Question

我的Spring启动有问题。我创建了Entity和Repository，但Repository中的方法findByName不起作用。我的网址：

http://localhost:8080/student/search/findByName?name=Artem 在Google Chrome中：找不到localhost，但搜索已映射。

实体：

@Getter @Setter
@Entity @Table(name = "Student")

public class Student extends BaseEntity{
    private String name;
    private String dateOfBirthDay;
    private String sex;
    private String phoneNumber;
}

BaseEntity：

@Getter
@Setter
@MappedSuperclass
public class BaseEntity {
    @Id @GeneratedValue(strategy = GenerationType.SEQUENCE) @Column protected Long id;

我的存储库：

@RepositoryRestResource(collectionResourceRel = "student", path = "student")
public interface StudentRepository extends PagingAndSortingRepository<Student, Long> {
    Student findByName(@Param("name") String name);
}

应用：

@SpringBootApplication
@EnableTransactionManagement

public class Application {
    public static void main(String[] args) {
        SpringApplication.run(Application.class, args);
    }
}

Application.yaml：

spring:
    application:
        name: students
    datasource:
        driverClassName: org.postgresql.Driver
        url: jdbc:postgresql://localhost:5432/students
        username: postgres
        password: postgres
    jpa:
      hibernate:
        ddl-auto: update
server:
    port: 8080

Answer 1

我建议你创建一个资源类，就像控制器一样。这是一个简单的例子：

@RestController
@RequestMapping("/yourPath") //students, whatever
public class StudentsResource {

@Autowired
private StudentRepository studentRepository;

//type media that you want to show (json, xml...in this case is JSON)
@RequestMapping(method = RequestMethod.GET, produces = { MediaType.APPLICATION_JSON_VALUE })

// <Student> is the entity, object
@RequestMapping(value = "/yourPath/{studentName}")
public ResponseEntity<Student> findByName(@pathVariable("studentName") String name) {
    Student student = studentRepository.findByName(name);

    if(student == null){
     //handler your own exception here
    }

    //show the student as json object
    return ResponseEntity.status(HttpStatus.OK).body(student);
}

注意：这是资源类。但是你的问题是关于localhost，所以如果你使用Spring Boot，看看你的＆＃34; application.properties＆＃34;是正确的。这是我的例子：

spring.datasource.url=jdbc:mysql://localhost:3306/yourDataBase
spring.datasource.username=yourUser
spring.datasource.password=yourPassword

spring.jpa.hibernate.ddl-auto=update //makes the spring create the database automatic!

findByName不起作用[Spring boot]

1 个答案: