从Spring MVC作为JSON发送时动态忽略Java对象中的字段
我有这样的模型类，用于休眠
@Entity
@Table(name = "user", catalog = "userdb")
@JsonIgnoreProperties(ignoreUnknown = true)
public class User implements java.io.Serializable {
private Integer userId;
private String userName;
private String emailId;
private String encryptedPwd;
private String createdBy;
private String updatedBy;
@Id
@GeneratedValue(strategy = IDENTITY)
@Column(name = "UserId", unique = true, nullable = false)
public Integer getUserId() {
return this.userId;
}
public void setUserId(Integer userId) {
this.userId = userId;
}
@Column(name = "UserName", length = 100)
public String getUserName() {
return this.userName;
}
public void setUserName(String userName) {
this.userName = userName;
}
@Column(name = "EmailId", nullable = false, length = 45)
public String getEmailId() {
return this.emailId;
}
public void setEmailId(String emailId) {
this.emailId = emailId;
}
@Column(name = "EncryptedPwd", length = 100)
public String getEncryptedPwd() {
return this.encryptedPwd;
}
在Spring MVC控制器中，使用DAO，我可以获取对象。并返回为JSON对象。
@Controller
public class UserController {
@Autowired
private UserService userService;
@RequestMapping(value = "/getUser/{userId}", method = RequestMethod.GET)
@ResponseBody
public User getUser(@PathVariable Integer userId) throws Exception {
User user = userService.get(userId);
user.setCreatedBy(null);
user.setUpdatedBy(null);
return user;
}
}
视图部分是使用AngularJS完成的，因此它将获得像这样的JSON
{
"userId" :2,
"userName" : "john",
"emailId" : "john@gmail.com",
"encryptedPwd" : "Co7Fwd1fXYk=",
"createdBy" : null,
"updatedBy" : null
}
如果我不想设置加密的密码，则将该字段也设置为null。
但是我不想这样，我不想将所有字段发送到客户端。如果我不希望发送密码，updatedby，createdby字段，则我的结果JSON应该像
{
"userId" :2,
"userName" : "john",
"emailId" : "john@gmail.com"
}
我不想发送给其他数据库表的客户端的字段列表。因此，它将根据登录的用户进行更改。我该怎么做？
海绵宝宝撒3回答
-
梵蒂冈之花
将@JsonIgnoreProperties("fieldname")注释添加到您的POJO。或者，您可以@JsonIgnore在反序列化JSON时在要忽略的字段名称之前使用。例：@JsonIgnore@JsonProperty(value = "user_password")public java.lang.String getUserPassword() { return userPassword;}
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