為什么 Criteria API 會拋出 StackOverFlow 異常？

Question

我有 2 個實體，它們之間存在一對一的關系：客戶和購物車。 Customer 是 cart 的所有者，因此 FK 的一列是指向 Cart PK 的 Customer 表。

這是購物車 class：

@Entity
@Table(name = "testCart_cart")
public class Cart {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "cart_id", unique = true, nullable = false)
    private long id;

    @OneToMany(cascade = CascadeType.ALL,
            orphanRemoval = true,
            fetch = FetchType.LAZY)
    @JoinColumn(name = "cart_id")
    private List<CartItem> listOfCartItems;

    @OneToOne(mappedBy = "cart",
            fetch = FetchType.LAZY)
    private Customer customer;

    public long getId() {
        return id;
    }

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

    public List<CartItem> getListOfCartItems() {
        return listOfCartItems;
    }

    public void setListOfCartItems(List<CartItem> listOfCartItems) {
        this.listOfCartItems = listOfCartItems;
    }

    public void setCustomer(Customer customer) { this.customer = customer; }

    public Customer getCustomer() { return customer; }

    @Override
    public String toString() {
        return "Cart{" +
                "id=" + id +
                ", listOfCartItems=" + listOfCartItems +
                ", customer=" + customer +
                '}';
    }
}

這是我的客戶 class：

@Entity
@Table(name = "testCart_customer")
public class Customer implements Serializable {

    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Id
    @Column(name = "id", unique = true, nullable = false)
    private long id;

    @Column(name = "username", unique = true, nullable = false)
    private String username;

    @OneToOne(cascade = CascadeType.ALL,
            fetch = FetchType.LAZY)
    @JoinColumn(name = "cart_id")
    private Cart cart;

    public long getId() {
        return id;
    }

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

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    public Cart getCart() {
        return cart;
    }

    public void setCart(Cart cart) {
        this.cart = cart;
    }

    @Override
    public String toString() {
        return "Costumer name: " + username;
    }
}

所以，我用標准 API 代替 JPQL（對我來說更容易）。
這是我的代碼：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Cart> query = cb.createQuery(Cart.class);
Root<Customer> customer = query.from(Customer.class);
ParameterExpression<String> strignParameter = cb.parameter(String.class);
Join<Customer, Cart> carts = customer.join("cart");
query.select(carts).where(cb.equal(customer.get("username"),name));
Cart cart = entityManager.createQuery(query).getSingleResult();

return cart;

這是在控制台中：

Exception in thread "main" java.lang.StackOverflowError
    at org.hibernate.query.sqm.tree.domain.AbstractSqmAttributeJoin.getNodeJavaTypeDescriptor(AbstractSqmAttributeJoin.java:71)
    at org.hibernate.query.sqm.tree.domain.SqmSingularJoin.getJavaTypeDescriptor(SqmSingularJoin.java:52)
    at org.hibernate.query.sqm.tree.domain.AbstractSqmAttributeJoin.getNodeJavaTypeDescriptor(AbstractSqmAttributeJoin.java:71)
    at org.hibernate.query.sqm.tree.domain.SqmSingularJoin.getJavaTypeDescriptor(SqmSingularJoin.java:52)
    at org.hibernate.query.sqm.tree.domain.AbstractSqmAttributeJoin.getNodeJavaTypeDescriptor(AbstractSqmAttributeJoin.java:71)
    at org.hibernate.query.sqm.tree.domain.SqmSingularJoin.getJavaTypeDescriptor(SqmSingularJoin.java:52)

總共有 1024 個電話。
嘗試了來自多個來源的不同方法，並嘗試使用調試器 go 以查看發生了什么，機器人沒有真正解釋。 當調試器從query.select...跳轉到Cart cart...時會拋出錯誤。

Answer 1

當你想要購物車時，為什么要從客戶開始。

試試這個：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Cart> query = cb.createQuery(Cart.class);
Root<Cart> cart = query.from(Cart.class);
ParameterExpression<String> strignParameter = cb.parameter(String.class);
Join<Cart, Customer> customer = cart.join("customer");
query.select(cart).where(cb.equal(customer.get("username"), name));
Cart cart = entityManager.createQuery(query).getSingleResult();