How do I update an entity using spring-data-jpa?

Question

Well the question pretty much says everything. Using JPARepository how do I update an entity?

JPARepository has only a save method, which does not tell me if it's create or update actually. For example, I insert a simple Object to the database User, which has three fields: firstname, lastname and age:

 @Entity
 public class User {

  private String firstname;
  private String lastname;
  //Setters and getters for age omitted, but they are the same as with firstname and lastname.
  private int age;

  @Column
  public String getFirstname() {
    return firstname;
  }
  public void setFirstname(String firstname) {
    this.firstname = firstname;
  }

  @Column
  public String getLastname() {
    return lastname;
  }
  public void setLastname(String lastname) {
    this.lastname = lastname;
  }

  private long userId;

  @Id
  @GeneratedValue(strategy=GenerationType.AUTO)
  public long getUserId(){
    return this.userId;
  }

  public void setUserId(long userId){
    this.userId = userId;
  }
}

Then I simply call save(), which at this point is actually an insert into database:

 User user1 = new User();
 user1.setFirstname("john"); user1.setLastname("dew");
 user1.setAge(16);

 userService.saveUser(user1);// This call is actually using the JPARepository: userRepository.save(user);

So far so good. Now I want to update this user, say change his age. For this purpose I could use a Query, either QueryDSL or NamedQuery, whatever. But, considering I just want to use spring-data-jpa and the JPARepository, how do I tell it that instead of an insert I want to do an update?

Specifically, how do I tell spring-data-jpa that users with the same username and firstname are actually EQUAL and that the existing entity supposed to be updated? Overriding equals did not solve this problem.

Answer 1

Identity of entities is defined by their primary keys. Since firstname and lastname are not parts of the primary key, you cannot tell JPA to treat Users with the same firstnames and lastnames as equal if they have different userIds.

So, if you want to update a User identified by its firstname and lastname, you need to find that User by a query, and then change appropriate fields of the object your found. These changes will be flushed to the database automatically at the end of transaction, so that you don't need to do anything to save these changes explicitly.

EDIT:

Perhaps I should elaborate on overall semantics of JPA. There are two main approaches to design of persistence APIs:

• insert/update approach. When you need to modify the database you should call methods of persistence API explicitly: you call insert to insert an object, or update to save new state of the object to the database.

• Unit of Work approach. In this case you have a set of objects managed by persistence library. All changes you make to these objects will be flushed to the database automatically at the end of Unit of Work (i.e. at the end of the current transaction in typical case). When you need to insert new record to the database, you make the corresponding object managed. Managed objects are identified by their primary keys, so that if you make an object with predefined primary key managed, it will be associated with the database record of the same id, and state of this object will be propagated to that record automatically.

JPA follows the latter approach. save() in Spring Data JPA is backed by merge() in plain JPA, therefore it makes your entity managed as described above. It means that calling save() on an object with predefined id will update the corresponding database record rather than insert a new one, and also explains why save() is not called create().

Answer 2

Since the answer by @axtavt focuses on JPA not spring-data-jpa

To update an entity by querying then saving is not efficient because it requires two queries and possibly the query can be quite expensive since it may join other tables and load any collections that have fetchType=FetchType.EAGER

Spring-data-jpa supports update operation.
You have to define the method in Repository interface.and annotated it with @Query and @Modifying.

@Modifying
@Query("update User u set u.firstname = ?1, u.lastname = ?2 where u.id = ?3")
void setUserInfoById(String firstname, String lastname, Integer userId);

@Query is for defining custom query and @Modifying is for telling spring-data-jpa that this query is an update operation and it requires executeUpdate() not executeQuery().

You can specify the return type as int, having the number of records being updated.

---

Note: Run this code in a Transaction.

Answer 3

You can simply use this function with save() JPAfunction, but the object sent as parameter must contain an existing id in the database otherwise it will not work, because save() when we send an object without id, it adds directly a row in database, but if we send an object with an existing id, it changes the columns already found in the database.

public void updateUser(Userinfos u) {
    User userFromDb = userRepository.findById(u.getid());
    // crush the variables of the object found
    userFromDb.setFirstname("john"); 
    userFromDb.setLastname("dew");
    userFromDb.setAge(16);
    userRepository.save(userFromDb);
}

Answer 4

As what has already mentioned by others, the save() itself contains both create and update operation.

I just want to add supplement about what behind the save() method.

Firstly, let's see the extend/implement hierarchy of the CrudRepository<T,ID>,

Ok, let's check the save() implementation at SimpleJpaRepository<T, ID>,

@Transactional
public <S extends T> S save(S entity) {

    if (entityInformation.isNew(entity)) {
        em.persist(entity);
        return entity;
    } else {
        return em.merge(entity);
    }
}

As you can see, it will check whether the ID is existed or not firstly, if the entity is already there, only update will happen by merge(entity) method and if else, a new record is inserted by persist(entity) method.

Answer 5

spring data save() method will help you to perform both: adding new item and updating an existed item.

Just call the save() and enjoy the life :))

Answer 6

Using spring-data-jpa save(), I was having same problem as @DtechNet. I mean every save() was creating new object instead of update. To solve this I had to add version field to entity and related table.

Answer 7

This is how I solved the problem:

User inbound = ...
User existing = userRepository.findByFirstname(inbound.getFirstname());
if(existing != null) inbound.setId(existing.getId());
userRepository.save(inbound);

Answer 8

With java 8 you can use repository's findById in UserService

@Service
public class UserServiceImpl {

    private final UserRepository repository;

    public UserServiceImpl(UserRepository repository) {
        this.repository = repository;
    }

    @Transactional
    public void update(User user) {
        repository
                .findById(user.getId()) // returns Optional<User>
                .ifPresent(user1 -> {
                    user1.setFirstname(user.getFirstname);
                    user1.setLastname(user.getLastname);

                    repository.save(user1);
                });
    }

}

Answer 9

public void updateLaserDataByHumanId(String replacement, String humanId) {
    List<LaserData> laserDataByHumanId = laserDataRepository.findByHumanId(humanId);
    laserDataByHumanId.stream()
            .map(en -> en.setHumanId(replacement))
            .collect(Collectors.toList())
            .forEach(en -> laserDataRepository.save(en));
}

Answer 10

Specifically how do I tell spring-data-jpa that users that have the same username and firstname are actually EQUAL and that it is supposed to update the entity. Overriding equals did not work.

For this particular purpose one can introduce a composite key like this:

CREATE TABLE IF NOT EXISTS `test`.`user` (
  `username` VARCHAR(45) NOT NULL,
  `firstname` VARCHAR(45) NOT NULL,
  `description` VARCHAR(45) NOT NULL,
  PRIMARY KEY (`username`, `firstname`))

Mapping:

@Embeddable
public class UserKey implements Serializable {
    protected String username;
    protected String firstname;

    public UserKey() {}

    public UserKey(String username, String firstname) {
        this.username = username;
        this.firstname = firstname;
    }
    // equals, hashCode
}

Here is how to use it:

@Entity
public class UserEntity implements Serializable {
    @EmbeddedId
    private UserKey primaryKey;

    private String description;

    //...
}

JpaRepository would look like this:

public interface UserEntityRepository extends JpaRepository<UserEntity, UserKey>

Then, you could use the following idiom: accept DTO with user info, extract name and firstname and create UserKey, then create a UserEntity with this composite key and then invoke Spring Data save() which should sort everything out for you.

Answer 11

You can see the example below:

private void updateDeliveryStatusOfEvent(Integer eventId, int deliveryStatus) {
    try {
        LOGGER.info("NOTIFICATION_EVENT updating with event id:{}", eventId);
        Optional<Event> eventOptional = eventRepository.findById(eventId);
        if (!eventOptional.isPresent()) {
            LOGGER.info("Didn't find any updatable notification event with this eventId:{}", eventId);
        }
        Event event = eventOptional.get();
        event.setDeliveryStatus(deliveryStatus);
        event = eventRepository.save(event);
        if (!Objects.isNull(event)) {
            LOGGER.info("NOTIFICATION_EVENT Successfully Updated with this id:{}", eventId);
        }
    } catch (Exception e) {
        LOGGER.error("Error :{} while updating NOTIFICATION_EVENT of event Id:{}", e, eventId);
    }
}

Or Update Using Native Query:

public interface YourRepositoryName extends JpaRepository<Event,Integer>{
@Transactional
    @Modifying
    @Query(value="update Event u set u.deliveryStatus = :deliveryStatus where u.eventId = :eventId", nativeQuery = true)
    void setUserInfoById(@Param("deliveryStatus")String deliveryStatus, @Param("eventId")Integer eventId);
}

Answer 12

As mentioned by others answer, method save() is dual function. It can both do save or update, it's automatically update if you provide the id.

for update method in controller class I suggested to use @PatchMapping. below is the example.

#Save method POST

{
    "username": "jhon.doe",
    "displayName": "Jhon",
    "password": "xxxyyyzzz",
    "email": "jhon.doe@mail.com"
}
@PostMapping("/user")
public void setUser(@RequestBody User user) {
    userService.save(user);
}

#Update method PATCH

{
    "id": 1, // this is important. Widly important
    "username": "jhon.doe",
    "displayName": "Jhon",
    "password": "xxxyyyzzz",
    "email": "jhon.doe@mail.com"
}

@PatchMapping("/user")
public void patchUser(@RequestBody User user) {
    userService.save(user);
}

Maybe you're wondering where the id's come from. It comes from the database of course, you want to update the existing data right?

Answer 13

If your primary key is autoincrement then, you have to set the value for the primary key. for the save(); method to work as a update().else it will create a new record in db.

if you are using jsp form then use hidden filed to set primary key.

Jsp:

<form:input type="hidden" path="id" value="${user.id}"/>

Java:

@PostMapping("/update")
public String updateUser(@ModelAttribute User user) {
    repo.save(user);
    return "redirect:userlist";
}

also look at this:

@Override
  @Transactional
  public Customer save(Customer customer) {

    // Is new?
    if (customer.getId() == null) {
      em.persist(customer);
      return customer;
    } else {
      return em.merge(customer);
    }
  }

Answer 14

Use @DynamicUpdate annotation. it is cleaner and you don't have to deal with querying the database in order to get the saved values.

Answer 15

I did this for my Entity UserModel:

In the Controller:

        @PutMapping("/{id}")
        public Optional<UserModel> update(@RequestBody UserModel user, @PathVariable Long id) {
            return this.userService.update(user, id);
        }

And in the Service:

    public Optional<UserModel> update(UserModel req, Long id){
        Optional<UserModel> user = userRepository.findById(id);
        if (user != null) {
            userRepository.save(req);
        }
        return user;
    }

Example with postman: Postman method PUT example

Answer 16

just use

userRepository.save(user)

if any user of this user's primary-key (i.e. id in your case) is found then this statement will update existing user otherwise this statement will save new user.