How can I hide some sensitive data on this example. I'm testing APIs in rest client (Postman), when I call Api List of Bills, I want to hide some data. In BillsDto I want to hide username, password and user age fields. Is it possible to do this in my BillsDto class (not in UserDto). I know I can hide some fields using #JsonProperty but how to do it for some fields belonging to another class?
***BillsDto***
public class BillsDto {
private String numberBills;
private double amount;
private Date deadlinePayment
private UserDto user; // try to hide username, password, age from BillsDto
}
***UserDto***
public class UserDto {
private String number_id;
private String username;
private String password;
private String firstName;
private String lastName;
private String age;
}
I know I can hide some fields using #JsonProperty but how to do it for some fields belonging to another class?
The fact that you're using UserDto as a nested object somewhere, doesn't change the serialization policy that you can express through data binding annotations in the UserDto.
If you can change UserDto, apply #JsonProperty with it's property access set to JsonProperty.Access.WRITE_ONLY on the fields want to hide during serialization.
public class UserDto {
private String number_id;
#JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
private String username;
#JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
private String password;
private String firstName;
private String lastName;
#JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
private String age;
}
If for some reason, you want to achieve this by editing BillsDto only, then you can implement a custom serializer for UserDto and apply it by making use of the #JsonSerialize. But to ensure that you're not disclosing the sensitive data somewhere, it would be better to apply this policy in one place - in the UserDto, because you or one of your colleagues might simply forget to #JsonSerialize in some of the classes which uses UserDto.
Related
I am Java web developer, usually develop Spring MVC.
I have been using #RequestMapping or #RequestParam for mapping to hashMap at Controller.
It is a terrible way. I should always cast type when using value.
But nowadays I try to use #ModelAttribute to write clean code at Controller.
However, there are some problem.
case 1) make DTO for each EndPoint.
We can make DTO for each EndPoint, but DTO will have many duplicated property.
#Getter
#Setter
#ToString
class GetUserInfoDTO {
private String id;
private String name;
}
#Getter
#Setter
#ToString
class PostUserInfoDTO {
private String name;
private Integer age;
private String address;
private String gender;
private String email;
private Date joinDate;
}
in controller,
#GetMapping("/user")
public ResultDTO getUserInfo (#ModelAttribute GetUserInfoDTO){
...
return ResultDTO;
}
#PostMapping("/user")
public ResultDTO postUserInfo (#ModelAttribute PostUserInfoDTO){
...
return ResultDTO;
}
In this case, we can apply independent validation strategy for each End-Point.
for example..
#Getter
#Setter
#ToString
class GetUserInfoDTO {
#NotNull
private String id;
private String name;
}
#Getter
#Setter
#ToString
class PostUserInfoDTO {
#NotNull
private String name;
#NotNull
private Integer age;
#NotEmpty
private String address;
private String gender;
private String email;
private Date joinDate;
}
like this.
But so many model classes made, and so many duplicated property exists.
case 2. make common DTO for each Controller.
We can make DTO for each Controller, and reuse them.
#Getter
#Setter
#ToString
class UserInfoDTO {
private String id;
private String name;
private Integer age;
private String address;
private String gender;
private String email;
private Date joinDate;
}
#GetMapping("/user")
public ResultDTO getUserInfo (#ModelAttribute UserInfoDTO){
//I want only id, name
...
return ResultDTO;
}
#PostMapping("/user")
public ResultDTO postUserInfo (#ModelAttribute UserInfoDTO){
...
return ResultDTO;
}
But In this case, we can only pass specific properties.
If someone send other parameter than id and name, we can't notice. ( 400 error not occur )
Code assistance can't recommend us specific properties that use at single end-point.
I don't like these cases.
In first case, I should make so many models and It's management will be so hard.
Second case, unnecessary properties exists and it hard to validate for each end-point.
Which way is the best?
Or Can you recommend another way for mapping request parameter to model object?
I'm building a rest API using Spring Boot rest services.
I have a Java class:
class Person{
int id;
#notNull
String name;
#notNull
String password;
}
And I want to make an API to create a Person object. I will recieve a POST request with json body like:
{
"name":"Ahmad",
"password":"myPass",
"shouldSendEmail":1
}
As you can see there are an extra field "shouldSendEmail" that I have to use it to know if should I send an email or not after I create the Person Object.
I am using the following API:
#RequestMapping(value = "/AddPerson", method = RequestMethod.POST)
public String savePerson(
#Valid #RequestBody Person person) {
personRepository.insert(person);
// Here I want to know if I should send an email or Not
return "success";
}
Is there a method to access the value of "shouldSendEmail" while I using the autoMapping in this way?
There's many options for you solve. Since you don't want to persist the shouldSendEmail flag and it's ok to add into you domain class, you can use the #Transient annotation to tell JPA to skip the persistence.
#Entity
public class Person {
#Id
private Integer id;
#NotNull
private String name;
#NotNull
private String password;
#Transient
private Boolean shouldSendEmail;
}
If you want more flexible entity personalizations, I recommend using DTO`s.
MapStruct is a good library to handle DTO`s
You will need an intermediary DTO, or you will otherwise have to modify person to include a field for shouldSendEmail. If that is not possible, the only other alternative is to use JsonNode and manually select the properties from the tree.
For example,
#Getter
public class PersonDTO {
private final String name;
private final String password;
private final Integer shouldSendEmail;
#JsonCreator
public PersonDTO(
#JsonProperty("name") final String name,
#JsonProperty("password") final String password,
#JsonProperty("shouldSendEmail") final Integer shouldSendEmail
) {
this.name = name;
this.password = password;
this.shouldSendEmail = shouldSendEmail;
}
}
You can use #RequestBody and #RequestParam together as following
.../addPerson?sendEmail=true
So send the “sendEmail” value as request param and person as request body
Spring MVC - Why not able to use #RequestBody and #RequestParam together
You have mutli solutions
1 - You can put #Column(insertable=false, updatable=false) above this property
2 - send it as request param #RequestParam
#RequestMapping(value = "/AddPerson", method = RequestMethod.POST)
public String savePerson(
#Valid #RequestBody Person person, #RequestParam boolean sendMail) {}
3- use DTO lets say PersonModel and map it to Person before save
I am making a simple CRUD application using Spring boot and MongoDB, the problem that I am facing is that I don't know how to define the model classes.
My application should be like this:
A site has some characteristics such as an ID, region, city, ... and contains 4 parts (cellulars) that each has its own characteristics. Any help would be appreciated.
This is what I have so far:
public class Site {
#Id
String siteId;
String projectPhase;
String region;
String city;
String siteName;
String newSiteName;
String clusterName ;
String longitude ;
String lattitude ;
#OneToMany(mappedBy = "siteId")
List L;
What I want to know is how do I associate another class inside this one.
Annotations like #OneToMany are typically used within JPA-context, and are unnecessary when using Spring Data MongoDB. This is also mentioned by the documentation:
There’s no need to use something like #OneToMany because the mapping framework sees that you want a one-to-many relationship because there is a List of objects.
You have a few options when you want to define one-to-many relations when using MongoDB. The first of them is to define them as embedded objects within the same document:
#Document
public class Site {
#Id
private String id;
private String city;
private String region;
private List<Part> cellulars;
}
public class Part {
private String characteristic1;
private String characteristic2;
}
This means that the parts do not exist on their own, so they don't need their own ID either.
Another possibility is to reference to another document:
#Document
public class Site {
#Id
private String id;
private String city;
private String region;
#DBRef
private List<Part> cellulars;
}
#Document
public class Part {
#Id
private String id;
private String characteristic1;
private String characteristic2;
}
In this case, parts are also separate documents, and a site simply contains a reference to the part.
Spring LdapRepository save() method throws exception when I'm trying to update an existing object in LDAP database.
org.apache.directory.api.ldap.model.exception.LdapEntryAlreadyExistsException: ERR_250_ENTRY_ALREADY_EXISTS
What method should I use to update existing ldap objects?
Person class:
#Entry(objectClasses = { "inetOrgPerson", "organizationalPerson", "person", "top" })
public class Person implements Serializable {
public Person() {
}
#Id
private Name dn;
#Attribute(name = "cn")
#DnAttribute(value = "cn")
#JsonProperty("cn")
private String fullName;
#Attribute(name = "uid")
private String uid;
private String mail;
#Attribute(name = "sn")
private String surname;
//setters and getters
}
Person repo interface:
public interface PersonRepo extends LdapRepository<Person> {
}
That's how I'm updating person:
personRepo.save(person);
Default implementation for Spring LDAP repositories is SimpleLdapRepository, that checks the property annotated with #Id to determine if the objects is new - and perform create, or old - and perform update.
I'm guessing that Person.dn is null when you're trying to perform update.
You also can take the control over this by implementing org.springframework.data.domain.Persistable and place your logic in the isNew() method.
See the implementation details.
I have a user object that is sent to and from the server. When I send out the user object, I don't want to send the hashed password to the client. So, I added #JsonIgnore on the password property, but this also blocks it from being deserialized into the password that makes it hard to sign up users when they don't have a password.
How can I only get #JsonIgnore to apply to serialization and not deserialization? I'm using Spring JSONView, so I don't have a ton of control over the ObjectMapper.
Things I've tried:
Add #JsonIgnore to the property
Add #JsonIgnore on the getter method only
Exactly how to do this depends on the version of Jackson that you're using. This changed around version 1.9, before that, you could do this by adding #JsonIgnore to the getter.
Which you've tried:
Add #JsonIgnore on the getter method only
Do this, and also add a specific #JsonProperty annotation for your JSON "password" field name to the setter method for the password on your object.
More recent versions of Jackson have added READ_ONLY and WRITE_ONLY annotation arguments for JsonProperty. So you could also do something like:
#JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
private String password;
Docs can be found here.
In order to accomplish this, all that we need is two annotations:
#JsonIgnore
#JsonProperty
Use #JsonIgnore on the class member and its getter, and #JsonProperty on its setter. A sample illustration would help to do this:
class User {
// More fields here
#JsonIgnore
private String password;
#JsonIgnore
public String getPassword() {
return password;
}
#JsonProperty
public void setPassword(final String password) {
this.password = password;
}
}
Since version 2.6: a more intuitive way is to use the com.fasterxml.jackson.annotation.JsonProperty annotation on the field:
#JsonProperty(access = Access.WRITE_ONLY)
private String myField;
Even if a getter exists, the field value is excluded from serialization.
JavaDoc says:
/**
* Access setting that means that the property may only be written (set)
* for deserialization,
* but will not be read (get) on serialization, that is, the value of the property
* is not included in serialization.
*/
WRITE_ONLY
In case you need it the other way around, just use Access.READ_ONLY.
In my case, I have Jackson automatically (de)serializing objects that I return from a Spring MVC controller (I am using #RestController with Spring 4.1.6). I had to use com.fasterxml.jackson.annotation.JsonIgnore instead of org.codehaus.jackson.annotate.JsonIgnore, as otherwise, it simply did nothing.
Another easy way to handle this is to use the argument allowSetters=truein the annotation. This will allow the password to be deserialized into your dto but it will not serialize it into a response body that uses contains object.
example:
#JsonIgnoreProperties(allowSetters = true, value = {"bar"})
class Pojo{
String foo;
String bar;
}
Both foo and bar are populated in the object, but only foo is written into a response body.
"user": {
"firstName": "Musa",
"lastName": "Aliyev",
"email": "klaudi2012#gmail.com",
"passwordIn": "98989898", (or encoded version in front if we not using https)
"country": "Azeribaijan",
"phone": "+994707702747"
}
#CrossOrigin(methods=RequestMethod.POST)
#RequestMapping("/public/register")
public #ResponseBody MsgKit registerNewUsert(#RequestBody User u){
root.registerUser(u);
return new MsgKit("registered");
}
#Service
#Transactional
public class RootBsn {
#Autowired UserRepository userRepo;
public void registerUser(User u) throws Exception{
u.setPassword(u.getPasswordIn());
//Generate some salt and setPassword (encoded - salt+password)
User u=userRepo.save(u);
System.out.println("Registration information saved");
}
}
#Entity
#JsonIgnoreProperties({"recordDate","modificationDate","status","createdBy","modifiedBy","salt","password"})
public class User implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
private String country;
#Column(name="CREATED_BY")
private String createdBy;
private String email;
#Column(name="FIRST_NAME")
private String firstName;
#Column(name="LAST_LOGIN_DATE")
private Timestamp lastLoginDate;
#Column(name="LAST_NAME")
private String lastName;
#Column(name="MODIFICATION_DATE")
private Timestamp modificationDate;
#Column(name="MODIFIED_BY")
private String modifiedBy;
private String password;
#Transient
private String passwordIn;
private String phone;
#Column(name="RECORD_DATE")
private Timestamp recordDate;
private String salt;
private String status;
#Column(name="USER_STATUS")
private String userStatus;
public User() {
}
// getters and setters
}
You can use #JsonIgnoreProperties at class level and put variables you want to igonre in json in "value" parameter.Worked for me fine.
#JsonIgnoreProperties(value = { "myVariable1","myVariable2" })
public class MyClass {
private int myVariable1;,
private int myVariable2;
}
You can also do like:
#JsonIgnore
#JsonProperty(access = Access.WRITE_ONLY)
private String password;
It's worked for me
I was looking for something similar. I still wanted my property serialized but wanted to alter the value using a different getter. In the below example, I'm deserializing the real password but serializing to a masked password. Here's how to do it:
public class User() {
private static final String PASSWORD_MASK = "*********";
#JsonIgnore
private String password;
#JsonProperty(access = JsonProperty.Access.WRITE_ONLY)
public String setPassword(String password) {
if (!password.equals(PASSWORD_MASK) {
this.password = password;
}
}
public String getPassword() {
return password;
}
#JsonProperty("password")
public String getPasswordMasked() {
return PASSWORD_MASK;
}
}
The ideal solution would be to use DTO (data transfer object)