Question

我使用spring数据JPA开发了一个Spring启动应用程序有人有这个问题的解决方案吗？：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'

上收到此错误

我的数据库是正确生成的，它对bean的生成是正确的！我真的不明白这是什么问题！所以请有人帮我解决这个问题吗？谢谢你们，如果你需要任何其他信息，请问我

Answer 1

正如另一个answer所指出的那样：

默认情况下Spring使用   生成org.springframework.boot.orm.jpa.SpringNamingStrategy   表名。这是一个非常薄的扩展   org.hibernate.cfg.ImprovedNamingStrategy。其中的tableName方法   class传递一个源String值，但它不知道它是否来   来自@Column.name属性或是否已隐式生成   来自字段名称。

ImprovedNamingStrategy将CamelCase转换为SNAKE_CASE位置   因为EJB3NamingStrategy只使用表名不变。

如果您不想更改命名策略，您可以随时使用   以小写字母指定列名称：

@Column(name="testname")

另外，你确定专栏是＆＃34; employeeadress＆＃34;而不是＆＃34; employeeaddress＆＃34;？

使用Spring数据进行Spring启动JPA：无法提取ResultSet

1 个答案: