I use GraphQL SPQR with the entity
#Entity
public class MyEntity {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private UUID id;
#GraphQLNonNull
#GraphQLQuery(name = "a", description = "Any field")
private String a;
// Getters and Setters
}
and the service
#Service
#Transactional
public class MyService {
#Autowired
private MyRepository myRepository;
#GraphQLMutation(name = "createEntity")
public MyEntity createEntity(#GraphQLArgument(name = "entity") MyEntity entity) {
myRepository.save(entity);
return entity;
}
}
In GraphiQL I am allowed to set the id:
mutation {
createEntity(entity: {
id: "11111111-2222-3333-4444-555555555555"
a: "any value"
}) {
id
}
}
But the id shall not be made editable to the user because it will be overwritten by the DB. It shall only be shown at the queries. I tried and added #GraphQLIgnore, but the id is shown all the same.
How can I hide the id at creation?
In GraphQL-SPQR version 0.9.9 and earlier, the private members are not scanned at all, so annotations on the private fields don't normally do anything. Incidentally, Jackson (or Gson, if so configured) is used to discover the deserializable fields on input types, and those libraries do look at private fields, so some annotations will appear to be working for input types. This is what is happening in your case. But, #GraphQLIgnore is not among the annotations that will work on a private field.
What you need to do is move the annotations to getters and setters.
#Entity
public class MyEntity {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private UUID id;
#GraphQLIgnore //This will prevent ID from being mapped on the input type
//#JsonIgnore would likely work too
public void setId(UUID id) {...}
}
There's other ways to achieve this, but this is the most straight-forward.
Note: In the future versions of SPQR (post 0.9.9), it will be possible to place the annotations on private fields as well, but mixing (placing some annotations on a field and some on the related getter/setter) will not work.
Related
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();
}
I know there has been multiple questions on bidirectional relations using spring jpa in the past but my case is a little bit different because i am using 3 entities with 2 relationships to implement a medical system
I have 3 entities : doctor/patient/appointment
here is the code for the 3 entities
please note all setters , getters and constructors implemented but ommited here for clarity
Patient class
#Entity
public class resPatient {
#Id
#GeneratedValue( strategy= GenerationType.IDENTITY )
private long code;
private String name;
private String gender;
private String email;
private String mobile;
private int age;
private String notes;
#OneToMany(mappedBy = "patient")
List<resPackageMembership> memberships;
#OneToMany(mappedBy = "patient")
List<resAppointment> appointments;
#OneToMany(fetch = FetchType.LAZY,mappedBy = "patient")
List<resMedImage> medImages;
Doctor class
#Entity
public class resDoctor {
#Id
#GeneratedValue( strategy= GenerationType.IDENTITY )
private long code;
private String name;
private String mobile;
private String email;
private String gender;
private int age;
private String speciality;
#OneToMany(mappedBy = "doctor")
List<resAppointment> appointments;
Appointment class
#Entity
public class resAppointment {
#Id
#GeneratedValue( strategy= GenerationType.IDENTITY )
private long code;
private String speciality;
#Basic
#Temporal(TemporalType.TIMESTAMP)
private Date dateCreated;
#Basic
#Temporal(TemporalType.TIMESTAMP)
private Date dateToVisit;
private String status;
private String notes;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "doctorCode")
private resDoctor doctor;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "patientCode")
private resPatient patient;
the way my medical system should work is that when i get a patient using my restful controller i want all the patient data including his appointments but this leads to an infinite loop as the appointment has the doctor which also has appointments and so on.
i cannot user #JSONIGNORE as there are 2 relationships i want to get the patient with his appointments which should have the doctor without the appointments array and should not have any patient data as i already am in the patient object
As a general best-practice, it's recommended to separate the entities from the data transfer objects used for the rest controllers. With DTO's in place, you have more control on which data to include and serialize within them to avoid the circlular references.
If you like check out https://bootify.io, it generates the DTOs from your database schema, but the custom endpoint you still need to define/build.
I develop an annotation processor called beanknife recently, it support generate DTO from any class. You need config by annotation. But you don't need change the original class. This library support configuring on a separate class. Of course you can choose which property you want and which you not need. And you can add new property by the static method in the config class. For your question:
// this will generate a DTO class named "resPatientView".
// You can change this name using genName attribute.
#ViewOf(value=resPatient.class, includePattern = ".*")
public class PatientViewConfigure {
// here tell the processor to automatically convert the property appointments from List<resAppointment> to List<resAppointmentWithoutPatient>.
// resAppointmentWithoutPatient is the generated class configured at the following.
// Note, although at this moment it not exists and your idea think it is an error.
// this code really can be compiled, and after compiled, all will ok.
#OverrideViewProperty("appointments")
private List<resAppointmentWithoutPatient> appointments;
}
// here generated a class named resAppointmentWithoutPatient whick has all properties of resAppointment except patient
#ViewOf(value=resAppointment.class, genName="resAppointmentWithoutPatient", includePattern = ".*", excludes={"patient"})
public class AppointmentWithoutPatientViewConfigure {
// the doctor property will be converted to its dto version which defined by the configure class DoctorWithoutAppointmentsViewConfigure.
#OverrideViewProperty("doctor")
private resDoctorWithoutAppointments doctor;
}
// here we generate a class which has all properties of resDoctor except appointments
#ViewOf(value=resDoctor.class, genName="resDoctorWithoutAppointments", includePattern = ".*", excludes={"appointments"})
public class DoctorWithoutAppointmentsViewConfigure {}
// in you rest controller. return the dto instead of the entities.
resPatient patient = ...
resPatientView dto = resPatientView.read(patient);
List<resPatient> patients = ...
List<resPatientView> dto = resPatientView.read(patients);
At the end, the class resPatientView will has the same shap with resPatient except its appointments not having patient property and its doctor property is replaced with a version without appointments property.
Here are more examples.
The version 1.10 is ready. Will fix some bug and support the configure bean to be managed by spring.
I have a Springboot Application with Repositories having Spring Data JPA Queries like findOne, findAll and also derived ones like findByID or findByName etc.
What I want to achieve is multitenancy. All entities have an "account_id" column which holds the tenant.
How do I add a filter like "account_id" to all the queries metioned above without using derived queries that contains those name slike findIdAndAccountid (which would be findone)
#Repository
public interface CategoryRepository extends JpaRepository<Category, Long> {
Category findByName(String name);
}
Here's the corresponding entity
#Entity
#Table(name = "unit")
#Data
public class Unit {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
private String name;
#Column(name = "account_id")
private Long account_id;
}
I know most people use schemas as tenant separation but that's impossible for me. Is there a way (I didn't find one) to add such a tenant filter condition on those queries without writing NamedQueries or using DerivedQueries. An elegeant solution like annotate the repository or entity or maybe the queries that all queries should add the additional filter "account_id"?
You can add Where clause on your Entity classes (Didnt had time to test )
#Entity
#Table(name = "unit")
#Data
#Where(clause = "account_id= :account_id")
public class Unit {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
private String name;
#Column(name = "account_id")
private Long account_id;
}
Update and Solution
1. Create a Filter & FilterDef on the entity like so
#FilterDef(name="accountFilter", parameters=#ParamDef( name="accountId", type="long" ) )
#Filters( {
#Filter(name="accountFilter", condition=":accountId = account_id")
} )
public class Category {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
private String name;
#Column(name = "account_id")
private Long account_id;
}
enable filtering in the controller by autowiring entitymanager, writing a method to enable the filter and activate the filter in #ModelAttribute for each request
#RestController
#RequestMapping(path = "/categories",produces = MediaType.APPLICATION_JSON_VALUE )
public class CategoryController {
private final CategoryRepository repository;
#Autowired
private EntityManager entityManager;
CategoryController(CategoryRepository repository) {
this.repository = repository;
}
private void activateFilter() {
Session session = entityManager.unwrap(Session.class);
Filter filter = session.enableFilter("accountFilter");
filter.setParameter("accountId", Long.valueOf(TenantContext.getCurrentTenant()));
}
#ModelAttribute
public void initFilter() {
activateFilter();
}
... your rest methods here
}
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.