java - 无法在 hibernate 状态下保存实体

Question

我创建了简单的 CRUD 服务。有 4 个实体:客户、提供商、产品、交易。

Customer 和 Provider 实体已将 id AppId 组成为具有以下结构:

@Getter
@Setter
@Embeddable
@NoArgsConstructor
public class AppId implements Serializable {
    private String app;
    private String id;
    //...
}

这是我想要的业务逻辑: 提供者实体级联并创建产品实体。 当客户与提供商进行交易时，我需要创建实体 Deal，它不会级联任何其他实体。 它只有涉及交易的提供商、客户和产品的字段。

我创建了一些提供商和客户。 然后我尝试创建交易，但我得到了客户和提供商字段 null。

这是我的实体定义:

<小时/> 提供者:

@Entity
@Getter
@Setter
@ToString
@NoArgsConstructor
@Table(name = "provider")
public class Provider implements Serializable {

    @EmbeddedId
    @Column(name = "appid")
    private AppId appId;

    @Column(name = "name")
    private String name;

    @Column(name = "firstname")
    private String firstName;

    @Column(name = "lastname")
    private String lastName;

    @Column(name = "latitude")
    private float latitude;

    @Column(name = "longitude")
    private float longitude;

    @Column(name = "work_date")
    private Date workDate;

    @ManyToMany(cascade = CascadeType.ALL)
    @JoinTable(name = "provider_product"
            , joinColumns = {
                @JoinColumn(name = "provider_app"),
                @JoinColumn(name = "provider_id")
            }
            , inverseJoinColumns = @JoinColumn(name="product_id"))
    private Set<Product> products;

    @OneToMany(cascade = CascadeType.ALL)
    @JoinColumns({
            @JoinColumn(name = "app", referencedColumnName = "app", updatable = false, insertable = false),
            @JoinColumn(name = "id", referencedColumnName = "id", updatable = false, insertable = false)
    })
    private List<Deal> dealList = new ArrayList<>();
}

<小时/> 客户:

@Entity
@Getter
@Setter
@ToString
@NoArgsConstructor
@Table(name = "customer")
public class Customer implements Serializable {

    @EmbeddedId
    @Column(name = "appid")
    private AppId appId;

    @Column(name = "firstname")
    private String firstName;

    @Column(name = "lastname")
    private String lastName;

    public Customer(AppId appId, String firstName, String lastName) {
        this.appId = appId;
        this.firstName = firstName;
        this.lastName = lastName;
    }
}

<小时/> 产品:

@Entity
@Getter
@Setter
@ToString
@NoArgsConstructor
@Table(name = "product")
public class Product implements Serializable {

    @Id
    @GeneratedValue
    private long id;

    @Column(name = "name")
    private String name;

    @Column(name = "cost")
    private long cost;
}

<小时/> 交易:

@Entity
@Getter
@Setter
@ToString
@NoArgsConstructor
@Table(name = "deal")
public class Deal implements Serializable {

    @Id
    @GeneratedValue
    private long id;

    @ManyToOne
    @JoinColumns({
            @JoinColumn(name = "provider_app", referencedColumnName = "app", insertable = false, updatable = false),
            @JoinColumn(name = "provider_id", referencedColumnName = "id", insertable = false, updatable = false)
    })
    private Provider provider;

    @ManyToOne
    @JoinColumns({
            @JoinColumn(name = "customer_app", insertable = false, updatable = false),
            @JoinColumn(name = "customer_id", insertable = false, updatable = false)
    })
    private Customer customer;

    @ManyToMany
    @JoinTable(name = "deal_product"
            , joinColumns = @JoinColumn(name="deal_id", insertable = false, updatable = false)
            , inverseJoinColumns = @JoinColumn(name="product_id", insertable = false, updatable = false))
    private Set<Product> product;

    // deal is complete when provider entered deal id
    @Column(name = "closed")
    private boolean closed = false;
}

Answer 1

通过删除 insertable = false对于 customer和provider Deal 中的字段实体，一切正常。

    {
    "id": 5,
    "provider": {
        "appId": {
            "app": "vk",
            "id": "123"
        },
        "name": null,
        "firstName": null,
        "lastName": null,
        "latitude": 0,
        "longitude": 0,
        "workDate": null,
        "products": null,
        "dealList": []
    },
    "customer": {
        "appId": {
            "app": "vk",
            "id": "123"
        },
        "firstName": null,
        "lastName": null
    },
    "product": [
        {
            "id": 2,
            "name": "Temp",
            "cost": 100
        }
    ],
    "closed": false
}

我可以得到以下响应。

insertable = false在字段上意味着当您保存实体时，您不会保存该字段的值，而是会在某处显式设置该字段。

insertable = true并不意味着您将创建一个新的 Customer或Provider ，由 CascadeType 处理