Hibernate with JPA annotation problem - lazy object - java

I have a USER table associated with many other tables, in general, star topology.
Like this:
#Entity
#Table(name = "user")
public class User implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#SequenceGenerator(name = "user_USERID_GENERATOR", sequenceName = "user_SEQ")
#GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "userR_USERID_GENERATOR")
#Column(name = "user_id")
private long userId;
#Basic
#Column(name = "password_hex")
private String password;
#Basic
#Column(name = "language")
private String language;
#Temporal(TemporalType.DATE)
private Date created;
#Temporal(TemporalType.DATE)
private Date modyfied;
#Basic
#Column(name = "first_name")
private String firstName;
#Basic
#Column(name = "last_name")
private String lastName;
#Basic
#Column(name = "passport")
private String passport;
#Basic
#Column(name = "pesel")
private String pesel;
#Basic
#Column(name = "phone_nr1")
private String phoneNr1;
#Basic
#Column(name = "phone_nr2")
private String phoneNr2;
#Column(name = "hash")
private String hash;
// uni-directional many-to-one association to DictUserType
#ManyToOne
#JoinColumn(name = "status")
private DictUserStatus status;
#OneToMany(fetch = FetchType.LAZY, mappedBy = "user", cascade = { CascadeType.ALL })
private Set<Email> emails = new HashSet<Email>(0);
#OneToMany(fetch = FetchType.LAZY, mappedBy = "user", cascade = { CascadeType.ALL })
private Set<Address> address = new HashSet<Address>(0);
#OneToMany(fetch = FetchType.LAZY, mappedBy = "user", cascade = { CascadeType.ALL })
private Set<ArchivePasswords> archivePasswords = new HashSet<ArchivePasswords>(
0);
#OneToMany(fetch = FetchType.LAZY, mappedBy = "user", cascade = { CascadeType.ALL })
private Set<HostsWhitelist> hostsWhitelist = new HashSet<HostsWhitelist>(0);
....
I have a DAO layer, the method of search by user ID.
public User findUser(long userId) throws UserNotFoundException {
User user = userDao.findUser(userId);
if (user == null) {
throw new UserNotFoundException("Could not find user with id = "
+ userId);
}
return user;
}
Why lazy fetching does not work?

You should post the stack trace you are receiving. Where is the LazyLoadingException occurring? On the user Object? Are you trying to access it from another Object?
Is this the notorious LazyInitializationException? If so, then you need to either traverse the Object graph manually in the service (assuming you DAO code snippet is actually a Service method and not the DAO itself), or research the OpenSessionInViewFilter (assuming you are using Spring).

If you want to fetch the user with emails.
#Transactional
public List getUserWithEmails(long userId){
User user = userDao.findUser(userId);
if (user == null) {
throw new UserNotFoundException("Could not find user with id = "
+ userId);
}
for(Email email:user.getEmails()){
email.getId();
}
return user;
}
The same procedure apply to other one-to-many sets. Just like others have stated, you need to add OpenSessionInView (Hibernate) filter or OpenEntityManagerInView (JPA)filter in web.xml

If not specified, lazy fet hing will not take place defaults to EAGER.
public #interface Basic
The simplest type of mapping to a database column. The Basic annotation can be applied to a persistent property or instance variable of any of the following types: Java primitive types, wrappers of the primitive types, String, java.math.BigInteger, java.math.BigDecimal, java.util.Date, java.util.Calendar, java.sql.Date, java.sql.Time, java.sql.Timestamp, byte[], Byte[], char[], Character[], enums, and any other type that implements java.io.Serializable.
The use of the Basic annotation is optional for persistent fields and properties of these types. If the Basic annotation is not specified for such a field or property, the default values of the Basic annotation will apply.
Example 1:
#Basic
protected String name;
Example 2:
#Basic(fetch=LAZY)
protected String getName() { return name; }
fetch
public abstract FetchType fetch
(Optional) Defines whether the value of the field or property should be lazily loaded or must be eagerly fetched. The EAGER strategy is a requirement on the persistence provider runtime that the value must be eagerly fetched. The LAZY strategy is a hint to the persistence provider runtime. If not specified, defaults to EAGER.
Default:
javax.persistence.FetchType.EAGER
optional
public abstract boolean optional
(Optional) Defines whether the value of the field or property may be null. This is a hint and is disregarded for primitive types; it may be used in schema generation. If not specified, defaults to true.
Default:
true

Related

Hibernate: How to update parent_id field without retrieving parent from db?

In a Spring Boot app, I have the following entities that have one-to-many relationship (Category is the parent of Recipe):
#Entity
public class Recipe {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(nullable = false, length = 50)
private String title;
#ManyToOne(optional = true, fetch = FetchType.LAZY)
#JoinColumn(name = "category_id", referencedColumnName = "id")
private Category category;
}
#Entity
public class Category {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(unique = true, nullable = false, length = 50)
private String name;
#OneToMany(mappedBy = "category", cascade = CascadeType.ALL)
private Set<Recipe> recipes = new HashSet<>();
public void addRecipe(Recipe recipe) {
recipes.add(recipe);
recipe.setCategory(this);
}
public void removeRecipe(Recipe recipe) {
recipes.remove(recipe);
recipe.setCategory(null);
}
}
When I create a Recipe, I send categoryId that is selected from Dropdown list and create Recipe by retrieving and adding category to the recipe as shown below:
#Transactional
public void update(RecipeRequest request) {
final Category category = categoryRepository.findById(request.getCategoryId())
.orElseThrow(() -> new NoSuchElementFoundException(NOT_FOUND_CATEGORY));
/* instead of retrieving category, I want to set the categoryId field of Recipe,
but there is not such kind of setter */
recipe.setCategoryId(request.getCategoryId());
recipe.setTitle(capitalizeFully(request.getTitle()));
recipe.setCategory(category);
recipeRepository.save(recipe);
}
Instead of retrieving category, I want to set the categoryId field of Recipe, but there is not such kind of setter:
recipe.setCategoryId(request.getCategoryId());
So, what is the most proper way for just setting the categoryId of the recipe and then saving it without requiring the category from db? Do I need a setter for categoryId field to the Recipe (I thought it, but does not seem elegant way)?
I would just add a categoryId field along with the corresponding getter and settter methods to the Recipe class.
#Entity
public class Recipe {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(nullable = false, length = 50)
private String title;
#ManyToOne(optional = true, fetch = FetchType.LAZY)
#JoinColumn(name = "category_id", referencedColumnName = "id")
private Category category;
#Column(name = "category_id", nullable = false)
private Integer categoryId;
// getters/setters
}
Having a categoryId field means that when we don't have to create an instance of Category when adding new Recipes. Sure, Recipe.category will be null but that's ok if we're just adding new Recipes. This approach could also prove beneficial if we later decide that we need to add many Recipes simultaneously.
If your repository implements org.springframework.data.jpa.repository.JpaRepository you may take advantage of using JpaRepository#getReferenceById method, in that case Hibernate instead of querying DB for data will return proxy object. However, such implementation may cause issues in some cases, for example:
// this call typically returns entity
// or null if entity wasn't found
repository.findById(id);
but:
// this call returns proxy object
repository.getReferenceById(id);
// now instead of returning entity
// repository either returns initialized proxy
// object or throws EntityNotFoundException
// if entity wasn't found
repository.findById(id);

JPA: How to handle versioned entities?

I have a versioning on an entity as part of its primary key. The versioning is done via a timestamp of the last modification:
#Entity
#Table(name = "USERS")
#IdClass(CompositeKey.class)
public class User {
#Column(nullable = false)
private String name;
#Id
#Column(name = "ID", nullable = false)
private UUID id;
#Id
#Column(name = "LAST_MODIFIED", nullable = false)
private LocalDateTime lastModified;
// Constructors, Getters, Setters, ...
}
/**
* This class is needed for using the composite key.
*/
public class CompositeKey {
private UUID id;
private LocalDateTime lastModified;
}
The UUID is translated automatically into a String for the database and back for the model. The same goes for the LocalDateTime. It gets automatically translated to a Timestamp and back.
A key requirement of my application is: The data may never update or be deleted, therefore any update will result in a new entry with a younger lastModified. This requirement is satisfied with the above code and works fine until this point.
Now comes the problematic part: I want another object to reference on a User. Due to versioning, that would include the lastModified field, because it is part of the primary key. This yields a problem, because the reference might obsolete pretty fast.
A way to go might be depending on the id of the User. But if I try this, JPA tells me, that I like to access a field, which is not an Entity:
#Entity
#Table(name = "USER_DETAILS")
public class UserDetail {
#Id
#Column(nullable = false)
private UUID id;
#OneToOne(optional = false)
#JoinColumn(name = "USER_ID", referencedColumnName = "ID")
private UUID userId;
#Column(nullable = false)
private boolean married;
// Constructors, Getter, Setter, ...
}
What would be the proper way of solving my dilemma?
Edit
I got a suggestion by JimmyB which I tried and failed too. I added the failing code here:
#Entity
#Table(name = "USER_DETAILS")
public class UserDetail {
#Id
#Column(nullable = false)
private UUID id;
#OneToMany
#JoinColumn(name = "USER_ID", referencedColumnName = "ID")
private List<User> users;
#Column(nullable = false)
private boolean married;
public User getUser() {
return users.stream().reduce((a, b) -> {
if (a.getLastModified().isAfter(b.getLastModified())) {
return a;
}
return b;
}).orElseThrow(() -> new IllegalStateException("User detail is detached from a User."));
}
// Constructors, Getter, Setter, ...
}
What you seem to require seems to be on the lines of a history table, to keep track of the changes. See https://wiki.eclipse.org/EclipseLink/Examples/JPA/History on how EclipseLink can handle this for you while using normal/traditional JPA mappings and usage.
What you have here is a logical 1:1 relationship which, due to versioning, becomes a technical 1:n relationship.
You have basically three options:
Clean JPA way: Declare an 'inverse' #ManyToOne relationship from user to the "other object" and make sure you always handle it whenever a new User record is created.
'Hack-ish' way: Declare a #OneToMany relationship in the "other object" and force it to use a specific set of columns for the join using #JoinColumn. The problem with this is that JPA always expects unique reference over the join columns so that reading the UserDetail plus referenced User records should work, whereas writing UserDetail should not cascade onto User to avoid unwanted/undocumented effects.
Just store the user's UUID in the "other object" and resolve the reference yourself whenever you need it.
The added code in your question is wrong:
#JoinColumn(name = "USER_ID", referencedColumnName = "ID")
private UUID userId;
More correct, albeit not with the result you want, would be
#JoinColumn(name = "USER_ID", referencedColumnName = "ID")
private User user;
This won't work though, because, as I said above, you may have more than one user record per UserDetail, so you'd need a #OneToMany relationship here, represented by a Collection<User>.
Another 'clean' solution is to introduce an artificial entity with a 1:1 cardinality w.r.t. to the logical User to which you can refer, like
#Entity
public class UserId {
#Id
private UUID id;
#OneToMany(mappedBy="userId")
private List<User> users;
#OneToOne(mappedBy="userId")
private UserDetail detail;
}
#Entity
public class User {
#Id
private Long _id;
#ManyToOne
private UserId userId;
}
#Entity
public class UserDetail {
#OneToOne
private UserId userId;
}
This way, you can somewhat easily navigate from users to details and back.
I came to a solution, that is not really satisfying, but works. I created a UUID field userId, which is not bound to an Entity and made sure, it is set only in the constructor.
#Entity
#Table(name = "USER_DETAILS")
public class UserDetail {
#Id
#Column(nullable = false)
private UUID id;
#Column(nullable = false)
// no setter for this field
private UUID userId;
#Column(nullable = false)
private boolean married;
public UserDetail(User user, boolean isMarried) {
this.id = UUID.randomUUID();
this.userId = user.getId();
this.married = isMarried;
}
// Constructors, Getters, Setters, ...
}
I dislike the fact, that I cannot rely on the database, to synchronize the userId, but as long as I stick to the no setter policy, it should work pretty well.

How to insert in to multiple tables via Hibernate

I have two tables I need to insert in to in Hibernate - I have a User and every user belongs is a Member. Therfore when creating a new user I need a new entry in the Member table. I have attempted this via creating a Member object which maps to my member table and then having that as a field in my User object which maps to the user table
#Entity
#Table(name = "USER")
public class User
{
#Id
#GeneratedValue
#Column(name = "id")
private int id;
#Column(name = "username")
private String username;
#Column(name = "password")
private String password;
#Column(name = "fullName")
private String fullName;
//other fields ommited
#OneToOne
#JoinColumn(name = "id")
private Member member;
My member pojo looks as follows
#Entity
#Table(name = "MEMBER")
public class Member
{
#Id
#GeneratedValue
#Column(name = "id")
private int id;
#Column(name = "sgpid")
private int sgpid;
#Column(name = "username")
private String username;
Trying to save the object i do as follows;
#Override
public boolean addUser(User user)
{
if (user == null)
{
throw new IllegalArgumentException("Unable to add null user");
}
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
session.save(user);
session.getTransaction().commit();
return true;
}
This gives me the row saved in my user table but the entry is not inserted in to the member table. I think my linking annotations are probably incorrect but I am not too sure - please could someone provide some assistance.
Thanks
Try to set the cascade value of the #OneToOne annotation:
#OneToOne(cascade = CascadeType.PERSIST)
#JoinColumn(name = "id")
private Member member;
First thing in your user class you should change the joinColumn to member_id.
As mentioned in another answer to persist a related entity you need to set the cascade to persist, i would recommend using cascade All which will involve the related entity in all operations check the doc
https://docs.oracle.com/cd/E19798-01/821-1841/bnbqm/index.html
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = "member_id")
private Member member;

Hibernate Lazy Loading makes criteria slow to run

I am experiencing a problem with hibernate and lazy loading of objects.
basically I want to load an class which has an eagerly loaded field and not load the lazy fields of child classes
Take the following QuestionVO class
#Entity
#Table(name = "question")
public class QuestionVO extends BaseDAOVO implements Serializable {
/**
*
*/
private static final long serialVersionUID = -5867047752936216092L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", unique = true, nullable = false)
private Integer id;
#Column(name = "questionText", unique = false, nullable = false, length = 4000)
#Size(min = 3, max = 4000)
#Pattern(regexp = MobileAppsRegexConstants.GENERAL_ALLOWED_CHARCHTERS, message = "Question Text Not valid.")
private String questionText;
#ManyToOne(fetch = FetchType.EAGER)
#Cascade({ CascadeType.SAVE_UPDATE })
#JoinColumn(name = "MENU_STYLE_ID", nullable = true)
private MenuStyleVO menuStyle;
}
Take the following MenuStyleVO class
#Entity
#Table(name = "menu_style")
public class MenuStyleVO extends BaseDAOVO implements Serializable{
/**
*
*/
private static final long serialVersionUID = 3697798179195096156L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", unique = true, nullable = false)
private Integer id;
#Column(name = "menuStyleName", unique = false, nullable = false, length = 200)
private String menuStyleName;
#Column(name = "menuTemplate", unique = false, nullable = false, length = 200)
private String menuTemplate;
#OneToOne(fetch = FetchType.LAZY, optional=false)
#Cascade({ CascadeType.SAVE_UPDATE })
#JoinColumn(name="logo_id")
#JsonProperty("logo")
private ApplicationImageVO logo;
}
And this ApplicationImageVO class
#Entity
#Table(name = "application_image")
public class ApplicationImageVO extends BaseDAOVO implements Serializable {
/**
*
*/
private static final long serialVersionUID = -9158898930601867545L;
#OneToOne(fetch = FetchType.LAZY, mappedBy = "image1242x2208")
#Cascade({ CascadeType.ALL })
#JsonIgnore
private SubmissionLauncherImagesVO launcherImage1242x2208;
#OneToOne(fetch = FetchType.LAZY, mappedBy = "image1536x2048")
#Cascade({ CascadeType.ALL })
#JsonIgnore
private SubmissionLauncherImagesVO launcherImage1536x2048;
#OneToOne(fetch = FetchType.LAZY, mappedBy = "image2048x1536")
#Cascade({ CascadeType.ALL })
#JsonIgnore
private SubmissionLauncherImagesVO launcherImage2048x1536;
#OneToOne(fetch = FetchType.LAZY, mappedBy = "logo")
#Cascade({ CascadeType.ALL })
#JsonIgnore
private MenuStyleVO menuStyleLogo;
}
If L load the QuestionVO class from the database using the following hibernate criteria code - all the lazy fields of MenuStyleVO and ApplicationImageVO are also loaded.
On complicated use cases, this results in this query getting very slow
public QuestionVO findMasterAppQuestionById(int id) {
Criteria criteria = currentSession().createCriteria(QuestionVO.class);
criteria.add(Restrictions.eq("id", id));
QuestionVO questionVO = (QuestionVO) criteria.uniqueResult();
return questionVO;
}
What I am wondering is - would it be possible to load the QuestionVO class and its eager fields and tell hibernate to ignore lazy fields from the other classes bar those that are needed?
Cheers
Damien
Last time we faced an issue like this we used a constructor on parent class, which use only the desired fields of determined query.
I can't remember in fully how constructor inside a jpql query works, but it must be something like this:
select new com.package.class(c.field1, c.field2) from com.package.class c
Remember, a constructor with same arguments must be present on the desired entity.
Pros:
- Better query perfomance;
- Can be replicated with other arguments;
Cons:
- Pretty limited, you can only use this hack on the main entity you are querying;
- Includes a constructor only for determined query, poor design;
Also, you should take a look on EnttyGraphs of JPA. Seems quite promising, but didn't work as desired in our project.
Btw, Hibernate has put us many times on performance issues, hope this hack help you, good luck!
Edit:
Why this pattern would help in performance issues?
Basically, with the example i've showed before, you are not loading everything via Hibernate, only the two fields (field1 and field2) of the main entity. Without using a constructor you shoudn't be able to do that, because your query would not result in a collection of the desired entity, but in a collection of two objects each iteration (Object[]). Using the constructor pattern you are creating instances of the desired entity, but only selecting a few fields from database, and that's why this pattern can help you, you are returning a collection of the desired entity with only a few fields.

JPA Compound key with #EmbeddedId

In a legacy database, I have three tables: Users, Workgroups, and UsersWorkgroup. UsersWorkgroup stores what role a user has in a workgroup.
Here are the relevant code snippets:
#Entity
#Table(name = "users_workgroup")
public class UsersWorkgroup implements Serializable {
private static final long serialVersionUID = 1L;
#EmbeddedId
protected UsersWorkgroupPK usersWorkgroupPK;
#JoinColumn(name = "idworkgroup", referencedColumnName = "idworkgroup")
#ManyToOne(optional = false)
private Workgroup workgroup;
#JoinColumn(name = "user_name", referencedColumnName = "user_name")
#ManyToOne(optional = false)
private Users users;
#Column(name = "role")
private Integer role;
#Embeddable
public class UsersWorkgroupPK implements Serializable {
#Basic(optional = false)
#Column(name = "idworkgroup", insertable=false, updatable=false)
private int idworkgroup;
#Basic(optional = false)
#Column(name = "user_name", insertable=false, updatable=false)
private String userName;
#Entity
#Table(name = "workgroup")
public class Workgroup implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Basic(optional = false)
#Column(name = "idworkgroup")
private Integer idworkgroup;
#Column(name = "name")
private String name;
#OneToMany(cascade = CascadeType.ALL, mappedBy = "idworkgroup")
private Collection<UsersWorkgroup> usersWorkgroupCollection;
And of course, problem is, it doesn't work.
Currently I get this exception:
Exception Description: An incompatible
mapping has been encountered between
[class entity.Workgroup] and [class
entity.UsersWorkgroup]. This usually
occurs when the cardinality of a
mapping does not correspond with the
cardinality of its backpointer.
Which I don't understand since OneToMany should match ManyToOne... Or is it a ManyToMany relationship? If I switch to #ManyToMany, I get this:
Exception Description: The target
entity of the relationship attribute
[workgroup] on the class [class
com.ericsson.rsg.ejb.entity.UsersWorkgroup]
cannot be determined. When not using
generics, ensure the target entity is
defined on the relationship mapping.
I'm trying to understand compound keys (embedded), but all the examples I could find have only simple columns that are not foreign keys (but that's the whole point of a compound key, isn't it?). Can the UsersWorkgroup table secretly be a join table?
Should I declare the PK class as a strict POJO class? Or should I put the #JoinColumn annotations in the PK class? How do I refer to the columns within the compound key from another table? Should I initialize the PK object in the refering class constructor, or is it not necessary?
I feel stuck completely.
First of all, I think your relation is a Many To Many, as a user can be in many groups, and a group can have many users (or I would assume so).
Second, as far as I know you have to reference both id_workgroup and user_name as JoinColumns, because they are part of the PK and a unit, so both should be referenced.
Also, I see the "equals" and "hashCode" methods missing from your embedded PK, as well as the getters/setters. I believe they are mandatory.
Your mapping looks fine except for mappedBy - it should be a property name, not a column name:
#OneToMany(cascade = CascadeType.ALL, mappedBy = "workgroup")
private Collection<UsersWorkgroup> usersWorkgroupCollection;

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