Question

据我所知，DB需要一个compareTo方法才能理解这些是同一个对象，所以我为下面的对象写了一个与methd的比较

@JsonIgnoreProperties(ignoreUnknown = true)
@Entity
@Table(name = "networkServiceConfig")
public class NetworkServiceConfig implements Serializable{

    String endpoint;
    String url;
    String host;
    int port;
    String networkServiceName;
    String uuid;

    @JsonIgnore
    @OneToOne(mappedBy="networkServiceConfig")
    Service service;

    private static final long serialVersionUID = 1L;
    @JsonProperty
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    long id;

    public NetworkServiceConfig() {

    }

    public NetworkServiceConfig(String endpoint, String url, String host, int port, String networkServiceName) {
    this.endpoint = endpoint;
    this.url = url;
    this.host = host;
    this.port = port;
    this.networkServiceName = networkServiceName;
    }



    public String getEndpoint() {
    return this.endpoint;
    }

    public void setEndpoint(String endpoint) {
    this.endpoint = endpoint;
    }

    public String getUrl() {
    return this.url;
    }

    public void setUrl(String url) {
    this.url = url;
    }

    public String getHost() {
    return this.host;
    }

    public void setHost(String host) {
    this.host = host;
    }

    public int getPort() {
    return this.port;
    }

    public void setPort(int port) {
    this.port = port;
    }

    public void setNetworkServiceName(String name) { 
    this.networkServiceName = name;

    }

    public String getNetworkServiceName() {
    return this.networkServiceName;
    }

    public String getUuid() {
    return this.uuid;
    }

    public void setUuid(String uuid) {
    this.uuid = uuid;
    }

    /**
     * Implement comparable to compare two NetworkServiceConfig objects for equality
     */
    public int compareTo(NetworkServiceConfig o) {
    if (o == null) {
        return 1;
    }

    if (this.networkServiceName.equals(o.networkServiceName)) {
        return 0;
    }

    int value = (int) (this.id - o.id);

    if (value == 0) {
        return this.networkServiceName.compareTo(o.networkServiceName);
    }

    return value;
    }

    @Override
    public boolean equals(Object obj) {
    NetworkServiceConfig o = (NetworkServiceConfig) obj;
    return this.networkServiceName.equals(o.networkServiceName);
    }




}

网络服务名称对于每个对象都是唯一的。但是每次我将这个对象保存到各种保存调用时，我都会看到20行。请注意，因此我们在不同对象上的属性这是关系

服务对象：

 @OneToOne (cascade= CascadeType.ALL)
    @JoinColumn(name="networkServiceConfig_id")
    @JsonProperty
    NetworkServiceConfig networkServiceConfig;

此外，我没有在表中看到networkServiceConfig_id列

Answer 1

我认为您有点困惑，与Java相比较或类似的实现，在您使用Collections Framework时用于排序对象，但它与Hibernate或JPA的数据库实现没有直接关系

Answer 2

最好的方法是在保存和更新networkServiceConfig之前检查networkServiceName表NetworkServiceConfig重复项，如果networkServiceName已存在则返回422 HTTP状态。这样检查更新有点棘手。

除了你应该在数据库级别添加保护，正如@ xiaofeng.li建议的那样。您可以为networkServiceName添加唯一约束，或将该字段作为主键。

如果您使用NetworkServiceConfig作为参考表，则应删除cascade= CascadeType.ALL。此外，如果多个@ManyToOne可以拥有相同的Service

，您应该使用NetworkServiceConfig并从Service移除NetworkServiceConfig关联

 @ManyToOne
 @JoinColumn(name="networkServiceConfig_id")
 @JsonProperty
 NetworkServiceConfig networkServiceConfig;

数据库正在创建对象的多行

2 个答案: