休眠选择查询不返回任何内容

Posted 2023-04-14

【中文标题】休眠选择查询不返回任何内容

【英文标题】：Hibernate select query returns nothing

【发布时间】：2012-02-23 15:15:37

【问题描述】：

我在使用 hibernate 时遇到了一点问题，因为 hibernate 不接受正常的 sql 查询语法。当我使用 select 语句发送查询时，该语句应该将 37 的精确整数返回到数据库，但我没有得到任何回报。这是 sql 语法中的查询： "select id from tbl_employee where bsn = '36372837'" 这将返回 37。但是当我在 hibernate 中使用所有对象引用执行此查询时，它不起作用。

请检查我的代码，看看你是否知道如何解决问题：

public void RegisterWorkHours(TimeRegistration object)
    
        EntityManagerFactory emf = javax.persistence.Persistence.createEntityManagerFactory("timereg");
        EntityManager em = emf.createEntityManager();
        try
        


            String get_employee_id = "SELECT emp.id FROM Employee as emp WHERE emp.bsn=:bsn";
            Query employee_query = em.createQuery(get_employee_id);
            employee_query.setParameter("bsn", object.getEmployee().getBsn());
            int id = employee_query.getFirstResult();
            System.out.println("query returns employee id: " + id);
            object.getEmployee().setId(id);


            String get_project_id = "SELECT p.projectID FROM Project as p WHERE p.projectname=:projectname";
            Query project_query = em.createQuery(get_project_id); 
            project_query.setParameter("projectname", object.getProject().getProjectname());
            int projectid = project_query.getFirstResult();
            System.out.println("query returns projectid: " + projectid);
            object.getProject().setProjectID(projectid);

            em.getTransaction().begin();



                em.persist(object);
                em.getTransaction().commit();

        
        catch (Exception ex)
        
            System.out.println(ex);
        
    

员工类：

@Entity
@Table(name = "tbl_employee")
public class Employee

    @Id
    @SequenceGenerator(name="employeeSequence", sequenceName="SEQ_EMPLOYEE", allocationSize =1)
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="employeeSequence")
    @Column(name="id")
    private int id;


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

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

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

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

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

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

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

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

    //Constructor
    protected Employee() 

    public Employee(String bsn, String firstname, String lastname)
    
        setBsn(bsn);
        setFirstname(firstname);
        setLastname(lastname);

    
    public Employee(String bsn, String firstname, String lastname, String address, String housenumber)
    
        setBsn(bsn);
        setFirstname(firstname);
        setLastname(lastname);
        setAddress(address);
        setHouseNumber(housenumber);
    
    public Employee(String bsn, String firstname, String lastname, String address, String housenumber, String zip, String city)
    
        setBsn(bsn);
        setFirstname(firstname);
        setLastname(lastname);
        setAddress(address);
        setHouseNumber(housenumber);
        setZip(zip);
        setCity(city);
    

    //The rest is one big list of getters and setters.

时间注册类

@Entity
@Table(name = "tbl_timeregtest")
@Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public class TimeRegistration

    @ManyToOne(cascade = CascadeType.PERSIST, CascadeType.MERGE)
    private Project project;

    @ManyToOne(cascade = CascadeType.PERSIST, CascadeType.MERGE)
    private Employee employee;

    @Id
    @SequenceGenerator(name="timeregSequence", sequenceName="SEQ_TIMEREG", allocationSize =1)
    @GeneratedValue(strategy=GenerationType.SEQUENCE, generator="timeregSequence")
    @Column(name="ID")
    private int ID;

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


    @Column(name="hours")
    private int hours;

    //Constructor
    protected TimeRegistration() 

    public TimeRegistration(Project project, Employee employee, String date, int hours  )
    
        setProject(project);
        setEmployee(employee);
        setDate(date);
        setHours(hours);
    

//the rest is all getter setter stuff

主要空白

public class Main


    public static void main(String [ ] args)
    
        Persistence persistence = new Persistence();
        Project project = new Project("AlphaMouse", "11-2-2013", "12-4-2019");
        Employee employee = new Employee("398723912", "Stoel",  "Stra");    
        TimeRegistration register = new TimeRegistration(project, employee, "21-2-2024", 8);
        persistence.RegisterWorkHours(register) ;

提前致谢， 本杰明

【问题讨论】：

可能需要查看您的 Employee 对象和您设置的映射（或注解）

oke Kevin Crowell 我添加了 Employee 和 TimeRegistration 类，我还添加了对象的主要 void。

【参考方案1】：

您正在使用方法getFirstResult()。 但是这个方法返回记录在表中的位置（整数）。

要检索记录（对象），您应该改用getSingleResult()。

见http://docs.oracle.com/javaee/6/api/javax/persistence/Query.html。

【参考方案2】：

我不是 JPA 人，如果没有意义，请尝试将您的查询包装在事务中

    em.getTransaction().begin();

    String get_employee_id = "SELECT emp.id FROM Employee as emp WHERE emp.bsn=:bsn";
    Query employee_query = em.createQuery(get_employee_id);
    employee_query.setParameter("bsn", object.getEmployee().getBsn());
    long id = employee_query.getFirstResult();
    System.out.println("query returns employee id: " + id);

    em.getTransaction().commit();

【讨论】：

感谢@Sajan 的反馈，这可能是有道理的，但它不起作用:(...我希望其他人可以帮助我。

@B Westra 尝试在查询中的等号和“：”之间添加空格。 "SELECT emp.id FROM Employee as emp WHERE emp.bsn = :bsn";

我试过了，但没用。它仍然返回 0 而应该返回 37。

【参考方案3】：

我找到了解决办法，

    Integer id = (Integer) em.createQuery("select id from Employee where bsn =:bsn")
.setParameter("bsn", object.getEmployee.getBsn())
.getSingleResult();

getSingleResult() 返回一个对象，所以你必须使用 Integer 而不是 int。