Retrieving foreign key attributes in DTO - spring

I am using java+Spring framework+Hibernate for creating rest api but I have stumbled upon retrieving details of a table using foreign key attributes.
I have the following tables::
https://i.stack.imgur.com/lG7UR.png
I am retrieving all the ratings given using product id and then mapping to DTO, now I also want to populate the username using idusers as this is my foreign key.
Same is the case when I try to retrieve ratings given by the users, instead of displaying idproducts I want to display the product name and product description as It is a foreign key.
Any advice on how to do so using DTO's.

This is a perfect use case for Blaze-Persistence Entity Views.
Blaze-Persistence is a query builder on top of JPA which supports many of the advanced DBMS features on top of the JPA model. I created Entity Views on top of it to allow easy mapping between JPA models and custom interface defined models, something like Spring Data Projections on steroids. The idea is that you define your target structure the way you like and map attributes(getters) via JPQL expressions to the entity model. Since the attribute name is used as default mapping, you mostly don't need explicit mappings as 80% of the use cases is to have DTOs that are a subset of the entity model.
Assuming you have an entity model like this
#Entity
public class User {
#Id
Integer id;
String role;
String username;
String password;
boolean enabled;
}
#Entity
public class Product {
#Id
Integer id;
String imageUrl;
String category;
int productPrice;
int productQuantity;
String productName;
String productDesc;
#OneToMany(mappedBy = "product")
Set<Rating> ratings;
}
#Entity
public class Rating {
#Id
Integer id;
int rating;
String review;
String ratingscol;
#ManyToOne(fetch = LAZY)
Product product;
#ManyToOne(fetch = LAZY)
User user;
}
A DTO mapping for your model could look as simple as the following
#EntityView(Rating.class)
interface RatingDto {
Integer getId();
UserDto getUser();
ProductDto getProduct();
}
#EntityView(User.class)
interface UserDto {
Integer getId();
String getUsername();
}
#EntityView(Rating.class)
interface ProductDto {
Integer getId();
String getProductName();
String getProductDesc();
}
Querying is a matter of applying the entity view to a query, the simplest being just a query by id.
RatingDto dto = entityViewManager.find(entityManager, RatingDto.class, id);
But the Spring Data integration allows you to use it almost like Spring Data Projections: https://persistence.blazebit.com/documentation/entity-view/manual/en_US/index.html#spring-data-features
It will only fetch the mappings that you tell it to fetch

You can use ModelMapper when converting a DTO to an Entity bean and back from Entity bean to a DTO.
Add ModelMapper to your project
<dependency>
<groupId>org.modelmapper</groupId>
<artifactId>modelmapper</artifactId>
<version>2.3.5</version>
</dependency>
Define the ModelMapper bean in your Spring configuration
#Bean
public ModelMapper modelMapper() {
return new ModelMapper();
}
Assuming the following models based on the given ER diagram you have given
public class UserDto {
Integer userId;
String role;
String username;
String password;
boolean enabled;
...default and parameterized constructor
...getter and setter methods
}
public class ProductDto {
Integer productId;
String imageUrl;
String category;
int productPrice;
int productQuantity;
String productName;
String productDesc;
...default and parameterized constructor
...getter and setter methods
}
public class RatingDto {
#Id
Integer id;
int rating;
String review;
String ratingscol;
ProductDto productDto;
UserDto userDto;
...default and parameterized constructor
...getter and setter methods
}
You can retrieve the ratings of a product using product id along with the user details by using the following method
#Repository
public interface RatingRepository extends JpaRepository<Rating, Integer>{
List<Rating> findByProduct_ProductId(Integer productId);
}
Then mapping rating objects to DTO
RatingDto ratingDto = modelMapper.map(rating, RatingDto.class);
Now you can retrieve username as following
ratingsDto.getUserDto().getUserName()
The same way you can retrieve the ratings by userId and access product details

Related

Spring Data JPA save child object with the ID of parent object

I have two objects, one parent and one child as follows :
#Entity
#Table(name="category")
public class CategoryModel {
private #Id #GeneratedValue Long id;
private String name;
#OneToMany(mappedBy="category", cascade=CascadeType.PERSIST)
private List<AttributeModel> attributes;
}
#Entity
#Table(name="attribute")
public class AttributeModel {
private #Id #GeneratedValue Long id;
private String name;
#ManyToOne(cascade = CascadeType.ALL)
#JoinColumn(name="category_id")
private CategoryModel category;
}
I also have dtos which maps to these model objects but I ommited them.
When I try to save a category object with this payload Attribute values are also created in the attribute table but with null category ids.
{
"name":"Chemicals",
"attributes":[
{"name": "volume"}, {"name":"humidity"}
]
}
What can I do to have my attribute values persisted into the database with the category id which is created before them?
First of all, this problem is not a "Spring Data JPA" problem, it is a JPA (probably Hibernate) problem.
Analysis
Since you left out the code for the controller and the JSON mapping, I have to guess a bit:
fact 1: The relationship between category and attributes is controlled by the attribute AttributeModel.category but not by CategoryModel.attributes. (That is how JPA works).
observation 2: Your JSON object define CategoryModel.attributes (i.e. opposite to how JPA works).
Without knowing your JSON mapping configuration and controller code, I would guess that the problem is: that your JSON mapper does not set the AttributeModel.category field when it deserialises the JSON object.
Solution
So you need to instruct the JSON mapper to set the AttributeModel.category field during deserialisation. If you use Jackson, you could use:
#JsonManagedReference and
#JsonBackReference
#Entity
#Table(name="category")
public class CategoryModel {
...
#JsonManagedReference
#OneToMany(mappedBy="category", cascade=CascadeType.PERSIST)
private List<AttributeModel> attributes;
}
#Entity
#Table(name="attribute")
public class AttributeModel {
...
#JsonBackReference
#ManyToOne(cascade = CascadeType.ALL)
#JoinColumn(name="category_id")
private CategoryModel category;
}
I solved this by manually setting child object's reference to the parent object as follows :
public Long createCategory(CategoryDto categoryDto) {
CategoryModel categoryModel = categoryDto.toModel(true,true);
categoryModel.getAttributes().forEach(a -> a.setCategory(categoryModel));
return categoryRepository.save(categoryModel).getId();
}

Object attributes combination query Mongodb Repository

I am trying to query based on Object attributes combination in MongoDB:
My Object looks like:
#Getter
#Setter
class Person {
private String name;
private String email;
private Address address;
}
Address entity looks like:
#Getter
#Setter
class Address {
private String city;
private String state;
private String postal;
private String country;
}
Assume that above classes are also annotated with necessary attributes required for MongoDB collections.
Now when I am about to query from a PersonRepo interface that extends MongoRepository, as:
public interface PersonRepo extends MongoRepository<Person, String> {
}
The objective is to fetch all result-set where Address matches any combination of city, state, postal or country. For example, I want to fetch all results where city=XYZCity, state = ABCState, Postal might not be provided and country might be DEFCountry.
Address can be in any combination and whatever combination matches, the result-set could be produced.

How to Inner Join using Spring Boot JPA / Hibernate in Repository?

I am currently learning spring boot , hibernate and Spring Boot JPA
I developing a Classroom App for coaching centers and institutes .
In it, students enrolled to multiple courses in a single institute
The Student Entity class :
#Entity
#Table(name = "student")
public class Student {
private String name;
private String dob;
private String gender;
private String address;
private String email;
private Integer mobile;
private String joined;
private Integer instID;
#Id
private String studentid;
getters and setters()....
}
Course Table Entity class
#Entity
#Table(name = "courses")
public class Course {
private String name;
private String description;
private String logo;
private String start;
private String end;
private Integer fee;
#Id
private String courseid;
private Integer instID;
getters and setters();
}
Enrolled Classes Table's Entity class
public class EnrolledCourses {
#Id
String enrollID;
String courseid;
String studentid;
Date joined;
getters and setters()...
}
JPA Repository
#Repository
public interface StudentRepository extends CrudRepository<Student, String> {
Student findTopByInstIDOrderByStudentidDesc(int instID);
}
#Repository
public interface CourseRepository extends CrudRepository<Course,String> {
}
#Repository
public interface EnrolledRepository extends CrudRepository<Course,String> {
}
My Need
Now I am retrieving enrolled students for a given course in a given institute... by using this MySQL query
SELECT
`enrolled_courses.enrollID`,
`student.name`, `student.studentid`
FROM `enrolled_courses`
INNER JOIN `student`
WHERE
`enrolled_courses.studentid` = `student.studentid`
AND
`student.instID` = 13
AND
`enrolled_courses.courseid` = '13I01C' ;
Now I need to implement this Inner join query in CourseRepository (or enrolledstudent repo)
How to achieve this ?
Please guide me
If we use hibernate mapping in EnrolledCourses entity like
#Entity
public class EnrolledCourses {
#Id
String enrolledId;
#ManyToOne
Student student;
#ManyToOne
Course course;
Date joined;
getters and setters()...
}
from the above mappings without using any SQL queries, you can retrieve all student who comes under a particular course
By using the method in the interface
#Repository
public interface EnrolledRepository extends CrudRepository<EnrolledCourses,String> {
List<EnrolledCourses> findByCourse_CourseId(String courseId);
}
if we have some relations between the entities we can easily retrieve all the fields using Jpa.

How to write query method for sorting the results based on the field which is part of parent entity spring data jpa

In spring data JPA we can write query methods to execute query. I have two entities
class A {
#Id
#Column
private String id;
private String name;
#ManyToOne
#JoinColumn(name = "b_field")
private B b;
}
class B {
#Id
#KeyField
#Column
private String id;
private String b_field
}
I want to write a query method "findByNameOrderByb_field" i.e find all the rows having name as provided and order the results on the basis of class B's field i.e b_field. Writing above function is not returning ordered results. Is there any way I can write this orderby query method in spring boot.
public interface ARepository extends CrudRepository<A, String> {
List<A> findByNameOrderByb_field(String name);
}
the above function is not returning ordered results based on B's b_field.
To access properties of B you have to include the field name of B in the declaration of the query method.
public interface ARepository extends CrudRepository<A, String> {
List<A> findByNameOrderByb_b_field(String name);
}
More information how SpringData resolves nested Properties can be found under:
Query Property Expressions

Passing parent id reference when creating child object through REST api

I am using spring boot (version - 2.1.1). I have a one to many database model exposed for CRUD operations through rest api's. The model looks as below. How do I configure the POST /departments api (that creates a department object) to accept just the organization id in the input json body?
#PostMapping
public Long createDepartment(#RequestBody Department Department) {
Department d = departmentService.save(Department);
return d.getId();
}
Note - I do not want to allow creating organization object when creating a department.
Model object mapping
#Entity
#Table(name="ORGANIZATIONS")
public class Organization{
#Id
#GeneratedValue
Private long id;
#Column(unique=true)
Private String name;
#OneToMany(mappedBy = "organization", fetch = FetchType.EAGER)
private List<Department> departments;
}
#Entity
#Table(name="DEPARTMENTS")
Public class Department{
#Id
#GeneratedValue
Private long id;
#Column(unique=true)
Private String name;
#ManyToOne(fetch = FetchType.EAGER)
private Organization organization;
}
Thanks!
The easiest and most sane way in my opinion is to utilize the DTO (Data Transfer Object) pattern.
Create a class that represent the model you want to get as your input:
public class CreateDepartmentRequest {
private long id;
// getters and setters
}
Then use it in your controller:
#PostMapping
public Long createDepartment(#RequestBody CreateDepartmentRequest request) {
Department d = new Department();
d.setId(request.getId());
Department d = departmentService.save(d);
return d.getId();
}
Side note, its better to ALWAYS return JSON through REST API (unless you use some other format across your APIs) so you can also utilize the same pattern as I mentioned above to return a proper model as a result of the POST operation or a simple Map if you don't want to create to many models.

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