java - Jersey JAX-RS 自动生成的网络服务和 MSSQL 身份字段

Question

我无法使用 Netbeans 自动生成的 Jersey RESTful 网络服务。

当像这样发布 JSON 时，

{"name":"Some New Site"}

POST 查询返回这个，

Internal Exception: com.microsoft.sqlserver.jdbc.SQLServerException: Cannot insert
explicit value for identity column in table 'site' when IDENTITY_INSERT is set to OFF.
Error Code: 544
Call: INSERT INTO site (id, name) VALUES (?, ?)
    bind => [2 parameters bound]

与 MySQL 不同，MSSQL 在 INSERT 查询中传递 NULL 值时不会生成新的自增 ID；它需要根本不提及的 ID 字段为其生成值。

(IDENTITY_INSERT 也不会生成 ID)

我如何教我的 Jersey 服务省略 INSERT 查询中的 id 字段？

编辑:这是 POJO，它是由 Netbeans 使用“来自数据库的 Restful Web 服务”向导为我自动生成的。 (我尽量少原创)

/*
 * To change this template, choose Tools | Templates
 * and open the template in the editor.
 */
package ru.fedd.entities;

import java.io.Serializable;
import java.util.Collection;
import javax.persistence.Basic;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.NamedQueries;
import javax.persistence.NamedQuery;
import javax.persistence.OneToMany;
import javax.persistence.Table;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlTransient;

/**
 *
 * @author fkravchenko
 */
@Entity
@Table(name = "site")
@XmlRootElement
@NamedQueries({
    @NamedQuery(name = "Site.findAll", query = "SELECT s FROM Site s"),
    @NamedQuery(name = "Site.findById", query = "SELECT s FROM Site s WHERE s.id = :id"),
    @NamedQuery(name = "Site.findByName", query = "SELECT s FROM Site s WHERE s.name = :name")})
public class Site implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Basic(optional = false)
    @Column(name = "id")
    private Long id;
    @Basic(optional = false)
    @Column(name = "name")
    private String name;
    @OneToMany(cascade = CascadeType.ALL, mappedBy = "siteId")
    private Collection<Document> documentCollection;

    public Site() {
    }

    public Site(Long id) {
        this.id = id;
    }

    public Site(Long id, String name) {
        this.id = id;
        this.name = name;
    }

    public Long getId() {
        return id;
    }

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

    public String getName() {
        return name;
    }

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

    @XmlTransient
    public Collection<Document> getDocumentCollection() {
        return documentCollection;
    }

    public void setDocumentCollection(Collection<Document> documentCollection) {
        this.documentCollection = documentCollection;
    }

    @Override
    public int hashCode() {
        int hash = 0;
        hash += (id != null ? id.hashCode() : 0);
        return hash;
    }

    @Override
    public boolean equals(Object object) {
        // TODO: Warning - this method won't work in the case the id fields are not set
        if (!(object instanceof Site)) {
            return false;
        }
        Site other = (Site) object;
        if ((this.id == null && other.id != null) || (this.id != null && !this.id.equals(other.id))) {
            return false;
        }
        return true;
    }

    @Override
    public String toString() {
        return "ru.fedd.entities.Site[ id=" + id + " ]";
    }

}

Answer 1

你能在这里分享你的 POJO 类吗？

您可能必须使用 @GeneratedValue 注释 POJO 类的 id 以通知它是自动生成的。然后 JPA 将负责插入。