SpringREST应用程序中的“内容类型”application/json；charset=UTF-8“不受支持”

Question

当我在localhost:8080/api/users上做一个POST请求来创建一个新用户时，我会得到以下错误：

{
    "timestamp": "2018-05-28T09:44:55.704+0000",
    "status": 415,
    "error": "Unsupported Media Type",
    "message": "Content type 'application/json;charset=UTF-8' not supported",
    "path": "/api/users/"
}

是请求的主体，选择JSON（应用程序/JSON）。即使我删除角色并将其保持为空，它也会给出相同的错误。

标题的内容类型也是application/json。

这是我的控制器：

@PostMapping("/api/users" )
public User createUser(@Valid @RequestBody User user) {
    securityService.autologin(user.getUsername(), user.getPassword());
    return userService.createUser(user);
}

UserService中的createUser函数：

public User createUser(@Valid @RequestBody User user) {
    user.setPassword(bCryptPasswordEncoder.encode(user.getPassword()));
    user.setRoles(new HashSet<>(roleRepository.findAll()));
    return userRepository.save(user);
}

这是我的用户类：

@Entity
@Table(name = "user")
@EntityListeners(AuditingEntityListener.class)
@JsonIgnoreProperties(value = {"createdAt", "updatedAt"}, 
                      allowGetters = true)
public class User implements Serializable{

    private static final long serialVersionUID = 1L;


    public User() {
        super();
        // TODO Auto-generated constructor stub
    }

    @Id
    @Column(name = "user_id")
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private long id;

    @Column(name = "user_name")
    private String name;

    @Column(name = "user_email")
    private String email;

    @Column(name = "user_password")
    @NotBlank
    private String password;

    @Column(name = "user_status")
    private String status;

    @Column(name = "user_tel")
    private String tel;

    @Column(name = "user_confirmation")
    private String confirmation;

    @Column(name = "user_birth_date")
    @Temporal(TemporalType.DATE)
    private Date birth_date;

    @Column(nullable = false, updatable = false)
    @Temporal(TemporalType.TIMESTAMP)
    @CreatedDate
    private Date createdAt;

    @Column(nullable = false)
    @Temporal(TemporalType.TIMESTAMP)
    @LastModifiedDate
    private Date updatedAt;

    @JsonManagedReference
    @ManyToMany
    @JoinTable(name = "user_role", joinColumns = @JoinColumn(name = "user_id"), inverseJoinColumns = @JoinColumn(name = "role_id"))
    private Set<Role> roles;

    @Column(name = "username")
    @NotBlank
    private String username;

    public long getId() {
        return id;
    }

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

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getUsername() {
        return username;
    }

    public void setUsername(String username) {
        this.username = username;
    }

    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 String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    public String getTel() {
        return tel;
    }

    public void setTel(String tel) {
        this.tel = tel;
    }

    public String getConfirmation() {
        return confirmation;
    }

    public void setConfirmation(String confirmation) {
        this.confirmation = confirmation;
    }

    public Date getBirth_date() {
        return birth_date;
    }

    public void setBirth_date(Date birth_date) {
        this.birth_date = birth_date;
    }

    public Date getCreatedAt() {
        return createdAt;
    }

    public Date getUpdatedAt() {
        return updatedAt;
    }

    public void setUpdatedAt(Date updatedAt) {
        this.updatedAt = updatedAt;
    }

    public Set<Role> getRoles() {
        return roles;
    }

    public void setRoles(Set<Role> roles) {
        this.roles = roles;
    }
}

Answer 1

你可以试试类似的东西

@PostMapping（value=“/rest/account/json”，使用={“application/json”}）

Answer 2

在我的案例中，有一个Jackson失败，记录为警告：

无法评估类型[[简单类型，类***]]：com的Jackson反序列化。fasterxml。杰克逊。数据绑定。JsonMappingException:属性[…]的setter定义冲突

我意外地超载了一个二传手，所以杰克逊无法解决，Spring扔了415。

Answer 3

我能够通过删除@JsonManaged参考来解决它。