50)
    public String getNom() {
        return this.nom;
    }

    public void setNom(String nom) {
        this.nom = nom;
    }

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "sickness")
    public Set<Rats> getRatses() {
        return this.ratses;
    }

    public void setRatses(Set<Rats> ratses) {
        this.ratses = ratses;
    }

    @Override
    public String toString() {
        return getNom()
                +           (   getRatses() != null ? (", getRatses()=" + getRatses() + "]"): "" );
    }



}

我错过了什么？ 提前致谢。

Answer 1

在 Rats 实体上，Sickness 实体属性为：

 private Sickness sickness;

因此，您的关联必须使用相同的名称。

session.createCriteria(Rats.class)
    .createCriteria("sickness")
    .list();

另一种解决方案，应该是更改 Rats 以使用 EAGER Fetch：

@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "Sickness_Id")
public Sickness getSickness() {
    return this.sickness;
}

public void setSickness(Sickness sickness) {
    this.sickness = sickness;
}

然后使用：

session.createCriteria(Rats.class)
    .list();