I'm working with Spring App, so to work with DB I use Spring Data JPA. Firstly I saved an object. And after some time I need to update this object in the table. But at this moment my object contains one field which is null. But I don't want to update this field with null. So my question is how to prevent updating fields with null? Maybe there is an annotation or some property to solve my problem.My entity:
#Entity
#Table(name = "users")
#Getter
#Setter
#NoArgsConstructor
#AllArgsConstructor
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "user_id")
private Long id;
#NotNull
#Column(name = "user_name")
#Field
private String username;
#NotNull
#Column(name = "user_identity")
private String identity;
#Column(name="user_image")
private String image;
#Column(name="user_joined")
private String date;
#Column(name="user_origin")
private String origin;
#Column(name="user_sub")
private String sub;
I save and update this entity with implementation of JpaRepository:
#Repository
public interface UserRepository extends JpaRepository<User, Long>
it looks like this:
#Autowired
private UserRepository userRepository;
....
userRepository.save(user);
I've saved my object with not null sub-field. And now I want to update some fields of saved entity, but not sub field, which is null in current object. I wonder if there is any possibility to avoid changing user_sub field to null?
You can add #DynamicUpdate annotation to your User class. This will ignore the fields whose values are null. You can simply do like:
//other annotations
#DynamicUpdate
public class User {
// other codes inside class
}
You can follow a good example from Mkyong's site.
Thanks, guys. I found the solution: #Query will help to update fields that I need
Related
I have an entity class and repository. Here I'm trying to execute update query but not working.
How to update Lob column in native query or any another solution on jpa query to update Lob column.
#Entity
#Table(name = "comment")
public class Comment implements Serializable {
#Basic
#Lob
#Column(name="Article_COMMENT", columnDefinition="TEXT")
private String articleComment;
#Basic
#Column(name = "ID_ARTICLE")
private Long articleId;
}
#Repository
public interface commentRepository extends JpaRepository<Comment, Long> {
#Query(value = "UPDATE comment set articleComment=: articleComment WHERE articleId =: articleId", nativeQuery=true)
void updateComment(#Param("articleComment") String articleComment, #Param("articleId") Long articleId );
}
Error:
No results were returned by query.
JpaSystemException thrown with message: could not extract ResultSet; nested exception is org.hibernate.exception.GenericJDBCException: could not extract ResultSet
Your question is very vague so I can answer on assumptions only. I think You want to update the articalComment field of your Entity. You can simply use .save() method of JpaRepository. Your code should be as follows. Here I am also assuming that your articleId is unique identifier to your entity class.
#Entity
#Table(name = "comment")
public class Comment implements Serializable {
#Basic
#Lob
#Column(name="Article_COMMENT", columnDefinition="TEXT")
private String articleComment;
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "ID_ARTICLE")
private Long articleId;
}
Now your Id should be unique and has a #Id Annotation to identify it inside spring data JPA.
You don't have to add any code inside of your JPA repository. Simply call commentRepository.save(commentObject) method. If commentObject has an ID as 0 then a new Comment will be created. If the ID is a positive value and is present in your table that particular row will be updated not created.
remove the space try this way
UPDATE comment set articleComment=:articleComment WHERE articleId =:articleId
I have a super Entity class like this:
#Getter
#Setter
#NoArgsConstructor
public class GenericEntity {
#Id
private Long id;
#JsonIgnore
#CreatedBy
private Long createdBy;
#JsonIgnore
#CreatedDate
private Long createdDate;
#JsonIgnore
#LastModifiedBy
private Long updatedBy;
#JsonIgnore
#LastModifiedDate
private Long updatedDate;
#JsonIgnore
#Version
private Integer version = 0;
}
and a Role class extends from GenericEntity like this:
#Getter
#Setter
#NoArgsConstructor
public class Role extends GenericEntity {
private String name;
private String desc;
private Integer sort;
}
And after that I have interface RoleRepo like this:
#Repository
public interface RoleRepo extends ReactiveCrudRepository<Role, Long>;
In Router function, I have 2 handler methods
private Mono<ServerResponse> findAllHandler(ServerRequest request) {
return ok()
.contentType(MediaType.APPLICATION_JSON)
.body(roleRepo.findAll(), Role.class);
}
private Mono<ServerResponse> saveOrUpdateHandler(ServerRequest request) {
return ok()
.contentType(MediaType.APPLICATION_JSON_UTF8)
.body(request.bodyToMono(Role.class).flatMap(role -> {
return roleRepo.save(role);
}), Role.class);
}
The method findAllHandler works fine, but the saveOrUpdateHandler throw exception like this:
java.lang.IllegalStateException: Required identifier property not found for class org.sky.entity.system.Role!
at org.springframework.data.mapping.PersistentEntity.getRequiredIdProperty(PersistentEntity.java:105) ~[spring-data-commons-2.2.0.M2.jar:2.2.0.M2]
at org.springframework.data.r2dbc.function.convert.MappingR2dbcConverter.lambda$populateIdIfNecessary$0(MappingR2dbcConverter.java:85) ~[spring-data-r2dbc-1.0.0.M1.jar:1.0.0.M1]
But when I move
#Id
private Long id;
from GenericEntity class to Role class, the two methods work fine.
Are there any Annations #MappedSuperclass/JPA in Spring Reactive Data like that
I wish the id field in GenericEntity for all extends class
Thanks for your help
Sorry, my English so bad
I had a similar problem and after some search, I didn't find an answer to your question, so I test it by writing code and the answer is spring data R2DBC doesn't need #Mappedsuperclass. it aggregates Role class properties with Generic class properties and then inserts all into the role table without the need to use any annotation.
I have a logic that saves some data and I use spring boot + spring data jpa.
Now, I have to save one object, and after moment, I have to save another objeect.
those of object consists of three primary key properties.
- partCode, setCode, itemCode.
let's say first object has a toString() returning below:
SetItem(partCode=10-001, setCode=04, itemCode=01-0021, qty=1.0, sortNo=2, item=null)
and the second object has a toString returning below:
SetItem(partCode=10-001, setCode=04, itemCode=01-0031, qty=1.0, sortNo=2, item=null)
there is a difference on itemCode value, and itemCode property is belonged to primary key, so the two objects are different each other.
but in my case, when I run the program, the webapp saves first object, and updates first object with second object value, not saving objects seperately.
(above image contains different values from this post question)
Here is my entity information:
/**
* The persistent class for the set_item database table.
*
*/
#Data
#DynamicInsert
#DynamicUpdate
#Entity
#ToString(includeFieldNames=true)
#Table(name="set_item")
#IdClass(SetGroupId.class)
public class SetItem extends BasicJpaModel<SetItemId> {
private static final long serialVersionUID = 1L;
#Id
#Column(name="PART_CODE")
private String partCode;
#Id
#Column(name="SET_CODE")
private String setCode;
#Id
#Column(name="ITEM_CODE")
private String itemCode;
private Double qty;
#Column(name="SORT_NO")
private int sortNo;
#Override
public SetItemId getId() {
if(BooleanUtils.ifNull(partCode, setCode, itemCode)){
return null;
}
return SetItemId.of(partCode, setCode, itemCode);
}
#ManyToMany(fetch=FetchType.LAZY)
#JoinColumns(value = {
#JoinColumn(name="PART_CODE", referencedColumnName="PART_CODE", insertable=false, updatable=false)
, #JoinColumn(name="ITEM_CODE", referencedColumnName="ITEM_CODE", insertable=false, updatable=false)
})
private List<Item> item;
}
So the question is,
how do I save objects separately which the objects' composite primary keys are partially same amongst them.
EDIT:
The entity extends below class:
#Setter
#Getter
#MappedSuperclass
#DynamicInsert
#DynamicUpdate
public abstract class BasicJpaModel<PK extends Serializable> implements Persistable<PK>, Serializable {
#Override
#JsonIgnore
public boolean isNew() {
return null == getId();
}
}
EDIT again: embeddable class.
after soneone points out embeddable class, I noticed there are only just two properties, it should be three of it. thank you.
#Data
#NoArgsConstructor
#RequiredArgsConstructor(staticName="of")
#Embeddable
public class SetGroupId implements Serializable {
//default serial version id, required for serializable classes.
private static final long serialVersionUID = 1L;
#NonNull
private String partCode;
#NonNull
private String setCode;
}
Check howto use #EmbeddedId & #Embeddable (update you might need to use AttributeOverrides in id field, not sure if Columns in #Embeddable works).
You could create class annotated #Embeddable and add all those three ID fields there.
#Embeddable
public class MyId {
private String partCode;
private String setCode;
private String itemCode;
}
Add needed getters & setters.
Then set in class SetItem this class to be the id like `#EmbeddedId´.
public class SetItem {
#EmbeddedId
#AttributeOverrides({
#AttributeOverride(name="partCode",
column=#Column(name="PART_CODE")),
#AttributeOverride(name="setCode",
column=#Column(name="SET_CODE"))
#AttributeOverride(name="itemCode",
column=#Column(name="ITEM_CODE"))
})
MyId id;
Check also Which annotation should I use: #IdClass or #EmbeddedId
Be sure to implement equals and hashCode in SetGroupId.
Can you provide that class?
Hi I am new to Spring Data JPA and I am wondering even though I pass the Id to the entity, the Spring data jpa is inserting instead of merge. I thought when I implement the Persistable interface and implement the two methods:
public Long getId();
public Boolean isNew();
It will automatically merge instead of persist.
I have an entity class called User like:
#Entity
#Table(name = "T_USER")
public class User implements Serializable, Persistable<Long> {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "USER_ID")
private Long id;
#Column(name = "CREATION_TIME", nullable = false)
private Date creationTime;
#Column(name = "FIRST_NAME", nullable = false)
private String firstName;
#Column(name = "LAST_NAME", nullable = false)
private String lastName;
#Column(name = "MODIFICATION_TIME", nullable = false)
private Date modificationTime;
And have another class
#Entity
#Table(name = "T_USER_ROLE")
public class UserRole implements Serializable, Persistable<Long> {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long roleId;
#Column(name = "ROLE_NAME")
private String userRole;
}
I have a custom repository called UserRepostory extending JpaReopistory. I am hitting the save for merge and persist as I see the implementation demonstrate that Spring Data Jpa uses above two methods to either update or insert.
#Repository
public interface UserRepository extends JpaRepository<User, Long> {
}
I have been trying to figure out but didn't get any clue. Maybe you
guys can help.
I ran into this issue, tried to implement Persistable to no avail, and then looked into the Spring Data JPA source. I don't necessarily see this in your example code, but I have a #Version field in my entity. If there is a #Version field Spring Data will test that value to determine if the entity is new or not. If the #Version field is not a primitive and is null then the entity is considered new.
This threw me for a long time in my tests because I was not setting the version field in my representation but only on the persisted entity. I also don't see this documented in the otherwise helpful Spring Data docs (which is another issue...).
Hope that helps someone!
By default Spring Data JPA inspects the identifier property of the given entity. If the identifier property is null, then the entity will be assumed as new, otherwise as not new. It's Id-Property inspection Reference
If you are using Spring JPA with EntityManager calling .merge() will update your entity and .persist() will insert.
#PersistenceContext
private EntityManager em;
#Override
#Transactional
public User save(User user) {
if (user.getId() == null) {
em.persist(user);
return user;
} else {
return em.merge(user);
}
}
There is no need to implement the Persistable interface.
I have two entity bean :
#Entity
public class User {
#Id
#GeneratedValue
private Long id;
#OneToMany(mappedBy="user", cascade = CascadeType.ALL)
private List<Comment> comments = new ArrayList<Comment>();
//SOME OTHER CLASS VARIABLES
//GETTERS AND SETTERS
}
and my Comment class is like this :
#Entity
public class Comment {
#Id
#GeneratedValue
private Long id;
private String title;
private String content;
#ManyToOne
private User user
//SOME OTHER CLASS VARIABLES
//GETTERS AND SETTERS
}
now I know that I can get the User Object from session and set the user for my comment like this in order to be able to use the join feature in JPA:
commentObject.setUser(TheSessionGrabedUserObject/UserObjectWhichHasFetchedFromDbUsingUserId);
but as long as I have the userId for my user Object I do not need to do this.
I'm looking for a way to insert this foreignKey into my comment table without getting the User Object from session or maybe query to database to fetch it first !
how I'm gonna do it using JPQL ?
You can use the entityManager.getReference() method. In your case:
entityManager.getReference(User.class, userId);
This will not perform any DB query, but will give you a User instance with only the ID populated, and you can pass that to commentObject.setUser().