Spring Repository issue - spring

I seem to be baffled on how JPA Repositories are suppose to work.
In a nut-shell
#Entity
public class User extends AbstractEntity {
protected final static String FK_NAME = "USER_ID";
#Column(nullable = false)
private String firstName;
#OneToMany(cascade = ALL, fetch = FetchType.LAZY, orphanRemoval = true)
#JoinColumn(name = "userId")
private List<Detail> details = new ArrayList<Detail>();
}
#Entity
public class Detail extends AbstractEntity {
Long userId;
String hello;
}
#Repository
public interface UserRepository extends CrudRepository<User, Long> {
User findByFirstName(#Param("firstName") String firstName);
}
And here is the only controller in the app:
#RestController
public class Home {
#Autowired
UserRepository userRepository;
#Autowired
DetailsRepository loanRepository;
#RequestMapping(value = "")
public HttpEntity home() {
User user = userRepository.findByFirstName("John");
if (user == null) {
user = new User();
user.setFirstName("John");
}
Detail detail = new Detail();
detail.setHello("Hello Msh");
user.getDetails().add(detail);
userRepository.save(user);
return new ResponseEntity("hi", HttpStatus.OK);
}
}
Below a screenshot from debugging session where the app just started and the get request to home() method creates new user, new detail, adds detail to user.
Below example - when the user is saved, the detail entity gets updated
Now on the next request, the old user John is found and has been added a new instance of detail.
The old user has been saved but now the newly created detail does not get updated outside.
How come this only works first time ?

Basically theres so much fail going on so that I would advise you to go a step backwards. If youre wana go the short path of getting a solution for exactly this problem continue reading ;)
First part related to the answer of Jaiwo99:
As I can see in the gradle view of intellij, your using Spring Boot. So it is necessary to place #EnableTransactionManagement on top of your configuration class. Otherwise the #Transacion annotation does not have any effect.
Second part your JPA/Hibernate model mapping. Theres so much bad practise on the net that it is no wonder that most beginners have troubles starting with it.
A correct version could look like (not tested)
#Entity
public class User extends AbstractEntity {
#Column(nullable = false)
private String firstName;
#OneToMany(cascade = ALL, fetch = FetchType.LAZY, orphanRemoval = true, mappedBy="user")
private List<Detail> details = new ArrayList<Detail>();
public void addDetail(Detail detail) {
details.add(detail);
detail.setUser(user);
}
}
#Entity
public class Detail extends AbstractEntity {
#ManyToOne
private User user;
private String hello;
public void setUser(User user){
this.user = user;
}
}
Some general advice related to creating a model mapping:
avoid bi-directional mappings whenever possible
cascade is a decision made on the service level and not at the model level and can have huge drawbacks. So for beginners avoid it.
I have no idea why people like to put JoinColumn, JoinTable and whatever join annotation on top of fields. The only reason to do this is when you have a legacy db (my opinion). When you do not like the names created by your jpa provider, provide a different naming strategy.
I would provide a custom name for the user class, because this is in some databases a reserved word.

Very simple, the first time you saved a new entity outside of hibernate session, the second time, the user object you got is a detached object, by default hibernate will not consider it is changed in this case.
*solution *
Move this logic to another service class, which annotated with #transactional
Or
Annotate your controller with transactional
Or
Override equals and hashCode method on user class may also help

Related

Spring #PreAuthorize throws LazyInitializationException accesing lazy data from Hibernate

I need to check whether user is a member of specific group for security reasons.
However, I'm getting LazyInitializationException all the time and can't find a solution.
org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: User.groups, could not initialize proxy - no Session
I've tried to use #Transactional annotation everywhere it is possible, that doesn't help. It works fine with Eager initialization for sure, but I would prefer not to use it.
Controller:
#MessageMapping("/group/chat")
#PreAuthorize("#decider.checkIfGroupMember(#user, #chatMessageRequest.getGroupId())")
public ChatMessageRequest processMessage(
#Payload ChatMessageRequest chatMessageRequest,
#AuthenticationPrincipal User user) {
return chatMessageService.processMessage(chatMessageRequest, user);
}
Decider class:
#Component
public class Decider {
#Transactional
public boolean checkIfGroupMember(Object principal, Integer groupId) {
User user = (User) principal;
Set<Group> groups = user.getGroups();
for (Group group : groups) {
if(group.getId().equals(groupId)) {
return true;
}
}
return false;
}
}
That's what is important about user:
#Data
#Builder
#Accessors(chain = true)
#Table(name = "users")
public class User extends BasicEntity {
#Column(nullable = false)
private String name;
#Column(nullable = false, unique = true)
private String email;
#Column
private String password;
#ManyToMany(mappedBy = "users", fetch = FetchType.LAZY)
#JsonIgnore
private Set<Group> groups = new HashSet<>();
}
Somebody could help me? Thanks in advance!
UPD: Transactional works fine in other parts of application, so I don't think this is configuration issue or it might be. Also, even initializing directly with Hibernate.initialize() inside decider method still fails.
I guess, this happening because #PreAuthorize doesn't access the method directly, so Spring is not able to create proxy. However, I still don't know how to fix it and even not sure if I am right here.
You need to open transaction before authentication. You can do that by specifying the order on both EnableTransactionManagement and EnableGlobalMethodSecurity. By default, they both have lowest precedence.
#EnableTransactionManagement(order = 0)
#EnableGlobalMethodSecurity(prePostEnabled = true, order = 1)
Credit goes to this answer.
Use fetch = FetchType.EAGER instead of FetchType.LAZY.

Why do I need #Transactional for saving an OneToOne mapped Entity

I have a simple straight forward demo application with spring-boot, spring-data-jpa and a h2-DB.
I have build two entities which are mapped by an OneToOne relationship.
Post.java
#Entity
public class Post {
#Id
#GeneratedValue
private Long id;
private String title;
#OneToOne(mappedBy = "post", cascade = CascadeType.ALL, fetch = FetchType.LAZY)
private PostDetail postDetail;
}
PostDetail.java
#Entity
public class PostDetail {
#Id
#GeneratedValue
private Long id;
private String message;
#OneToOne(fetch = FetchType.LAZY)
#MapsId
#JoinColumn(name = "id")
private Post post;
}
I try to create and save a new Post. Then I try to create a new PostDetail, set the previous generated Post to it and save it. In the one controller sample I dont have a #Transactional annotation and in the seconde sample I do annotate the method with #Transactional
#RestController
public class TestController {
#Autowired
PostRepository postRepository;
#Autowired
PostDetailRepository postDetailRepository;
#GetMapping("/test1")
public String test1() {
Post post = new Post();
post.setId(2L);
post.setTitle("Post 1");
postRepository.save(post);
PostDetail detail = new PostDetail();
detail.setMessage("Detail 1");
detail.setPost(post);
postDetailRepository.save(detail);
return "";
}
#Transactional
#GetMapping("/test2")
public String test2() {
Post post = new Post();
post.setId(2L);
post.setTitle("Post 1");
postRepository.save(post);
PostDetail detail = new PostDetail();
detail.setMessage("Detail 1");
detail.setPost(post);
postDetailRepository.save(detail);
return "";
}
}
Why do I get in the first sample a org.hibernate.PersistentObjectException: detached entity passed to persist: com.example.demo.jpa.model.Post exception and in the other sample not?
Can anyone explain why this happens?
You use bidirectional #OneToOne association. As hibernate documentation states:
Whenever a bidirectional association is formed, the application developer must make sure both sides are in-sync at all times.
So, you should rewrite your test method in this way:
#GetMapping("/test1")
public String test1() {
Post post = new Post();
post.setId(2L);
post.setTitle("Post 1");
PostDetail detail = new PostDetail();
detail.setMessage("Detail 1");
// synchronization of both sides of #OneToOne association
detail.setPost(post);
post.setDetail(detail);
// thanks to CascadeType.ALL on Post.postDetail
// postDetail will be saved too
postRepository.save(post);
return "";
}
You shouldn’t be saving those 2 entities separately — you should set PostDetail inside of post object and save only the Post object. Hibernate will take care of saving the aggregated PostDetail.
That is why you are getting PersistentObjectException which you are able to workaround by keeping it inside of the same transaction.
we do not always need a bidirectional mapping when we are mapping two entities
you can simple have a unidirection most of the time
Post post = new Post();
post.setId(2L);
post.setTitle("Post 1");
PostDetail detail = new PostDetail();
detail.setMessage("Detail 1");
detail.setPost(post);
postRepository.save(post);
as you have cascade.all ,so hibernate saves Post first and then it saves PostDetail, now as per the rule of Transaction behavior ,either it is completely done or not done,Hence we can not have the situation that Post is saved but PostDetail did not,Hence to avoid such ambiguity it is important to have #Transaction annotation ,at method level or may be class level as per your requirement

Spring Data Rest - sort by nested property

I have a database service using Spring Boot 1.5.1 and Spring Data Rest. I am storing my entities in a MySQL database, and accessing them over REST using Spring's PagingAndSortingRepository. I found this which states that sorting by nested parameters is supported, but I cannot find a way to sort by nested fields.
I have these classes:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
#ManyToOne
protected Address address;
#ManyToOne(targetEntity = Name.class, cascade = {
CascadeType.ALL
})
#JoinColumn(name = "NAME_PERSON_ID")
protected Name name;
#Id
protected Long id;
// Setter, getters, etc.
}
#Entity(name = "Name")
#Table(name = "NAME")
public class Name{
protected String firstName;
protected String lastName;
#Id
protected Long id;
// Setter, getters, etc.
}
For example, when using the method:
Page<Person> findByAddress_Id(#Param("id") String id, Pageable pageable);
And calling the URI http://localhost:8080/people/search/findByAddress_Id?id=1&sort=name_lastName,desc, the sort parameter is completely ignored by Spring.
The parameters sort=name.lastName and sort=nameLastName did not work either.
Am I forming the Rest request wrong, or missing some configuration?
Thank you!
The workaround I found is to create an extra read-only property for sorting purposes only. Building on the example above:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
// read only, for sorting purposes only
// #JsonIgnore // we can hide it from the clients, if needed
#RestResource(exported=false) // read only so we can map 2 fields to the same database column
#ManyToOne
#JoinColumn(name = "address_id", insertable = false, updatable = false)
private Address address;
// We still want the linkable association created to work as before so we manually override the relation and path
#RestResource(exported=true, rel="address", path="address")
#ManyToOne
private Address addressLink;
...
}
The drawback for the proposed workaround is that we now have to explicitly duplicate all the properties for which we want to support nested sorting.
LATER EDIT: another drawback is that we cannot hide the embedded property from the clients. In my original answer, I was suggesting we can add #JsonIgnore, but apparently that breaks the sort.
I debugged through that and it looks like the issue that Alan mentioned.
I found workaround that could help:
Create own controller, inject your repo and optionally projection factory (if you need projections). Implement get method to delegate call to your repository
#RestController
#RequestMapping("/people")
public class PeopleController {
#Autowired
PersonRepository repository;
//#Autowired
//PagedResourcesAssembler<MyDTO> resourceAssembler;
#GetMapping("/by-address/{addressId}")
public Page<Person> getByAddress(#PathVariable("addressId") Long addressId, Pageable page) {
// spring doesn't spoil your sort here ...
Page<Person> page = repository.findByAddress_Id(addressId, page)
// optionally, apply projection
// to return DTO/specifically loaded Entity objects ...
// return type would be then PagedResources<Resource<MyDTO>>
// return resourceAssembler.toResource(page.map(...))
return page;
}
}
This works for me with 2.6.8.RELEASE; the issue seems to be in all versions.
From Spring Data REST documentation:
Sorting by linkable associations (that is, links to top-level resources) is not supported.
https://docs.spring.io/spring-data/rest/docs/current/reference/html/#paging-and-sorting.sorting
An alternative that I found was use #ResResource(exported=false).
This is not valid (expecially for legacy Spring Data REST projects) because avoid that the resource/entity will be loaded HTTP links:
JacksonBinder
BeanDeserializerBuilder updateBuilder throws
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of ' com...' no String-argument constructor/factory method to deserialize from String value
I tried activate sort by linkable associations with help of annotations but without success because we need always need override the mappPropertyPath method of JacksonMappingAwareSortTranslator.SortTranslator detect the annotation:
if (associations.isLinkableAssociation(persistentProperty)) {
if(!persistentProperty.isAnnotationPresent(SortByLinkableAssociation.class)) {
return Collections.emptyList();
}
}
Annotation
#Retention(RetentionPolicy.RUNTIME)
#Target(ElementType.FIELD)
public #interface SortByLinkableAssociation {
}
At project mark association as #SortByLinkableAssociation:
#ManyToOne
#SortByLinkableAssociation
private Name name;
Really I didn't find a clear and success solution to this issue but decide to expose it to let think about it or even Spring team take in consideration to include at nexts releases.
Please see https://stackoverflow.com/a/66135148/6673169 for possible workaround/hack, when we wanted sorting by linked entity.

JPA Hibernate Spring Repository ensures transaction completes on save?

I am creating a simple spring application which is supposed to book seats in a seminar. Lets say Booking class looks like this
#Entity
#Table(name = "bookings")
#IdClass(BookingId.class)
public class Booking{
#Id
private Long seminarId;
#Id
private String seatNo;
// .. other fields like perticipant info
// .. getter setters
}
of course the BookingId class:
public class BookingId implements Serializable{
private static final long serialVersionUID = 1L;
private Long seminarId;
private String seatNo;
// .. constructors, getters, setters
}
And I have a repository
#Repository
public interface BookingsRepository extends JpaRepository<Booking, BookingId>{
}
in the controller when a booking request arrives I first check if a booking with same seminer id and seat number already exists, if it doesn't exist I create one
#RequestMapping(method = RequestMethod.POST)
public ResponseEntity<BaseCrudResponse> createNewBooking(#Valid #RequestBody NewBookingDao newBookingDao, BindingResult bindingResult){
logger.debug("Request for a new booking");
// .. some other stuffs
Booking newBooking = new Booking();
newBooking.setSeminarId(newBookingDao.getSeminarId());
newBooking.setSeatNumber(newBookingDao.getSeatNumber());
// .. set other fields
Booking existing = bookingsRepository.findOne(new BookingId(newBooking.getSeminarId(), newBooking.getSeatNumber());
if (existing == null)
bookingsRepository.save(newBooking);
return new ResponseEntity<>(new BaseCrudResponse(0), HttpStatus.CREATED);
}
return new ResponseEntity<>(response, HttpStatus.BAD_REQUEST);
}
Now what will happen if the save method of the repository didn't finish commiting transaction and another request already gets past the existence check ? There might be incorrect booking (the last commit will override the previous). Is this scenario likely to happen ? Will the repository ensures that it completes the transaction before another save call ?
Also is there any way to tell Jpa to throw some exception (for IntegrityConstraintException if the composite key (in this case seminerId and seatNumber) already exists ? Now in the present setting its just updating the row.
You can use javax.persistence.LockModeType.PESSIMISTIC_WRITE so other transactions except the one that got the lock cannot update the entity.
If you use spring-data > 1.6 you can annotate the repository method with #Lock :
interface BookingsRepository extends Repository<Booking, Long> {
#Lock(LockModeType.PESSIMISTIC_WRITE)
Booking findOne(Long id);
}
For sure you need to handle the locking exception that may be thron in the controller.

Proper Way to layer Spring JPA based DAO using Spring Boot Framework

Am new to Spring Boot & JPA...
Let's say I have two entities mapped to two tables which are joined in a database.
Student-1------<-Course
Also, lets presume that the database is already created and populated.
This depicts that one student has many courses...
My Student Entity:
#Entity
public class Student {
#OneToMany(mappedBy="student")
private List<Courses> courses;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "Student_Id")
private long studentId;
#Column(name = "Student_Name")
private String studentName;
protected Student() { }
// Getters & Setters
}
My Course Entity:
#Entity
public class Course {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "Course_Id")
private long courseId;
#Id
#Column(name = "Student_Id")
private long studentId;
#ManyToOne
#PrimaryKeyJoinColumn(name="Student_Id", referencedColumnName="Student_Id")
private Student student;
#Column(name = "Course_Name")
private String courseName;
// Getters & Setters
}
In Spring Boot's Tutorial Guides, it illustrates how to extend a CrudRepository interface, but
it doesn't specify how to setup a Spring based DAO which contains custom finder methods which use HQL and EntityManager inside it.
Is the following DAO and DaoImpl correct?
public interface CourseDao {
List<Course> findCoursesByStudentName(String studentName);
}
#Repository
public class CourseDaoImpl implements CourseDao {
#PersistenceContext
EntityManager em;
public List<Course> findCoursesByStudentName(String studentName) {
String sql = "select c.courseName" +
"from Course c, Student s " +
"where c.course_id = s.student_id " +
"and s.studentName = :studentName ";
Query query = em.createQuery(sql);
query.setParameter("studentName", studentName);
return query.getResultList();
}
}
And then in the client code, for example, in the main class:
public class Application {
#Autowired
CustomerDao dao;
public static void main (String args []) {
List<Course> courses = dao.findCoursesByStudentName("John");
}
}
Is this the standard way to use HQL inside Spring DAOs ? I've seend examples of the #Transactional annotation being prepended to the DAO class's impl (e.g. CustomerDAOImpl) ?
Please let me know if this is the write way to structure my Spring Boot app or am I supposed to extend / add to the CrudRepository only?
If someone could correct my example and point me to a URL which talks about HQL using Entities that are joined, I would be very grateful.
The Spring Boot guides didn't depict joins or DAOs - I just need to learn how to properly create finder methods which emulate select statement which return lists or data structures.
Thanks for taking the time to read this...
If I understood your question correct you do have two questions:
How to create a DAO and DAOImpl?
Where to put your Transaction annotations?
In regards to the first question I want to point out that this is a question in regards to spring-data-jpa using Hibernate as a JPA provider, not spring-boot.
Using Spring Data I usually skip completely to create a DAO but directly use a Custom Repository extending a standard one like CrudRepository. So in your case you don't even have to write more code than:
#Repository
public interface StudentRepository extends CrudRepository<Student, Long> {
List<Student> findByStudentName(String studentName);
}
Which will be sufficient and Spring Data will take care of filling it with the correct implementation if you use
#Autowired
StudentRepository studentRepo;
in your service class. This is where I also usually annotate my methods with #Transactional to make sure that everything is working as expected.
In regards to your question about HQL please look into the spring data jpa documentation, which points out that for most of the cases it should be sufficient to stick to proper named methods in the interface or go for named queries (section 3.3.3) or use the #Query annotation (section 3.3.4) to manually define the query, e.g. should work (didn't tried):
#Repository
public interface #CourseRepository extends CrudRepository<Course, Long> {
#Query("select c.courseName from Course c, Student s where c.course_id = s.student_id and s.studentName = :studentName")
public List<Course> findCoursesByStudentName(String studentName);
}
If you annotate your CourseDaoImpl with #Transactional (Assuming your have defined JpaTransactionManager correctly) You can just retrieve the Student with the matching name and call the getCourses() method to lazy load the Courses attached to that student. Since findCoursesByStudentName will run within a Transaction it will load the courses just fine.
#Repository
#Transactional(readOnly=true)
public class CourseDaoImpl implements CourseDao {
#PersistenceContext
EntityManager em;
public List<Course> findCoursesByStudentName(String studentName) {
String sql = "select s " +
"from Student s " +
"where s.studentName = :studentName ";
Query query = em.createQuery(sql);
query.setParameter("studentName", studentName);
User user = query.getSingleResult();
if(user != null) {
return user.getCourses();
}
return new ArrayList<Course>();
}
}

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