如何使用JPA通過外鍵搜索mySql數據庫表？

Question

我有2個類（BusinessAccount和Projects（如下所示）映射到MySql數據庫），其中BusinessAccounts和Projects之間存在1：M關系。 我成功地將數據插入數據庫，但在查詢數據庫時遇到了問題。 我遇到的問題是我沒有外鍵的getter或setter，在Projects類中沒有'contractor_id'。 我想要執行的查詢是通過在Projects表中搜索外鍵引用來返回給定BusinessAccount的所有項目的名稱列表。 我可以在mySQL中做到這一點沒有問題，但由於沒有引用contractor_id作為Projects類中的java實體，我不知道如何在我的java類中進行搜索。 （注意：我嘗試在Projects類中聲明外鍵以及getter和setter，但是因為我已經通過類中的1：Many關系映射了這些，所以它們不會被編譯，因為它們被標記為重復實體。）我確信這是顯而易見的，我很遺憾，但是非常感謝任何幫助！

public List<Projects> getProjectList() {
    factory = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
    EntityManager em = factory.createEntityManager();   
    List<Projects> projectList = new ArrayList<Projects>();

        em.getTransaction().begin();

        String sessionEmail=Util.getEmail();
        Query myQuery = em.createQuery("SELECT u FROM BusinessAccount u WHERE u.email=:email");
        myQuery.setParameter("email", sessionEmail);

        List<BusinessAccount> userList=myQuery.getResultList();
        BusinessAccount account =userList.get(0);

        Query myQuery2 = em.createQuery("SELECT distinct p.* FROM BusinessAccount u "
                + "INNER JOIN Projects p ON p.contractor_id=:userID");

/*Note p.contractor_id above refers to the entity in the 
mysql database (and won't work obviously), I want to refer 
to it's java equivalent but am not sure how to do that*/

        myQuery2.setParameter("userID", account.getId());
        projectList=myQuery2.getResultList();
        em.getTransaction().commit();
        em.close();

        return projectList;

    } 


@Entity
@Table(name = "business_accounts")
public class BusinessAccount {

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private int id;

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

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

        @OneToMany(mappedBy = "businessAccount", fetch = FetchType.EAGER, cascade = { CascadeType.ALL })
    private List<Projects> projects;

    public int getId() {
    return id;
    }

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

    public String getFirstName() {
    return firstName;
    }

    public void setFirstName(String firstName) {
    this.firstName = firstName;
    }

    public String getSurname() {
    return surname;
    }

       public List<Projects> getProjects()
    {
    if (projects == null)
    {
        projects = new ArrayList<Projects>();
    }

    return projects;
    }

   public void setProjects(List<Projects> projects)
    {
    this.projects = projects;
    }

    }


@Entity
@Table(name = "projects")
public class Projects {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int project_id;

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

@ManyToOne(fetch = FetchType.LAZY)
@JoinColumns({ @JoinColumn(name = "contractor_id", referencedColumnName="id") })
private BusinessAccount businessAccount;


public BusinessAccount getBusinessAccount() {
    if (businessAccount == null) {
        businessAccount = new BusinessAccount();
    }
    return businessAccount;
}

public void setBusinessAccount(BusinessAccount businessAccount) {
    this.businessAccount = businessAccount;
}

public int getProject_id() {
    return project_id;
}

public void setProject_id(int project_id) {
    this.project_id = project_id;
}

public String getProjectName() {
    return projectName;
}

public void setProjectName(String projectName) {
    this.projectName = projectName;
}

}

Answer 1

JPA查詢就像（你需要使用關系屬性，但不需要外鍵本身 - 請嘗試，它可能需要一些調整）：

SELECT p FROM BusinessAccount u, IN(u.projects) p WHERE u.id=:userId

但你真的需要查詢嗎？ 您可以從酒店獲取相關項目：

BusinessAccount account = ...
List<Projects> projectList = account.getProjects();

Answer 2

嘗試這個：

Query myQuery2 = em.createQuery("SELECT distinct p.* FROM BusinessAccount u "
            + "INNER JOIN Projects p ON p.businessAccount=:businessAccount");

myQuery2.setParameter("businessAccount", account);