参数值 [6] 与 Spring Boot 中的预期类型 [java.lang.Long (n/a)] 不匹配

Question

我是 Spring Boot 的初学者，我有两个具有一对一关系的表。问题是当我尝试通过其用户 ID 获取用户记录时，我遇到了如下异常。

我粘贴了所有代码。我的代码有什么错误？

用户

@Entity
@Table(name = "users")
public class User implements Serializable {
    @Id
    //@GeneratedValue(strategy = GenerationType.IDENTITY)
    @GeneratedValue(strategy= GenerationType.AUTO)
    private Long id;

    @NotNull
    @Size(max = 65)
    @Column(name = "first_name")
    private String firstName;

    @Size(max = 65)
    @Column(name = "last_name")
    private String lastName;

    @NotNull
    @Email
    @Size(max = 100)
    @Column(unique = true)
    private String email;

    @NotNull
    @Size(max = 128)
    private String password;

    @OneToOne(fetch = FetchType.LAZY,
            cascade =  CascadeType.ALL,
            mappedBy = "user")
    private UserProfile userProfile;

    // Hibernate requires a no-arg constructor
    public User() {

    }

    public User(String firstName, String lastName, String email, String password) {
        this.firstName = firstName;
        this.lastName = lastName;
        this.email = email;
        this.password = password;
    }

    // Getters and Setters (Omitted for brevity)
    
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    public String getPassword() {
        return password;
    }

    public void setPassword(String password) {
        this.password = password;
    }

    public UserProfile getUserProfile() {
        return userProfile;
    }

    public void setUserProfile(UserProfile userProfile) {
        this.userProfile = userProfile;
    }
}

用户资料

@Entity
@Table(name = "user_profiles")
public class UserProfile implements Serializable{
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Column(name = "phone_number")
    @Size(max = 15)
    private String phoneNumber;

    @Enumerated(EnumType.STRING)
    @Column(length = 10)
    private Gender gender;

    @Temporal(TemporalType.DATE)
    @Column(name = "dob")
    private Date dateOfBirth;

    @Size(max = 100)
    private String address1;

    @Size(max = 100)
    private String address2;

    @Size(max = 100)
    private String street;

    @Size(max = 100)
    private String city;

    @Size(max = 100)
    private String state;

    @Size(max = 100)
    private String country;

    @Column(name = "zip_code")
    @Size(max = 32)
    private String zipCode;

    @OneToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "user_id", nullable = false)
    private User user;

    public UserProfile() {

    }

    public UserProfile(String phoneNumber, Gender gender, Date dateOfBirth, 
                       String address1, String address2, String street, String city, 
                       String state, String country, String zipCode) {
        this.phoneNumber = phoneNumber;
        this.gender = gender;
        this.dateOfBirth = dateOfBirth;
        this.address1 = address1;
        this.address2 = address2;
        this.street = street;
        this.city = city;
        this.state = state;
        this.country = country;
        this.zipCode = zipCode;
    }
    // Getters and Setters (Omitted for brevity)
    
    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getPhoneNumber() {
        return phoneNumber;
    }

    public void setPhoneNumber(String phoneNumber) {
        this.phoneNumber = phoneNumber;
    }

    public Gender getGender() {
        return gender;
    }

    public void setGender(Gender gender) {
        this.gender = gender;
    }

    public Date getDateOfBirth() {
        return dateOfBirth;
    }

    public void setDateOfBirth(Date dateOfBirth) {
        this.dateOfBirth = dateOfBirth;
    }

    public String getAddress1() {
        return address1;
    }

    public void setAddress1(String address1) {
        this.address1 = address1;
    }

    public String getAddress2() {
        return address2;
    }

    public void setAddress2(String address2) {
        this.address2 = address2;
    }

    public String getStreet() {
        return street;
    }

    public void setStreet(String street) {
        this.street = street;
    }

    public String getCity() {
        return city;
    }

    public void setCity(String city) {
        this.city = city;
    }

    public String getState() {
        return state;
    }

    public void setState(String state) {
        this.state = state;
    }

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    public String getZipCode() {
        return zipCode;
    }

    public void setZipCode(String zipCode) {
        this.zipCode = zipCode;
    }

    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }

}

用户存储库

public interface UserRepository extends JpaRepository<User, Long> {
   User findById(int id);
}

控制器

@PostMapping(value = "/getUser")
@ResponseBody
public User getUSer(@RequestParam int userID) {
     User user = userRepository.findById(userID);
     return user;
}

错误

java.lang.IllegalArgumentException: Parameter value [6] did not match expected type [java.lang.Long (n/a)]
    at org.hibernate.query.spi.QueryParameterBindingValidator.validate(QueryParameterBindingValidator.java:54) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.spi.QueryParameterBindingValidator.validate(QueryParameterBindingValidator.java:27) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.internal.QueryParameterBindingImpl.validate(QueryParameterBindingImpl.java:90) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.internal.QueryParameterBindingImpl.setBindValue(QueryParameterBindingImpl.java:55) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.internal.AbstractProducedQuery.setParameter(AbstractProducedQuery.java:493) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.internal.AbstractProducedQuery.setParameter(AbstractProducedQuery.java:106) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.criteria.internal.compile.CriteriaQueryTypeQueryAdapter.setParameter(CriteriaQueryTypeQueryAdapter.java:385) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.hibernate.query.criteria.internal.compile.CriteriaQueryTypeQueryAdapter.setParameter(CriteriaQueryTypeQueryAdapter.java:59) ~[hibernate-core-5.3.9.Final.jar:5.3.9.Final]
    at org.springframework.data.jpa.repository.query.QueryParameterSetter$NamedOrIndexedQueryParameterSetter.lambda$setParameter$3(QueryParameterSetter.java:111) ~[spring-data-jpa-2.1.6.RELEASE.jar:2.1.6.RELEASE]
    at org.springframework.data.jpa.repository.query.QueryParameterSetter$ErrorHandling$1.execute(QueryParameterSetter.java:175) ~[spring-data-jpa-2.1.6.RELEASE.jar:2.1.6.RELEASE]
    at org.springframework.data.jpa.repository.query.QueryParameterSetter$NamedOrIndexedQueryParameterSetter.setParameter(QueryParameterSetter.java:111) ~[spring-data-jpa-2.1.6.RELEASE.jar:2.1.6.RELEASE]
    at org.springframework.data.jpa.repository.query.ParameterBinder.lambda$bind$0(ParameterBinder.java:79) ~[spring-data-jpa-2.1.6.RELEASE.jar:2.1.6.RELEASE]
    at java.util.ArrayList.forEach(Unknown Source) ~[na:1.8.0_191]

Answer 1

检查您的数据库并检查 Id 的数据类型，如果它是 int 则您无法发送 long