如何解决嵌套异常为java.lang.ClassCastException:无法强制转换java.lang.String
我是spring-data的新手,我遇到此错误java.lang.String cannot be cast to com.example.accessingdataMysqL.User
,不知道如何解决!我添加了代码的各个相关部分。
Main.Controller
package com.example.accessingdataMysqL;
import java.util.List;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Controller;
import org.springframework.transaction.annotation.Transactional;
import org.springframework.web.bind.annotation.GetMapping;
import org.springframework.web.bind.annotation.PostMapping;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestParam;
import org.springframework.web.bind.annotation.ResponseBody;
@Controller
@RequestMapping(path="/demo")
public class MainController {
@Autowired
private UserRepository userRepository;
@Transactional
@PostMapping(path="/add")
public @ResponseBody String addNewUser (@RequestParam String name,@RequestParam String email,@RequestParam String surname)
{
User n = new User();
n.setName(name);
n.setSurname(surname);
n.setEmail(email);
userRepository.create(n);
return "Saved";
}
@GetMapping(path="first")
User one(@RequestParam String name) {
return userRepository.findFirstByName(name);
}
}
User.java
package com.example.accessingdataMysqL;
import java.sql.Timestamp;
import java.time.Instant;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
@Entity
public class User {
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Integer id;
public String name;
private String email;
private String surname;
@Column(name="stmp",columnDeFinition = "TIMESTAMP (6)")
Timestamp timestamp = Timestamp.from(Instant.Now());
public void setTimestamp(Timestamp timestamp) {
this.timestamp = timestamp;
}
public Timestamp getTimestamp() {
return timestamp;
}
public String getSurname() {
return surname;
}
public void setSurname(String surname) {
this.surname = surname;
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getEmail() {
return email;
}
public void setEmail(String email) {
this.email = email;
}
}
RepoImpl.java
package com.example.accessingdataMysqL;
import java.util.List;
import javax.persistence.EntityManager;
import javax.persistence.TypedQuery;
import javax.persistence.criteria.CriteriaBuilder;
import javax.persistence.criteria.CriteriaQuery;
import javax.persistence.criteria.Root;
import org.springframework.stereotype.Component;
@Component
public class UserRepositoryImpl implements UserRepository {
private final EntityManager em;
public UserRepositoryImpl(EntityManager entityManager) {
this.em = entityManager;
}
@Override
public User findFirstByName(String name) {
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<User> criteria = builder.createquery(User.class);
Root<User> root = criteria.from(User.class);
criteria.select(root.get("name"));
criteria.orderBy(builder.asc(root.get("timestamp")));
TypedQuery<User> query = em.createquery(criteria).setMaxResults(1);
return query.getSingleResult();
}
@Override
// per la creazione//
public void create(User entity) {
em.persist(entity);
}
}
[Request processing Failed; nested exception is java.lang.classCastException: java.lang.String cannot be cast to com.example.accessingdataMysqL.User] with root cause
java.lang.classCastException: java.lang.String cannot be cast to com.example.accessingdataMysqL.User
at com.example.accessingdataMysqL.UserRepositoryImpl.findFirstByName(UserRepositoryImpl.java:35) ~[classes!/:0.0.1-SNAPSHOT]
解决方法
如果要按名称查找User
,则应该在criteria.select
中而不是criteria.where
中设置过滤器参数:
public User findFirstByName(String name) {
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<User> criteria = builder.createQuery(User.class);
Root<User> root = criteria.from(User.class); // root is User
criteria.select(root).where(builder.equal(root.get("name"),name));
criteria.orderBy(builder.asc(root.get("timestamp")));
TypedQuery<User> query = em.createQuery(criteria).setMaxResults(1);
return query.getSingleResult();
}
criteria.select(root.get("name"));
表示仅选择并返回列"name"
,即应返回第一个用户的名称。
如果需要此类信息,可以通过以下方式检索:
public String findFirstUserName() {
CriteriaBuilder builder = em.getCriteriaBuilder();
CriteriaQuery<String> criteria = builder.createQuery(String.class);
Root<User> root = criteria.from(User.class); // root is User
criteria.select(root.get("name")); // getting name
criteria.orderBy(builder.asc(root.get("timestamp"))); // of the first/earliest user
TypedQuery<String> query = em.createQuery(criteria).setMaxResults(1);
return query.getSingleResult();
}
,
按照错误提示进行操作-
CriteriaQuery<String> criteria = builder.createQuery(String.class);
代替
CriteriaQuery<User> criteria = builder.createQuery(User.class);
在此处Reference
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