我用Spring Data 开发了一个Spring引导应用程序JPA有人能解决这个问题吗我的豆子:
@Entity
@Table(name = "employee", catalog = "explorerrh")
public class Employee implements java.io.Serializable {
/**
*
*/
private static final long serialVersionUID = 1L;
private Integer idemployee;
private String employeeName;
private String employeeLastName;
private String employeeCin;
private String employeePhone;
private String employeeAdress;
private String employeePost;
private String employeeCnss;
private String employeeCv;
private String employeePhoto;
private String employeeSalaire;
public Employee() {
}
public Employee(String employeeName, String employeeLastName,
String employeeCin, String employeePhone, String employeeAdress,
String employeePost, String employeeCnss, String employeeCv,
String employeePhoto, String employeeSalaire) {
this.employeeName = employeeName;
this.employeeLastName = employeeLastName;
this.employeeCin = employeeCin;
this.employeePhone = employeePhone;
this.employeeAdress = employeeAdress;
this.employeePost = employeePost;
this.employeeCnss = employeeCnss;
this.employeeCv = employeeCv;
this.employeePhoto = employeePhoto;
this.employeeSalaire = employeeSalaire;
}
@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "idemployee", unique = true, nullable = false)
public Integer getIdemployee() {
return this.idemployee;
}
public void setIdemployee(Integer idemployee) {
this.idemployee = idemployee;
}
@Column(name = "employeeName", length = 45)
public String getEmployeeName() {
return this.employeeName;
}
public void setEmployeeName(String employeeName) {
this.employeeName = employeeName;
}
@Column(name = "employeeLastName", length = 45)
public String getEmployeeLastName() {
return this.employeeLastName;
}
public void setEmployeeLastName(String employeeLastName) {
this.employeeLastName = employeeLastName;
}
@Column(name = "employeeCIN", length = 45)
public String getEmployeeCin() {
return this.employeeCin;
}
public void setEmployeeCin(String employeeCin) {
this.employeeCin = employeeCin;
}
@Column(name = "employeePhone", length = 45)
public String getEmployeePhone() {
return this.employeePhone;
}
public void setEmployeePhone(String employeePhone) {
this.employeePhone = employeePhone;
}
@Column(name = "employeeAdress", length = 45)
public String getEmployeeAdress() {
return this.employeeAdress;
}
public void setEmployeeAdress(String employeeAdress) {
this.employeeAdress = employeeAdress;
}
@Column(name = "employeePost", length = 45)
public String getEmployeePost() {
return this.employeePost;
}
public void setEmployeePost(String employeePost) {
this.employeePost = employeePost;
}
@Column(name = "employeeCNSS", length = 45)
public String getEmployeeCnss() {
return this.employeeCnss;
}
public void setEmployeeCnss(String employeeCnss) {
this.employeeCnss = employeeCnss;
}
@Column(name = "employeeCV", length = 45)
public String getEmployeeCv() {
return this.employeeCv;
}
public void setEmployeeCv(String employeeCv) {
this.employeeCv = employeeCv;
}
@Column(name = "employeePhoto", length = 45)
public String getEmployeePhoto() {
return this.employeePhoto;
}
public void setEmployeePhoto(String employeePhoto) {
this.employeePhoto = employeePhoto;
}
@Column(name = "employeeSalaire", length = 45)
public String getEmployeeSalaire() {
return this.employeeSalaire;
}
public void setEmployeeSalaire(String employeeSalaire) {
this.employeeSalaire = employeeSalaire;
}
}
我的服务是这样的:
interface EmployeeRepository extends Repository<Employee, Long> {
Page<Employee> findAll(Pageable pageable);
Employee findByEmployeeNameAndEmployeeCin(String employeeName, String cin);
}
然后我在com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'employee0_.employee_adress' in 'field list'
上得到这个错误
我的数据库已正确生成,它与bean生成的相同!我真的不明白是什么问题!所以谁能帮我解决这个问题?谢谢你们,如果你还需要其他信息,尽管问我
2条答案
按热度按时间aamkag611#
正如另一篇ij0k1l中指出的那样:
默认情况下,Spring使用
org.springframework.boot.orm.jpa.SpringNamingStrategy
生成表名。这是org.hibernate.cfg.ImprovedNamingStrategy
的一个非常瘦的扩展。该类中的tableName方法传递了一个源String值,但它不知道它是来自@Column.name
属性还是从字段名隐式生成的。ImprovedNamingStrategy将CamelCase转换为SNAKE_CASE,因为EJB3Naming战略性只使用未更改的表名。
如果不想更改命名策略,您可以只指定小写的列名:
@Column(name="testname")
另外,您确定该列是“employeeaddress”而不是“emplyoeeaddress”吗?
5lhxktic2#
确保根据数据库列名正确给出
@Column(name="table_name")
的所有名称。