我正在使用hibernate从db获取json对象。我希望json对象应该包含最近创建的vgis中仅有的前两个的所有细节。但是我不知道当没有使用hql查询时,如何限制结果集,然后根据创建的日期进行排序。它正在从mysql数据库获取所有细节。
import java.util.Date;
import java.util.List;
import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.ManyToOne;
import javax.persistence.OneToMany;
import javax.persistence.OneToOne;
import javax.persistence.Table;
@Entity
@Table(name = "vgi", catalog = "coworkee5")
public class VGI {
@ManyToOne
@JoinColumn(name = "employee_id", referencedColumnName = "id", insertable = false, updatable = false)
private Employees employees;
public Employees getEmployees() {
return employees;
}
public void setEmployees(Employees employees) {
this.employees = employees;
}
@OneToMany(cascade = CascadeType.ALL, targetEntity = VgiGoals.class, mappedBy = "vgi")
private List<VgiGoals> vgiGoals;
public List<VgiGoals> getVgiGoals() {
return vgiGoals;
}
public void setVgi_goals(List<VgiGoals> vgiGoals) {
this.vgiGoals = vgiGoals;
}
public VGI() {
}
@Id
@Column(name = "id")
private String id;
@Column(name = "title")
private String vgi;
@Column(name = "employee_id")
private String employee_id;
@Column(name = "created_on")
private String created;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getvgi() {
return vgi;
}
public void setvgi(String vgi) {
this.vgi = vgi;
}
public String getEmployee_id() {
return employee_id;
}
public void setEmployee_id(String employee_id) {
this.employee_id = employee_id;
}
public String getCreated() {
return created;
}
public void setCreated(String created) {
this.created = created;
}
}
1条答案
按热度按时间a9wyjsp71#
添加
@OrderBy
注解排序如下。注意:用vgigoals类中的日期字段替换createdon。
并限制在查询中使用下面的方法。