java - java.lang.IllegalArgumentException：参数值 [1604438222] 与预期类型不匹配 [java.lang.Integer (n/a)

Question

我有一个独立的应用程序，在调用“ http://localhost:8080/user/person/1604438222 ”时出现错误。（试图从数据库中获取人员 ID 详细信息）

2019-03-01 11:23:34.296 错误 52000 --- [nio-8080-exec-1] oaccC[.[.[/].[dispatcherServlet]：Servlet.service() 用于 servlet [dispatcherServlet] path [] 抛出异常 [请求处理失败；嵌套异常是 org.springframework.dao.InvalidDataAccessApiUsageException: Parameter value [1604438222] did not match expected type [java.lang.Integer (n/a)]; 嵌套异常是 java.lang.IllegalArgumentException：参数值 [1604438222] 与预期类型 [java.lang.Integer (n/a)]] 不匹配，根本原因

java.lang.IllegalArgumentException：参数值 [1604438222] 与 org.hibernate.query.spi.QueryParameterBindingValidator.validate(QueryParameterBindingValidator.java:54) 处的预期类型 [java.lang.Integer (n/a)] 不匹配~[hibernate -core-5.3.7.Final.jar:5.3.7.Final] 在 org.hibernate.query.spi.QueryParameterBindingValidator.validate(QueryParameterBindingValidator.java:27) ~[hibernate-core-5.3.7.Final.jar: 5.3.7.决赛]

1. 控制器类：

   @RestController
   @RequestMapping("/user")
   public class UserController {
   
   
       @Autowired
       PersonRepository personRepository; 
   
        @GetMapping("/person/{prsId}")
        public Person getPerson(@PathVariable Long prsId) {
             return personRepository.findByPrsId(prsId);
        }
   }
   

2. 存储库：

   @Repository
   public interface PersonRepository extends  CrudRepository<Person, Long>
   {
       Person findByPrsId(Long Id);
   
   }
   

3. POJO

   @Entity
   @Table(name = "Person")
   public class Person {
   
       @Column(name = "Prs_Id")
       @Id
       @GeneratedValue(strategy = GenerationType.AUTO)
       private Integer prsId;
   
       @Column(name = "P_Email_Internet_Addr", nullable = true, length = 255)
       private String email;
   
   
       protected Person() {
       }
   
   
       public Integer getPrsId() {
           return prsId;
       }
   
       public void setPrsId(Integer prsId) {
           this.prsId = prsId;
       }
   
       public String getEmail() {
           return email;
       }
   
       public void setEmail(String email) {
           this.email = email;
       }
   
   
       @Override
           public String toString() {
                   return "User [prsId=" + prsId + ", email=" + email 
                    + "]";
           }
       }
   

标签： javaspringspring-bootspring-mvcspring-boot-jpa

Answer 1

2,147,483,647是值Integer.MAX_VALUE

2,304,438,636太大而不能包含在Integer变量中

Long一般用于您的 ID 。或者甚至更好，long因为您的 ID 理想情况下不应该是原语null