JPA #Id annoation - jpa

Question

i want to insert into a table with a specified value,but it just don't work,
here is my code:
#Id
#Column(insertable=true,updatable=true)
public Long getS_id() {
return s_id;
}
#Resource(name="studentService")
private StudentService stus;
Student student = new Student();
student.setS_id(123213L);
student.setName("vincent");
stus.add(student);
If I change:
#Id
#Column(insertable=true,updatable=true)
public Long getS_id() {
return s_id;
}
to this:
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(insertable=true,updatable=true)
public Long getS_id() {
return s_id;
}
and don't set s_id manualy it works well.
here my student class
#Entity()
#Table(name="stu_info")
public class Student implements Serializable{
private static final long serialVersionUID = 1L;
/**
* 学生的学号
*/
private Long s_id;
/**
* 学生姓名
*/
private String name;
/**
* 学生性别
*/
private String sex;
/**
* 学生生日
*/
private Date birthday;
/**
* 学生电话号码
*/
private String telephone;
/**
* 学生所在年级
*/
private String grade;
/**
* 学生所在班级
*/
private String classes;
/**
* 学生编号
*/
private int number;
/**
* 学生父亲姓名
*/
private String father_name;
/**
* 学生母亲姓名
*/
private String mother_name;
/**
* 学生个人疾病史
*/
private String diseases_history;
#Id
#Column(insertable=true,updatable=true)
public Long getS_id() {
return s_id;
}
public void setS_id(Long s_id) {
this.s_id = s_id;
}
#Column(length=32)
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Column(length=12)
public String getSex() {
return sex;
}
public void setSex(String sex) {
this.sex = sex;
}
#Temporal(TemporalType.DATE)
public Date getBirthday() {
return birthday;
}
public void setBirthday(Date birthday) {
this.birthday = birthday;
}
#Column(length=12)
public String getTelephone() {
return telephone;
}
public void setTelephone(String telephone) {
this.telephone = telephone;
}
#Column(length=32)
public String getGrade() {
return grade;
}
public void setGrade(String grade) {
this.grade = grade;
}
#Column(length=32)
public String getClasses() {
return classes;
}
public void setClasses(String classes) {
this.classes = classes;
}
#Column(length=32)
public int getNumber() {
return number;
}
public void setNumber(int number) {
this.number = number;
}
#Column(length=32)
public String getFather_name() {
return father_name;
}
public void setFather_name(String father_name) {
this.father_name = father_name;
}
#Column(length=32)
public String getMother_name() {
return mother_name;
}
public void setMother_name(String mother_name) {
this.mother_name = mother_name;
}
#Column(length=32)
public String getDiseases_history() {
return diseases_history;
}
public void setDiseases_history(String diseases_history) {
this.diseases_history = diseases_history;
}
}

Answer 1

From the limited information posted I would guess you are using SQL Server and to insert a record into an SQLServer table with an explicit value defined for an Identity column requires you to turn on identity inserts for that table.
https://msdn.microsoft.com/en-gb/library/ms188059.aspx
So if you run the above against your table you should then be able to persist using a specific value.
So not really anything to do with JPA.